Class 11

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 7 Conic Sections Ex 7.2 Questions and Answers.

11th Maths Part 1 Conic Sections Exercise 7.2 Questions And Answers Maharashtra Board

Question 1.
Find the
(i) lengths of the principal axes
(ii) co-ordinates of the foci
(iii) equations of directrices
(iv) length of the latus rectum
(v) distance between foci
(vi) distance between directrices of the ellipse:
(a) \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)
(b) 3x² + 4y² = 12
(c) 2x² + 6y² = 6
(d) 3x² + 4y² = 1
Solution:
(a) Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 25 and b² = 9
a = 5 and b = 3
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(5) = 10
Length of minor axis = 2b = 2(3) = 6
Lengths of the principal axes are 10 and 6.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)
= \(\frac{\sqrt{25-9}}{5}\)
= \(\frac{\sqrt{16}}{5}\)
= \(\frac{4}{5}\)
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S(5(\(\frac{4}{5}\)), 0) and S'(-5(\(\frac{4}{5}\)), 0)
i.e., S(4, 0) and S'(-4, 0)

(iii) Equations of the directrices are x = ±\(\frac{\mathrm{a}}{\mathrm{e}}\)
= \(\pm \frac{5}{\frac{4}{5}}\)
= \(\pm \frac{25}{4}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(3)^{2}}{5}=\frac{18}{5}\)

(v) Distance between foci = 2ae
= 2(5)(\(\frac{4}{5}\))
= 8

(vi) Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)
= \(\frac{2(5)}{\frac{4}{5}}\)
= \(\frac{25}{2}\)

(b) Given equation of the ellipse is 3x² + 4y² = 12
\(\frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 4 and b² = 3
a = 2 and b = √3
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(2) = 4
Length of minor axis = 2b = 2√3
Lengths of the principal axes are 4 and 2√3.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)
= \(\frac{\sqrt{4-3}}{2}\)
= \(\frac{1}{2}\)
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S(2(\(\frac{1}{2}\)), 0) and S'(-2(\(\frac{1}{2}\)), 0)
i.e., S(1, 0) and S'(-1, 0)

(iii) Equations of the directrices are x = ±\(\frac{\mathrm{a}}{\mathrm{e}}\)
= \(\pm \frac{2}{\frac{1}{2}}\)
= ±4

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(\sqrt{3})^{2}}{2}=3\)

(v) Distance between foci = 2ae = 2(2)(\(\frac{1}{2}\)) = 2

(vi) Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)
= \(\frac{2(2)}{\frac{1}{2}}\)
= 8

(c) Given equation of the ellipse is 2x² + 6y² = 6
\(\frac{x^{2}}{3}+\frac{y^{2}}{1}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 3 and b² = 1
a = √3 and b = 1
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2√3
Length of minor axis = 2b = 2(1) = 2
Lengths of the principal axes are 2√3 and 2.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)
= \(\frac{\sqrt{3-1}}{\sqrt{3}}\)
= \(\frac{\sqrt{2}}{\sqrt{3}}\)
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S(√3(\(\frac{\sqrt{2}}{\sqrt{3}}\)), o) and S'(-√3(\(\frac{\sqrt{2}}{\sqrt{3}}\)), 0)
i.e., S(√2, 0) and S'(-√2, 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\),
= \(\pm \frac{\sqrt{3}}{\frac{\sqrt{2}}{\sqrt{3}}}\)
= \(\pm \frac{3}{\sqrt{2}}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}=\frac{2(1)^{2}}{\sqrt{3}}=\frac{2}{\sqrt{3}}\)

(v) Distance between foci = 2ae
= \(2(\sqrt{3})\left(\frac{\sqrt{2}}{\sqrt{3}}\right)\)
= 2√2

(vi) Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)
= \(\frac{2 \sqrt{3}}{\frac{\sqrt{2}}{\sqrt{3}}}\)
= \(\frac{2 \times 3}{\sqrt{2}}\)
= 3√2

(d) Given equation of the ellipse is 3x² + 4y = 1.
\(\frac{x^{2}}{\frac{1}{3}}+\frac{y^{2}}{\frac{1}{4}}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = \(\frac{1}{3}\) and b² = \(\frac{1}{4}\)
a = \(\frac{1}{\sqrt{3}}\) and b = \(\frac{1}{2}\)
Since a > b,
X-axis is the major axis and Y-axis is the minor axis.

(i) Length of major axis = 2a = 2(\(\frac{1}{\sqrt{3}}\)) = \(\frac{2}{\sqrt{3}}\)
Length of minor axis = 2b = 2(\(\frac{1}{2}\)) = 1
Lengths of the principal axes are \(\frac{2}{\sqrt{3}}\) and 1.

(ii) We know that e = \(\frac{\sqrt{a^{2}-b^{2}}}{a}\)
e = \(\frac{\sqrt{\frac{1}{3}-\frac{1}{4}}}{\frac{1}{\sqrt{3}}}=\frac{\sqrt{\frac{1}{12}}}{\frac{1}{\sqrt{3}}}=\sqrt{\frac{3}{12}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., S\(\left(\frac{1}{\sqrt{3}}\left(\frac{1}{2}\right), 0\right)\) and S’\(\left(-\frac{1}{\sqrt{3}}\left(\frac{1}{2}\right), 0\right)\)
i.e., S(\(\frac{1}{2 \sqrt{3}}\), 0) and S'(-\(\frac{1}{2 \sqrt{3}}\), 0)

(iii) Equations of the directrices are x = ±\(\frac{a}{e}\),
= \(\pm \frac{\frac{1}{\sqrt{3}}}{\frac{1}{2}}\)
= \(\pm \frac{2}{\sqrt{3}}\)

(iv) Length of latus rectum = \(\frac{2 b^{2}}{a}\)
= \(\frac{2\left(\frac{1}{2}\right)^{2}}{\frac{1}{\sqrt{3}}}\)
= \(\frac{\sqrt{3}}{2}\)

(v) Distance between foci = 2ae
= \(2\left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{2}\right)\)
= \(\frac{1}{\sqrt{3}}\)

(vi) Distance between directrices = \(\frac{2 a}{e}\)
= \(\frac{2\left(\frac{1}{\sqrt{3}}\right)}{\frac{1}{2}}\)
= \(\frac{4}{\sqrt{3}}\)

Question 2.
Find the equation of the ellipse in standard form if
(i) eccentricity = \(\frac{3}{8}\) and distance between its foci = 6.
(ii) the length of the major axis is 10 and the distance between foci is 8.
(iii) distance between directrices is 18 and eccentricity is \(\frac{1}{3}\).
(iv) minor axis is 16 and eccentricity is \(\frac{1}{3}\).
(v) the distance between foci is 6 and the distance between directrices is \(\frac{50}{3}\).
(vi) the latus rectum has length 6 and foci are (±2, 0).
(vii) passing through the points (-3, 1) and (2, -2).
(viii) the distance between its directrices is 10 and which passes through (-√5, 2).
(ix) eccentricity is \(\frac{2}{3}\) and passes through (2, \(\frac{-5}{3}\)).
Solution:
(i) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Given, eccentricity (e) = \(\frac{3}{8}\)
Distance between foci = 2ae
Given, distance between foci = 6
2ae = 6

The required equation of ellipse is \(\frac{x^{2}}{64}+\frac{y^{2}}{55}=1\).

(ii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Length of major axis = 2a
Given, length of major axis = 10
2a = 10
a = 5
a² = 25
Distance between foci = 2ae
Given, distance between foci = 8
2ae = 8
2(5)e = 8

The required equation of ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\).

(iii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Given, eccentricity (e) = \(\frac{1}{3}\)
Distance between directrices = \(\frac{2a}{e}\)
Given, distance between directrices = 18

The required equation of ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{8}=1\)

(iv) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Length of minor axis = 2b
Given, length of minor axis = 16
2b = 16
b = 8
b² = 64
Given, eccentricity (e) = \(\frac{1}{3}\)
Now, b² = a² (1 – e²)

The required equation of ellipse is \(\frac{x^{2}}{72}+\frac{y^{2}}{64}=1\).

(v) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Distance between foci = 2ae
Given, distance between foci = 6
2ae = 6
ae = 3
a = \(\frac{3}{e}\) …….(i)
Distance between directrices = \(\frac{2a}{e}\)
Given, distance between directrices = \(\frac{50}{3}\)

The required equation of ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\).

(vi) Given, the length of the latus rectum is 6, and co-ordinates of foci are (±2, 0).
The foci of the ellipse are on the X-axis.
Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Length of latus rectum = \(\frac{2 b^{2}}{a}\)
\(\frac{2 b^{2}}{a}\) = 6
b² = 3a …..(i)
Co-ordinates of foci are (±ae, 0)
ae = 2
a²e² = 4 …..(ii)
Now, b² = a² (1 – e²)
b² = a² – a² e²
3a = a² – 4 …..[From (i) and (ii)]
a² – 3a – 4 = 0
a² – 4a + a – 4 = 0
a(a – 4) + 1(a – 4) = 0
(a – 4) (a + 1) = 0
a – 4 = 0 or a + 1 = 0
a = 4 or a = -1
Since a = -1 is not possible,
a = 4
a² = 16
Substituting a = 4 in (i), we get
b² = 3(4) = 12
The required equation of ellipse is \(\frac{x^{2}}{16}+\frac{y^{2}}{12}=1\).

(vii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
The ellipse passes through the points (-3, 1) and (2, -2).
Substituting x = -3 and y = 1 in equation of ellipse, we get

Equations (i) and (ii) become
9A + B = 1 …..(iii)
4A + 4B = 1 …..(iv)
Multiplying (iii) by 4, we get
36A + 4B = 4 …..(v)
Subtracting (iv) from (v), we get
32A = 3
A = \(\frac{3}{32}\)
Substituting A = \(\frac{3}{32}\) in (iv), we get
4(\(\frac{3}{32}\)) + 4B = 1
\(\frac{3}{8}\) + 4B = 1
4B = 1 – \(\frac{3}{8}\)
4B = \(\frac{5}{8}\)
B = \(\frac{5}{32}\)

(viii) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Distance between directrices = \(\frac{2 a}{e}\)
Given, distance between directrices = 10
\(\frac{2 a}{e}\) = 10
a = 5e …..(i)
The ellipse passes through (-√5, 2).
Substituting x = -√5 and y = 2 in equation of ellipse, we get

b² = 15(\(\frac{2}{5}\))
b² = 6
The required equation of ellipse is \(\frac{x^{2}}{15}+\frac{y^{2}}{6}=1\).

(ix) Let the required equation of ellipse be \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), where a > b.
Given, eccentricity (e) = \(\frac{2}{3}\)
The ellipse passes through (2, \(\frac{-5}{3}\)).
Substituting x = 2 and y = \(\frac{-5}{3}\) in equation of ellipse, we get

The required equation of ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\).

Question 3.
Find the eccentricity of an ellipse, if the length of its latus rectum is one-third of its minor axis.
Solution:
Let the equation of ellipse be \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), where a > b.
Length of latus rectum = \(\frac{2 b^{2}}{a}\)
Length of minor axis = 2b
According to the given condition,
Length of latus rectum = \(\frac{1}{3}\) (Minor axis)

Question 4.
Find the eccentricity of an ellipse, if the distance between its directrices is three times the distance between its foci.
Solution:
Let the required equation of ellipse be \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), where a > b.
Distance between directrices = \(\frac{2 \mathrm{a}}{\mathrm{e}}\)
Distance between foci = 2ae
According to the given condition,
distance between directrices = 3(distance between foci)
\(\frac{2 \mathrm{a}}{\mathrm{e}}\) = 3(2ae)
\(\frac{1}{\mathrm{e}}\) = 3e
\(\frac{1}{3}\) = e²
e = \(\frac{1}{\sqrt{3}}\) ……[∵ 0 < e < 1]
Eccentricity of the ellipse is \(\frac{1}{\sqrt{3}}\)

Question 5.
Show that the product of the lengths of the perpendicular segments drawn from the foci to any tangent line to the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) is equal to 16.
Solution:
Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\).
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 25, b² = 16
a = 5, b = 4

Question 6.
A tangent having slope \(\left(-\frac{1}{2}\right)\) the ellipse 3x² + 4y² = 12 intersects the X and Y axes in the points A and B respectively. If O is the origin, find the area of the triangle AOB.
Solution:
Given equation of the ellipse is 3x² + 4y² = 12.
\(\frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 4, b² = 3
Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
Here, m = \(-\frac{1}{2}\)
Equations of the tangents are
y = \(\frac{-1}{2} x \pm \sqrt{4\left(-\frac{1}{2}\right)^{2}+3}=\frac{-1}{2} x \pm 2\)
2y = -x ± 4
x + 2y ± 4 = 0
Consider the tangent x + 2y – 4 = 0
Let this tangent intersect the X-axis at A(x₁, 0) and Y-axis at B(0, y₁).
x₁ + 0 – 4 = 0 and 0 + 2y₁ – 4 = 0
x₁ = 4 and y₁ = 2
A = (4, 0) and B = (0, 2)
l(OA) = 4 and l(OB) = 2

Area of ∆AOB = \(\frac{1}{2}\) × l(OA) × l(OB)
= \(\frac{1}{2}\) × 4 × 2
= 4 sq.units

Question 7.
Show that the line x – y = 5 is a tangent to the ellipse 9x² + 16y² = 144. Find the point of contact.
Solution:
Given equation of the ellipse is 9x² + 16y² = 144
\(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 16 and b² = 9
Given equation of line is x – y = 5, i.e., y = x – 5
c² = a² m² + b²
Comparing this equation with y = mx + c, we get
m = 1 and c = -5
For the line y = mx + c to be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\)1, we must have
c² = a² m² + b²
c² = (-5)² = 25
a² m² + b² = 16(1)² + 9 = 16 + 9 = 25 = c²
The given line is a tangent to the given ellipse and point of contact
= \(\left(\frac{-\mathrm{a}^{2} \mathrm{~m}}{\mathrm{c}}, \frac{\mathrm{b}^{2}}{\mathrm{c}}\right)\)
= \(\left(\frac{(-16)(1)}{-5}, \frac{9}{-5}\right)\)
= \(\left(\frac{16}{5}, \frac{-9}{5}\right)\)

Question 8.
Show that the line 8y + x = 17 touches the ellipse x² + 4y² = 17. Find the point of contact.
Solution:
Given equation of the ellipse is x² + 4y² = 17.
\(\frac{x^{2}}{17}+\frac{y^{2}}{\frac{17}{4}}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 17 and b² = \(\frac{17}{4}\)
Given equation of line is 8y + x = 17,
y = \(\frac{-1}{8} x+\frac{17}{8}\)
Comparing this equation with y = mx + c, we get
m = \(\frac{-1}{8}\) and c = \(\frac{17}{8}\)
For the line y = mx + c to be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\)1, we must have

Question 9.
Determine whether the line x + 3y√2 = 9 is a tangent to the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\). If so, find the co-ordinates of the point of contact.
Solution:
Given equation of the ellipse is \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 9 and b² = 4
Given equation of line is x + 3y√2 = 9,
i.e., y = \(\frac{-1}{3 \sqrt{2}} x+\frac{3}{\sqrt{2}}\)
Comparing this equation with y = mx + c, we get
m = \(\frac{-1}{3 \sqrt{2}}\) and c = \(\frac{3}{\sqrt{2}}\)
For the line y = mx + c to be a tangent to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=\)1, we must have

Question 10.
Find k, if the line 3x + 4y + k = 0 touches 9x² + 16y² = 144.
Solution:
Given equation of the ellipse is 9x² + 16y² = 144.
\(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 16 and b² = 9
Given equation of line is 3x + 4y + k = 0,
i.e., y = \(-\frac{3}{4} x-\frac{k}{4}\)
Comparing this equation with y = mx + c, we get
m = \(\frac{-3}{4}\) and c = \(\frac{-\mathrm{k}}{4}\)
For the line y = mx + c to be a tangent to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\)1, we must have
c² = a² m² + b²
\(\left(\frac{-k}{4}\right)^{2}=16\left(\frac{-3}{4}\right)^{2}+9\)
\(\frac{\mathrm{k}^{2}}{16}\) = 9 + 9
\(\frac{\mathrm{k}^{2}}{16}\) = 18
k² = 288
k = ±12√2

Question 11.
Find the equations of the tangents to the ellipse:
(i) \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\) passing through the point (2, -2).
(ii) 4x² + 7y² = 28 from the point (3, -2).
(iii) 2x² + y² = 6 from the point (2, 1).
(iv) x² + 4y² = 9 which are parallel to the line 2x + 3y – 5 = 0.
(v) \(\frac{x^{2}}{25}+\frac{y^{2}}{4}=1\) which are parallel to the line x + y + 1 = 0.
(vi) 5x² + 9y² = 45 which are ⊥ to the line 3x + 2y + 1 = 0.
(vii) x² + 4y² = 20 which are ⊥ to the line 4x + 3y = 7.
Solution:
(i) Given equation of the ellipse is \(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\).
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 5 and b² = 4
Equations of tangents to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
Since (2, -2) lies on both the tangents,
-2 = 2m ±\(\sqrt{5 m^{2}+4}\)
-2 – 2m = ±\(\sqrt{5 m^{2}+4}\)
Squaring both the sides, we get
4m² + 8m + 4 = 5m² + 4
m² – 8m = 0
m(m – 8) = 0
m = 0 or m = 8
These are the slopes of the required tangents.
By slope point form y – y₁ = m(x – x₁),
the equations of the tangents are
y + 2 = 0(x – 2) and y + 2 = 8(x – 2)
y + 2 = 0 and y + 2 = 8x – 16
y + 2 = 0 and 8x – y – 18 = 0

(ii) Given equation of the ellipse is 4x² + 7y² = 28.
\(\frac{x^{2}}{7}+\frac{y^{2}}{4}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 7 and b² = 4
Equations of tangents to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
Since (3, -2) lies on both the tangents,
-2 = 3m ± \(\sqrt{7 \mathrm{~m}^{2}+4}\)
-2 – 3m = ±\(\sqrt{7 \mathrm{~m}^{2}+4}\)
Squaring both the sides, we get
9m² + 12m + 4 = 7m² + 4
2m² + 12m = 0
2m(m + 6) = 0
m = 0 or m = -6
These are the slopes of the required tangents.
By slope point form y – y₁ = m(x – x₁),
the equations of the tangents are
y + 2 = 0(x – 3) and y + 2 = -6(x – 3)
y + 2 = 0 and y + 2 = -6x + 18
y + 2 = 0 and 6x + y – 16 = 0

(iii) Given equation of the ellipse is 2x² + y² = 6.
\(\frac{x^{2}}{3}+\frac{y^{2}}{6}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 3 and b² = 6
Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
Since (2, 1) lies on both the tangents,
1 = 2m ± \(\sqrt{3 m^{2}+6}\)
1 – 2m = ±\(\sqrt{3 m^{2}+6}\)
Squaring both the sides, we get
1 – 4m + 4m² = 3m² + 6
m² – 4m – 5 = 0
(m – 5)(m + 1) = 0
m = 5 or m = -1
These are the slopes of the required tangents.
By slope point form y – y₁ = m(x – x₁),
the equations of the tangents are
y – 1 = 5(x – 2) and y – 1 = -1(x – 2)
y – 1 = 5x – 10 and y – 1 = -x + 2
5x – y – 9 = 0 and x + y – 3 = 0

(iv) Given equation of the ellipse is x² + 4y² = 9.
\(\frac{x^{2}}{9}+\frac{y^{2}}{\frac{9}{4}}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 9 and b² = \(\frac{9}{4}\)
Slope of the line 2x + 3y – 5 = 0 is \(\frac{-2}{3}\).
Since the given line is parallel to the required tangents, slope of the required tangents is
m = \(\frac{-2}{3}\)
Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are

(v) Given equation of the ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{4}=1\).
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 25 and b² = 4
Slope of the given line x + y + 1 = 0 is -1.
Since the given line is parallel to the required tangents,
the slope of the required tangents is m = -1.
Equations of tangents to the ellipse

(vi) Given equation of the ellipse is 5x² + 9y² = 45.
\(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 9 and b² = 5
Slope of the given line 3x + 2y + 1 = 0 is \(\frac{-3}{2}\)
Since the given line is perpendicular to the required tangents, slope of the required tangents is
m = \(\frac{2}{3}\)
Equations of tangents to the ellipse

(vii) Given equation of the ellipse is x² + 4y² = 20.
\(\frac{x^{2}}{20}+\frac{y^{2}}{5}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 20 and b² = 5
Slope of the given line 4x + 3y = 7 is \(\frac{-4}{3}\).
Since the given line is perpendicular to the required tangents,
slope of the required tangents is m = \(\frac{3}{4}\).
Equations of tangents to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
y = \(\frac{3}{4} x \pm \sqrt{20\left(\frac{3}{4}\right)^{2}+5}\)

Question 12.
Find the equation of the locus of a point, the tangents from which to the ellipse 3x² + 5y² = 15 are at right angles.
Solution:
Given equation of the ellipse is 3x² + 5y² = 15.
\(\frac{x^{2}}{5}+\frac{y^{2}}{3}=1\)
Comparing this equation with \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\), we get
a² = 5 and b² = 3
Equations of tangents to the ellipse \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
y = mx ± \(\sqrt{5 m^{2}+3}\)
y – mx =±\(\sqrt{5 m^{2}+3}\)
Squaring both the sides, we get
y² – 2mxy + m²x² = 5m² + 3
(x² – 5) m² – 2xym + (y² – 3) = 0
The roots m₁ and m₂ of this quadratic equation are the slopes of the tangents.
m₁m₂ = \(\frac{y^{2}-3}{x^{2}-5}\)
Since the tangents are at right angles,
m₁m₂ = -1
\(\frac{y^{2}-3}{x^{2}-5}=-1\)
y² – 3 = -x² + 5
x² + y² = 8, which is the required equation of the locus.

Alternate method:
The locus of the point of intersection of perpendicular tangents is the director circle of an ellipse.
The equation of the director circle of an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is x² + y² = a² + b²
Here, a² = 5 and b² = 3
x² + y² = 5 + 3
x² + y² = 8, which is the required equation of the locus.

Question 13.
Tangents are drawn through a point P to the ellipse 4x² + 5y² = 20 having inclinations θ₁ and θ₂ such that tan θ₁ + tan θ₂ = 2. Find the equation of the locus of P.
Solution:
Given equation of the ellipse is 4x² + 5y² = 20.
\(\frac{x^{2}}{5}+\frac{y^{2}}{4}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 5 and b² = 4
Since inclinations of tangents are θ₁ and θ₂,
m₁ = tan θ₁ and m₂ = tan θ₂
Equation of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
y = mx ± \(\sqrt{5 \mathrm{~m}^{2}+4}\)
y – mx = ± \(\sqrt{5 \mathrm{~m}^{2}+4}\)
Squaring both the sides, we get
y² – 2mxy + m²x² = 5m² + 4
(x² – 5)m² – 2xym + (y² – 4) = 0
The roots m₁ and m₂ of this quadratic equation are the slopes of the tangents.
m₁ + m₂ = \(\frac{-(-2 x y)}{x^{2}-5}=\frac{2 x y}{x^{2}-5}\)
Given, tan θ₁ + tan θ₂ = 2
m₁ + m₂ = 2
\(\frac{2 x y}{x^{2}-5}\) = 2
xy = x² – 5
x² – xy – 5 = 0, which is the required equation of the locus of P.

Question 14.
Show that the locus of the point of intersection of tangents at two points on an ellipse, whose eccentric angles differ by a constant, is an ellipse.
Solution:
Let P(θ₁) and Q(θ₂) be any two points on the given ellipse such that θ₁ – θ₂ = k, where k is a constant.
The equation of the tangent at point P(θ₁) is
\(\frac{x \cos \theta_{1}}{\mathrm{a}}+\frac{y \sin \theta_{1}}{\mathrm{~b}}=1\) ……(i)
The equation of the tangent at point Q(θ₂) is
\(\frac{x \cos \theta_{2}}{\mathrm{a}}+\frac{y \sin \theta_{2}}{\mathrm{~b}}=1\) ……(ii)
Multiplying equation (i) by cos θ₂ and equation (ii) by cos θ₁ and subtracting, we get
\(\frac{y}{b}\) (sin θ₁ cos θ₂ – sin θ₂ cos θ₁) = cos θ₂ – cos θ₁

Question 15.
P and Q are two points on the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) with eccentric angles θ₁ and θ₂. Find the equation of the locus of the point of intersection of the tangents at P and Q if θ₁ + θ₂ = \(\frac{\pi}{2}\).
Solution:
Given equation of the ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\).
θ₁ and θ₂ are the eccentric angles of a tangent.
Equation of tangent at point P is
\(\frac{x}{a} \cos \theta_{1}+\frac{y}{b} \sin \theta_{1}=1\) ……(i)
Equation of tangent at point Q is
\(\frac{x}{a} \cos \theta_{2}+\frac{y}{b} \sin \theta_{2}=1\) ………(ii)
θ₁ + θ₂ = \(\frac{\pi}{2}\) …..[Given]

i.e., points P and Q coincide, which is not possible, as P and Q are two different points.
cos θ₁ – sin θ₁ ≠ 0
Dividing equation (iii) by (cos θ₁ – sin θ₁), we get
\(\frac{x_{1}}{a}=\frac{y_{1}}{b}\)
bx₁ – ay₁ = 0
bx – ay = 0, which is the required equation of locus of point M.

Question 16.
The eccentric angles of two points P and Q of the ellipse 4x² + y² = 4 differ by \(\frac{2 \pi}{3}\). Show that the locus of the point of intersection of the tangents at P and Q is the ellipse 4x² + y² = 16.
Solution:
Given equation of the ellipse is 4x² + y² = 4.
\(\frac{x^{2}}{1}+\frac{y^{2}}{4}=1\)
Let P(θ₁) and Q(θ₂) be any two points on the given ellipse such that
θ₁ – θ₂ = \(\frac{2 \pi}{3}\)
The equation of the tangent at point P(θ₁) is
\(\frac{x \cos \theta_{1}}{1}+\frac{y \sin \theta_{1}}{2}=1\) ……(i)
The equation of the tangent at point Q(θ₂) is
\(\frac{x \cos \theta_{2}}{1}+\frac{y \sin \theta_{2}}{2}=1\)
Multiplying equation (i) by cos θ₂ and equation (ii) by cos θ₁ and subtracting, we get

Question 17.
Find the equations of the tangents to the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\), making equal intercepts on co-ordinate axes.
Solution:
Given equation of the ellipse is \(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)
Comparing this equation with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 16 and b² = 9
Since the tangents make equal intercepts on the co-ordinate axes,
m = -1.
Equations of tangents to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) having slope m are
y = mx ± \(\sqrt{a^{2} m^{2}+b^{2}}\)
y = -x ± \(\sqrt{16(-1)^{2}+9}\)
y = -x ± \(\sqrt{25}\)
x + y = ±5

Question 18.
A tangent having slope \(\left(-\frac{1}{2}\right)\) to the ellipse 3x² + 4y² = 12 intersects the X and Y axes in the points A and B respectively. If O is the origin, find the area of the triangle AOB.
Solution:
The equation of the ellipse is 3x² + 4y² = 12
\(\frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)
Comparing with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), we get
a² = 4, b² = 3
The equation of tangent with slope m is

It meets X axis at A
∴ for A, put y = 0 in equation (1), we get,
x = ±4
∴ A = (±4, 0)
Similarly, B = (0, ±2)
∴ OA = 4, OB = 2
∴ Area of ΔOAB = \(\frac{1}{2}\) × OA × OB
= \(\frac{1}{2}\) × 4 × 2
= 4 sq. units

Class 11 Maharashtra State Board Maths Solution 

• Exercise 6.1 Class 11 Maths 1
• Exercise 6.2 Class 11 Maths 1
• Exercise 6.3 Class 11 Maths 1
• Miscellaneous Exercise 6 Class 11 Maths 1
• Exercise 7.1 Class 11 Maths 1
• Exercise 7.2 Class 11 Maths 1
• Exercise 7.3 Class 11 Maths 1
• Miscellaneous Exercise 7 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 7 Conic Sections Ex 7.1 Questions and Answers.

11th Maths Part 1 Conic Sections Exercise 7.1 Questions And Answers Maharashtra Board

Question 1.
Find co-ordinates of focus, equation of directrix, length of latus rectum and the co-ordinates of end points of latus rectum of the parabola:
(i) 5y² = 24x
(ii) y² = -20x
(iii) 3x² = 8y
(iv) x² = -8y
(v) 3y² = -16x
Solution:
(i) Given equation of the parabola is 5y² = 24x.
⇒ y² = \(\frac{24}{5}\)x
Comparing this equation with y² = 4ax, we get
⇒ 4a = \(\frac{24}{5}\)
⇒ a = \(\frac{6}{5}\)
Co-ordinates of focus are S(a, 0), i.e., S(\(\frac{6}{5}\), 0)
Equation of the directrix is x + a = 0.
⇒ x + \(\frac{6}{5}\) = 0
⇒ 5x + 6 = 0
Length of latus rectum = 4a
= 4(\(\frac{6}{5}\))
= \(\frac{24}{5}\)
Co-ordinates of end points of latus rectum are (a, 2a) and (a, -2a),
⇒ \(\left(\frac{6}{5}, \frac{12}{5}\right)\) and \(\left(\frac{6}{5}, \frac{-12}{5}\right)\)

(ii) Given equation of the parabola is y² = -20x.
Comparing this equation with y² = -4ax, we get
⇒ 4a = 20
⇒ a = 5
Co-ordinates of focus are S(-a, 0), i.e., S(-5, 0)
Equation of the directrix is x – a = 0
⇒ x – 5 = 0
Length of latus rectum = 4a = 4(5) = 20
Co-ordinates of end points of latus rectum are (-a, 2a) and (-a, -2a),
⇒ (-5, 10) and (-5, -10).

(iii) Given equation of the parabola is 3x² = 8y
⇒ x² = \(\frac{8}{3}\) y
Comparing this equation with x² = 4by, we get
⇒ 4b = \(\frac{8}{3}\)
⇒ b = \(\frac{2}{3}\)
Co-ordinates of focus are S(0, b), i.e., S(0, \(\frac{2}{3}\))
Equation of the directrix is y + b = 0,
⇒ y + \(\frac{2}{3}\) = 0
⇒ 3y + 2 = 0
Length of latus rectum = 4b = 4(\(\frac{2}{3}\)) = \(\frac{8}{3}\)
Co-ordinates of end points of latus rectum are (2b, b) and (-2b, b),
⇒ \(\left(\frac{4}{3}, \frac{2}{3}\right)\) and \(\left(-\frac{4}{3}, \frac{2}{3}\right)\).

(iv) Given equation of the parabola is x² = -8y.
Comparing this equation with x² = -4by, we get
⇒ 4b = 8
⇒ b = 2
Co-ordinates of focus are S(0, -b), i.e., S(0, – 2)
Equation of the directrix is y – b = 0, i.e., y – 2 = 0
Length of latus rectum = 4b = 4(2) = 8
∴ Co-ordinates of end points of latus rectum are (2b, -b) and (-2b, -b), i.e., (4, -2) and (-4, -2).

(v) Given equation of the parabola is 3y² = -16x.
⇒ y² = \(-\frac{16}{3}\)x
Comparing this equation withy = -4ax, we get
⇒ 4a = \(\frac{16}{3}\)
⇒ a = \(\frac{4}{3}\)
Co-ordinates of focus are S(-a, 0), i.e., (\(-\frac{4}{3}\), 0)
Equation of the directrix is x – a = 0,
⇒ x – \(-\frac{4}{3}\) = 0
⇒ 3x – 4 = 0
Length of latus rectum = 4a = 4(\(\frac{4}{3}\)) = \(\frac{16}{3}\)
Co-ordinates of end points of latus rectum are (-a, 2a) and (-a, -2a),
i.e., \(\left(-\frac{4}{3}, \frac{8}{3}\right)\) and \(\left(-\frac{4}{3},-\frac{8}{3}\right)\)

Question 2.
Find the equation of the parabola with vertex at the origin, the axis along the Y-axis, and passing through the point (-10, -5).
Solution:
Vertex of the parabola is at origin (0, 0) and its axis is along Y-axis.
Equation of the parabola can be either x² = 4by or x² = -4by
Since the parabola passes through (-10, -5), it lies in 3rd quadrant.
Required parabola is x² = -4by.
Substituting x = -10 and y = -5 in x² = -4by, we get
⇒ (-10)² = -4b(-5)
⇒ b = \(\frac{100}{20}\) = 5
∴ The required equation of the parabola is x² = -4(5)y, i.e., x² = -20y.

Question 3.
Find the equation of the parabola with vertex at the origin, the axis along the X-axis, and passing through the point (3, 4).
Solution:
Vertex of the parabola is at the origin (0, 0) and its axis is along X-axis.
Equation of the parabola can be either y² = 4ax or y² = -4ax.
Since the parabola passes through (3, 4), it lies in the 1st quadrant.
Required parabola is y² = 4ax.
Substituting x = 3 and y = 4 in y² = 4ax, we get
⇒ (4)² = 4a(3)
⇒ a = \(\frac{16}{12}=\frac{4}{3}\)
The required equation of the parabola is
y² = 4(\(\frac{4}{3}\))x
⇒ 3y² = 16x

Question 4.
Find the equation of the parabola whose vertex is O(0, 0) and focus at (-7, 0).
Solution:
Focus of the parabola is S(-7, 0) and vertex is O(0, 0).
Since focus lies on X-axis, it is the axis of the parabola.
Focus S(-7, 0) lies on the left-hand side of the origin.
It is a left-handed parabola.
Required parabola is y = -4ax.
Focus is S(-a, 0).
a = 7
∴ The required equation of the parabola is y² =-4(7)x, i.e., y² = -28x.

Question 5.
Find the equation of the parabola with vertex at the origin, the axis along X-axis, and passing through the point
(i) (1, -6)
(ii) (2, 3)
Solution:
(i) Vertex of the parabola is at origin (0, 0) and its axis is along X-axis.
Equation of the parabola can be either y² = 4ax or y² = -4ax.
Since the parabola passes through (1, -6), it lies in the 4th quadrant.
Required parabola is y² = 4ax.
Substituting x = 1 and y = -6 in y² = 4ax, we get
⇒ (-6)² = 4a(1)
⇒ 36 = 4a
⇒ a = 9
∴ The required equation of the parabola is y² = 4(9)x, i.e., y² = 36x.

(ii) Vertex of the parabola is at origin (0, 0) and its axis is along X-axis.
Equation of the parabola can be either y² = 4ax or y² = -4ax.
Since the parabola passes through (2, 3), it lies in 1st quadrant.
∴ Required parabola is y² = 4ax.
Substituting x = 2 and y = 3 in y² = 4ax, we get
⇒ (3)² = 4a(2)
⇒ 9 = 8a
⇒ a = \(\frac{9}{8}\)
The required equation of the parabola is
y² = 4(\(\frac{9}{8}\))x
⇒ y² = \(\frac{9}{2}\) x
⇒ 2y² = 9x.

Question 6.
For the parabola 3y² = 16x, find the parameter of the point:
(i) (3, -4)
(ii) (27, -12)
Solution:
Given the equation of the parabola is 3y² = 16x.
⇒ y² = \(\frac{16}{3}\)x
Comparing this equation with y² = 4ax, we get
⇒ 4a = \(\frac{16}{3}\)
⇒ a = \(\frac{4}{3}\)
If t is the parameter of the point P on the parabola, then
P(t) = (at², 2at)
i.e., x = at² and y = 2at ………(i)
(i) Given point is (3, -4)
Substituting x = 3, y = -4 and a = \(\frac{4}{3}\) in (i), we get
3 = \(\frac{4}{3}\) t² and -4 = 2(\(\frac{4}{3}\)) t

∴ The parameter of the given point is \(\frac{-3}{2}\)

(ii) Given point is (27, -12)
Substituting x = 27, y = -12 and a = \(\frac{4}{3}\) in (i), we get

∴ The parameter of the given point is \(\frac{-9}{2}\)

Question 7.
Find the focal distance of a point on the parabola y² = 16x whose ordinate is 2 times the abscissa.
Solution:
Given the equation of the parabola is y² = 16x.
Comparing this equation with y² = 4ax, we get
⇒ 4a = 16
⇒ a = 4
Since ordinate is 2 times the abscissa,
y = 2x
Substituting y = 2x in y² = 16x, we get
⇒ (2x)² = 16x
⇒ 4x² = 16x
⇒ 4x² – 16x = 0
⇒ 4x(x – 4) = 0
⇒ x = 0 or x = 4
When x = 4,
focal distance = x + a = 4 + 4 = 8
When x = 0,
focal distance = a = 4
∴ Focal distance is 4 or 8.

Question 8.
Find coordinates of the point on the parabola. Also, find focal distance.
(i) y² = 12x whose parameter is \(\frac{1}{3}\)
(ii) 2y² = 7x whose parameter is -2
Solution:
(i) Given equation of the parabola is y² = 12x.
Comparing this equation with y² = 4ax, we get
⇒ 4a = 12
⇒ a = 3
If t is the parameter of the point P on the parabola, then
P(t) = (at², 2at)
i.e., x = at² and y = 2at ……..(i)
Given, t = \(\frac{1}{3}\)
Substituting a = 3 and t = \(\frac{1}{3}\) in (i), we get
x = 3(\(\frac{1}{3}\))² and y = 2(3)(\(\frac{1}{3}\))
x = \(\frac{1}{3}\) and y = 2
The co-ordinates of the point on the parabola are (\(\frac{1}{3}\), 2)
∴ Focal distance = x + a
= \(\frac{1}{3}\) + 3
= \(\frac{10}{3}\)

(ii) Given equation of the parabola is 2y² = 7x.
⇒ y² = \(\frac{7}{2}\)x
Comparing this equation with y² = 4ax, we get
⇒ 4a = \(\frac{7}{2}\)
⇒ a = \(\frac{7}{8}\)
If t is the parameter of the point P on the parabola, then
P(t) = (at², 2at)
i.e., x = at² and y = 2at …..(i)
Given, t = -2
Substituting a = \(\frac{7}{8}\) and t = -2 in (i), we get
x = \(\frac{7}{8}\)(-2)2 and y = 2(\(\frac{7}{8}\))(-2)
x = \(\frac{7}{2}\) and y = \(\frac{-7}{2}\)
The co-ordinates of the point on the parabola are (\(\frac{7}{2}\), \(\frac{-7}{2}\))
∴ Focal distance = x + a
= \(\frac{7}{2}\) + \(\frac{7}{8}\)
= \(\frac{35}{8}\)

Question 9.
For the parabola y² = 4x, find the coordinates of the point whose focal distance is 17.
Solution:
Given the equation of the parabola is y² = 4x.
Comparing this equation with y² = 4ax, we get
⇒ 4a = 4
⇒ a = 1
Focal distance of a point = x + a
Given, focal distance = 17
⇒ x + 1 = 17
⇒ x = 16
Substituting x = 16 in y² = 4x, we get
⇒ y² = 4(16)
⇒ y² = 64
⇒ y = ±8
∴ The co-ordinates of the point on the parabola are (16, 8) or (16, -8).

Question 10.
Find the length of the latus rectum of the parabola y² = 4ax passing through the point (2, -6).
Solution:
Given equation of the parabola is y² = 4ax and it passes through point (2, -6).
Substituting x = 2 and y = -6 in y² = 4ax, we get
⇒ (-6)² = 4a(2)
⇒ 4a = 18
∴ Length of latus rectum = 4a = 18 units

Question 11.
Find the area of the triangle formed by the line joining the vertex of the parabola x² = 12y to the endpoints of the latus rectum.
Solution:

Given the equation of the parabola is x² = 12y.
Comparing this equation with x² = 4by, we get
⇒ 4b = 12
⇒ b = 3
The co-ordinates of focus are S(0, b), i.e., S(0, 3)
End points of the latus-rectum are L(2b, b) and L'(-2b, b),
i.e., L(6, 3) and L'(-6, 3)
Also l(LL’) = length of latus-rectum = 4b = 12
l(OS) = b = 3
Area of ∆OLL’ = \(\frac{1}{2}\) × l(LL’) × l(OS)
= \(\frac{1}{2}\) × 12 × 3
Area of ∆OLL’ = 18 sq. units

Question 12.
If a parabolic reflector is 20 cm in diameter and 5 cm deep, find its focus.
Solution:

Let LOM be the parabolic reflector such that LM is the diameter and ON is its depth.
It is given that ON = 5 cm and LM = 20 cm.
LN = 10 cm
Taking O as the origin, ON along X-axis and a line through O ⊥ ON as Y-axis.
Let the equation of the reflector be y² = 4ax ……(i)
The point L has the co-ordinates (5, 10) and lies on parabola given by (i).
Substituting x = 5 and y = 10 in (i), we get
⇒ 10² = 4a(5)
⇒ 100 = 20a
⇒ a = 5
Focus is at (a, 0), i.e., (5, 0)

Question 13.
Find co-ordinates of focus, vertex, and equation of directrix and the axis of the parabola y = x² – 2x + 3.
Solution:
Given equation of the parabola is y = x² – 2x + 3
⇒ y = x² – 2x + 1 + 2
⇒ y – 2 = (x – 1)²
⇒ (x – 1)² = y – 2
Comparing this equation with X² = 4bY, we get
X = x – 1, Y = y – 2
⇒ 4b = 1
⇒ b = \(\frac{1}{4}\)
The co-ordinates of vertex are (X = 0, Y = 0)
⇒ x – 1 = 0 and y – 2 = 0
⇒ x = 1 and y = 2
The co-ordinates of vertex are (1, 2).
The co-ordinates of focus are S(X = 0, Y = b)
⇒ x – 1 = 0 and y – 2 = \(\frac{1}{4}\)
⇒ x = 1 and y = \(\frac{9}{4}\)
The co-ordinates of focus are (1, \(\frac{9}{4}\))
Equation of the axis is X = 0
x – 1 = 0, i.e., x = 1
Equation of directrix is Y + b = 0
⇒ y – 2 + \(\frac{1}{4}\) = 0
⇒ y – \(\frac{7}{4}\) = 0
⇒ 4y – 7 = 0

Question 14.
Find the equation of tangent to the parabola
(i) y² = 12x from the point (2, 5)
(ii) y² = 36x from the point (2, 9)
Solution:
(i) Given equation of the parabola is y² = 12x.
Comparing this equation with y² = 4ax, we get
⇒ 4a = 12
⇒ a = 3
Equation of tangent to the parabola y² = 4ax having slope m is
y = mx + \(\frac{a}{m}\)
Since the tangent passes through the point (2, 5)
⇒ 5 = 2m + \(\frac{3}{m}\)
⇒ 5m = 2m² + 3
⇒ 2m² – 5m + 3 = 0
⇒ 2m² – 2m – 3m + 3 = 0
⇒ 2m(m – 1) – 3(m – 1) = 0
⇒ (m- 1)(2m – 3) = 0
⇒ m = 1 or m = \(\frac{3}{2}\)
These are the slopes of the required tangents.
By slope point form, y – y₁ = m(x – x₁), the equations of the tangents are
⇒ y – 5 = 1(x – 2) and y – 5 = \(\frac{3}{2}\) (x – 2)
⇒ y – 5 = x – 2 and 2y – 10 = 3x – 6
⇒ x – y + 3 = 0 and 3x – 2y + 4 = 0

(ii) Given equation of the parabola is y² = 36x.
Comparing this equation with y² = 4ax, we get
⇒ 4a = 36
⇒ a = 9
Equation of tangent to the parabola y² = 4ax having slope m is
y = mx + \(\frac{a}{m}\)
Since the tangent passes through the point (2, 9),
⇒ 9 = 2m + \(\frac{9}{m}\)
⇒ 9m = 2m² + 9
⇒ 2m² – 9m + 9 = 0
⇒ 2m² – 6m – 3m + 9 = 0
⇒ 2m(m – 3) – 3(m – 3) = 0
⇒ (m – 3)(2m – 3) = 0
⇒ m = 3 or m = \(\frac{3}{2}\)
These are the slopes of the required tangents.
By slope point form, y – y₁ = m(x – x₁), the equations of the tangents are
⇒ y – 9 = 3(x – 2) and y – 9 = \(\frac{3}{2}\) (x – 2)
⇒ y – 9 = 3x – 6 and 2y – 18 = 3x – 6
⇒ 3x – y + 3 = 0 and 3x – 2y + 12 = 0

Question 15.
If the tangents drawn from the point (-6, 9) to the parabola y² = kx are perpendicular to each other, find k.
Solution:
Given equation of the parabola is y² = kx
Comparing this equation with y² = 4ax, we get
⇒ 4a = k
⇒ a = \(\frac{\mathrm{k}}{4}\)
Equation of tangent to the parabola y² = 4ax having slope m is
y = mx + \(\frac{a}{m}\)
Since the tangent passes through the point (-6, 9),
⇒ 9 = -6m + \(\frac{k}{4m}\)
⇒ 36m = -24m2 + k
⇒ 24m² + 36m – k = 0
The roots m₁ and m₂ of this quadratic equation are the slopes of the tangents.
m₁m₂ = \(\frac{-\mathrm{k}}{24}\)
Since the tangents are perpendicular to each other,
m₁m₂ = -1
⇒ \(\frac{-\mathrm{k}}{24}\) = -1
⇒ k = 24

Alternate method:
We know that, tangents drawn from a point on directrix are perpendicular.
(-6, 9) lies on the directrix x = -a.
⇒ -6 = -a
⇒ a = 6
Since 4a = k
⇒ k = 4(6) = 24

Question 16.
Two tangents to the parabola y² = 8x meet the tangents at the vertex in the points P and Q. If PQ = 4, prove that the equation of the locus of the point of intersection of two tangents is y² = 8(x + 2).
Solution:
Given equation of the parabola is y² = 8x
Comparing this equation with y² = 4ax, we get
⇒ 4a = 8
⇒ a = 2
Equation of tangent to given parabola at A(t₁) is y
t₁ = x + 2\(\mathrm{t}_{1}^{2}\) …….(i)
Equation of tangent to given parabola at B(t₂) is y
t₂ = x + 2\(\mathrm{t}_{2}^{2}\) …..(ii)

A tangent at the vertex is Y-axis whose equation is x = 0.
x-coordinate of points P and Q is 0.
Let P be(0, k₁) and Q be (0, k₂).
Then, from (i) and (ii), we get

∴ Equation of locus of R is y² = 8(x + 2).

Question 17.
Find the equation of common tangent to the parabolas y² = 4x and x² = 32y.
Solution:
Given equation of the parabola is y² = 4x
Comparing this equation with y² = 4ax, we get
⇒ 4a = 4
⇒ a = 1
Let the equation of common tangent be
y = mx + \(\frac{1}{m}\) …..(i)
Substituting y = mx + \(\frac{1}{m}\) in x² = 32y, we get
⇒ x² = 32(mx + \(\frac{1}{m}\)) = 32 mx + \(\frac{32}{m}\)
⇒ mx² = 32 m²x + 32
⇒ mx² – 32 m²x – 32 = 0 ……..(ii)
Line (i) touches the parabola x² = 32y.
The quadratic equation (ii) in x has equal roots.
Discriminant = 0
⇒ (-32m²)² – 4(m)(-32) = 0
⇒ 1024 m⁴ + 128m = 0
⇒ 128m (8m³ + 1) = 0
⇒ 8m³ + 1 = 0 …..[∵ m ≠ 0]
⇒ m³ = \(-\frac{1}{8}\)
⇒ m = \(-\frac{1}{2}\)
Substituting m = \(-\frac{1}{2}\) in (i), we get
⇒ \(y=-\frac{1}{2} x+\frac{1}{\left(-\frac{1}{2}\right)}\)
⇒ \(y=-\frac{1}{2} x-2\)
⇒ x + 2y + 4 = 0, which is the equation of the common tangent.

Question 18.
Find the equation of the locus of a point, the tangents from which to the parabola y² = 18x are such that sum of their slopes is -3.
Solution:
Given equation of the parabola is y² = 18x
Comparing this equation with y² = 4ax, we get
⇒ 4a = 18
⇒ a = \(\frac{9}{2}\)
Equation of tangent to the parabola y² = 4ax having slope m is
⇒ y = mx + \(\frac{a}{m}\)
⇒ y = mx + \(\frac{9}{2m}\)
⇒ 2ym = 2xm² + 9
⇒ 2xm² – 2ym + 9 = 0
The roots m₁ and m₂ of this quadratic equation are the slopes of the tangents.
m₁ + m₂ = \(-\frac{(-2 y)}{2 x}=\frac{y}{x}\)
But, m₁ + m₂ = -3
\(\frac{y}{x}\) = -3
y = -3x, which is the required equation of locus.

Question 19.
The towers of a bridge, hung in the form of a parabola, have their tops 30 metres above the roadway and are 200 metres apart. If the cable is 5 metres above the roadway at the centre of the bridge, find the length of the vertical supporting cable 30 metres from the centre.
Solution:
Let CAB be the cable of the bridge and X’OX be the roadway.
Let A be the centre of the bridge.
From the figure, vertex of parabola is at A(0, 5).
Let the equation of parabola be
x² = 4b(y – 5) …..(i)
Since the parabola passes through (100, 30).
Substituting x = 100 and y = 30 in (i), we get
⇒ 100² = 4b (30 – 5)
⇒ 100² = 4b(25)
⇒ 100² = 100b
⇒ b = 100
Substituting the value of b in (i), we get
x² = 400(y – 5) …..(ii)
Let l metres be the length of vertical supporting cable.
Then P(30, l) lies on (ii).
⇒ 30² = 400(l – 5)
⇒ 900 = 400(l – 5)
⇒ \(\frac{9}{4}\) = l – 5
⇒ l = \(\frac{9}{4}\) + 5
⇒ l = \(\frac{9}{4}\) m = 7.25 m
The length of the vertical supporting cable is 7.25 m.

Question 20.
A circle whose centre is (4, -1) passes through the focus of the parabola x² + 16y = 0. Show that the circle touches the directrix of the parabola.
Solution:
Given equation of the parabola is x² + 16y = 0.
⇒ x² = -16y
Comparing this equation with x² = -4by, we get
⇒ 4b = 16
⇒ b = 4
Focus = S(0, -b) = (0, -4)
Centre of the circle is C(4, -1) and it passes through focus S of the parabola.
Radius = CS
= \(\sqrt{(4-0)^{2}+(-1+4)^{2}}\)
= \(\sqrt{16+9}\)
= 5
Equation of the directrix is y – b = 0, i.e.,y – 4 = 0
Length of the perpendicular from centre C(4, -1) to the directrix
= \(\left|\frac{0(4)+1(-1)-4}{\sqrt{(0)^{2}+(1)^{2}}}\right|\)
= \(\left|\frac{-1-4}{1}\right|\)
= 5
= radius
∴ The circle touches the directrix of the parabola.

Class 11 Maharashtra State Board Maths Solution 

• Exercise 6.1 Class 11 Maths 1
• Exercise 6.2 Class 11 Maths 1
• Exercise 6.3 Class 11 Maths 1
• Miscellaneous Exercise 6 Class 11 Maths 1
• Exercise 7.1 Class 11 Maths 1
• Exercise 7.2 Class 11 Maths 1
• Exercise 7.3 Class 11 Maths 1
• Miscellaneous Exercise 7 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Miscellaneous Exercise 6 Questions and Answers.

11th Maths Part 1 Circle Miscellaneous Exercise 6 Questions And Answers Maharashtra Board

(I) Choose the correct alternative.

Question 1.
Equation of a circle which passes through (3, 6) and touches the axes is
(A) x² + y² + 6x + 6y + 3 = 0
(B) x² + y² – 6x – 6y – 9 = 0
(C) x² + y² – 6x – 6y + 9 = 0
(D) x² + y² – 6x + 6y – 3 = 0
Answer:
(C) x² + y² – 6x – 6y + 9 = 0

Question 2.
If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 sq. units, then find the equation of the circle.
(A) x² + y² – 2x + 2y = 40
(B) x² + y² – 2x – 2y = 47
(C) x² + y² – 2x + 2y = 47
(D) x² + y² – 2x – 2y = 40
Answer:
(C) x² + y² – 2x + 2y = 47
Hint:
Centre of circle = Point of intersection of diameters.
Solving 2x – 3y = 5 and 3x – 4y = 7, we get
x = 1, y = -1
Centre of the circle C(h, k) = C(1, -1)
∴ Area = 154
πr² = 154
\(\frac{22}{7} \times r^{2}\) = 154
r² = 154 × \(\frac{22}{7}\) = 49
∴ r = 7
equation of the circle is
(x – 1)² + (y + 1)² = 72
x² + y² – 2x + 2y = 47

Question 3.
Find the equation of the circle which passes through the points (2, 3) and (4, 5), and the center lies on the straight line y – 4x + 3 = 0.
(A) x² + y² – 4x – 10y + 25 = 0
(B) x² + y² – 4x – 10y – 25 = 0
(C) x² + y² – 4x + 10y – 25 = 0
(D) x² + y² + 4x – 10y + 25 = 0
Answer:
(A) x² + y² – 4x – 10y + 25 = 0

Question 4.
The equation(s) of the tangent(s) to the circle x² + y² = 4 which are parallel to x + 2y + 3 = 0 are
(A) x – 2y = 2
(B) x + 2y = ±2√3
(C) x + 2y = ±2√5
(D) x – 2y = ±2√5
Answer:
(C) x + 2y = ±2√5

Question 5.
If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find the radius of the circle.
(A) \(\frac{3}{4}\)
(B) \(\frac{4}{3}\)
(C) \(\frac{1}{4}\)
(D) \(\frac{7}{4}\)
Answer:
(A) \(\frac{3}{4}\)
Hint:
Tangents are parallel to each other.
The perpendicular distance between tangents = diameter

Question 6.
The area of the circle having centre at (1, 2) and passing through (4, 6) is
(A) 5π
(B) 10π
(C) 25π
(D) 100π
Answer:
(C) 25π
Hint:

Question 7.
If a circle passes through the points (0, 0), (a, 0), and (0, b), then find the co-ordinates of its centre.
(A) \(\left(\frac{-a}{2}, \frac{-b}{2}\right)\)
(B) \(\left(\frac{a}{2}, \frac{-b}{2}\right)\)
(C) \(\left(\frac{-a}{2}, \frac{b}{2}\right)\)
(D) \(\left(\frac{a}{2}, \frac{b}{2}\right)\)
Answer:
(D) \(\left(\frac{a}{2}, \frac{b}{2}\right)\)

Question 8.
The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is
(A) x² + y² = 9a²
(B) x² + y² = 16a²
(C) x² + y² = 4a²
(D) x² + y² = a²
Answer:
(C) x² + y² = 4a²
Hint:
Since the triangle is equilateral.
The centroid of the triangle is same as the circumcentre
and radius of the circumcircle = \(\frac{2}{3}\) (median) = \(\frac{2}{3}\)(3a) = 2a
Hence, the equation of the circumcircle whose centre is at (0, 0) and radius 2a is x² + y² = 4a²

Question 9.
A pair of tangents are drawn to a unit circle with centre at the origin and these tangents intersect at A enclosing an angle of 60. The area enclosed by these tangents and the arc of the circle is
(A) \(\frac{2}{\sqrt{3}}-\frac{\pi}{6}\)
(B) \(\sqrt{3}-\frac{\pi}{3}\)
(C) \(\frac{\pi}{3}-\frac{\sqrt{3}}{6}\)
(D) \(\sqrt{3}\left(1-\frac{\pi}{6}\right)\)
Answer:
(B) \(\sqrt{3}-\frac{\pi}{3}\)
Hint:

Question 10.
The parametric equations of the circle x² + y² + mx + my = 0 are
(A) x = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \cos \theta\), y = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \sin \theta\)
(B) x = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \cos \theta\), y = \(\frac{+m}{2}+\frac{m}{\sqrt{2}} \sin \theta\)
(C) x = 0, y = 0
(D) x = m cos θ, y = m sin θ
Answer:
(A) x = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \cos \theta\), y = \(\frac{-m}{2}+\frac{m}{\sqrt{2}} \sin \theta\)

(II) Answer the following:

Question 1.
Find the centre and radius of the circle x² + y² – x + 2y – 3 = 0.
Solution:
Given equation of the circle is x² + y² – x + 2y – 3 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -1, 2f = 2 and c = -3
g = \(\frac{-1}{2}\), f = 1 and c = -3
Centre of the circle = (-g, -f) = (\(\frac{1}{2}\), -1)
and radius of the circle

Question 2.
Find the centre and radius of the circle x = 3 – 4 sin θ, y = 2 – 4 cos θ.
Solution:
Given, x = 3 – 4 sin θ, y = 2 – 4 cos θ
⇒ x – 3 = -4 sin θ, y – 2 = -4 cos θ
On squaring and adding, we get
⇒ (x – 3)² + (y – 2)² = (-4 sin θ)² + (-4 cos θ)²
⇒ (x – 3)² + (y – 2)² = 16 sin² θ + 16 cos² θ
⇒ (x – 3)² + (y – 2)² = 16(sin² θ + cos² θ)
⇒ (x – 3)² + (y – 2)² = 16(1)
⇒ (x – 3)² + (y – 2)² = 16
⇒ (x – 3)² + (y – 2)² = 4²
Comparing this equation with (x – h)² + (y – k)² = r², we get
h = 3, k = 2, r = 4
∴ Centre of the circle is (3, 2) and radius is 4.

Question 3.
Find the equation of circle passing through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0 and whose centre is the point of intersection of lines x + y + 1 = 0 and x – 2y + 4 = 0.
Solution:
Required circle passes through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0.
x + 3y = 0
⇒ x = -3y ……..(i)
2x – 7y = 0 ……(ii)
Substituting x = -3y in (ii), we get
⇒ 2(-3y) – 7y = 0
⇒ -6y – 7y = 0
⇒ -13y = 0
⇒ y = 0
Substituting y = 0 in (i), we get
x = -3(0) = 0
Point of intersection is O(0, 0).
This point O(0, 0) lies on the circle.
Let C(h, k) be the centre of the required circle.
Since, point of intersection of lines x + y = -1 and x – 2y = -4 is the centre of circle.
∴ x = h, y = k
∴ Equations of lines become
h + k = -1 ……(iii)
h – 2k = -4 …..(iv)
By (iii) – (iv), we get
3k = 3
⇒ k = 1
Substituting k = 1 in (iii), we get
h + 1 = -1
⇒ h = -2
∴ Centre of the circle is C(-2, 1) and it passes through point O(0, 0).
Radius(r) = OC
= \(\sqrt{(0+2)^{2}+(0-1)^{2}}\)
= \(\sqrt{4+1}\)
= √5
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = -2, k = 1
the required equation of the circle is
(x + 2)² + (y – 1)² = (√5)²
⇒ x² + 4x + 4 + y² – 2y + 1 = 5
⇒ x² + y² + 4x – 2y = 0

Question 4.
Find the equation of the circle which passes through the origin and cuts off chords of lengths 4 and 6 on the positive side of the X-axis and Y-axis respectively.
Solution:

Let the circle cut the chord of length 4 on X-axis at point A and the chord of length 6 on the Y-axis at point B.
∴ the co-ordinates of point A are (4, 0) and co-ordinates of point B are (0, 6).
Since ∠BOA is a right angle.
AB represents the diameter of the circle.
The equation of a circle having (x₁, y₁) and (x₂, y₂) as endpoints of diameter is given by
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
Here, x₁ = 4, y₁ = 0, x₂ = 0, y₂ = 6
∴ the required equation of the circle is
⇒ (x – 4) (x – 0) + (y – 0) (y – 6) = 0
⇒ x² – 4x + y² – 6y = 0
⇒ x² + y² – 4x – 6y = 0

Question 5.
Show that the points (9, 1), (7, 9), (-2, 12) and (6, 10) are concyclic.
Solution:
Let the equation of circle passing through the points (9, 1), (7, 9), (-2, 12) be
x² + y² + 2gx + 2fy + c = 0 …….(i)
For point (9, 1),
Substituting x = 9 andy = 1 in (i), we get
81 + 1 + 18g + 2f + c = 0
⇒ 18g + 2f + c = -82 …..(ii)
For point (7, 9),
Substituting x = 7 andy = 9 in (i), we get
49 + 81 + 14g + 18f + c = 0
⇒ 14g + 18f + c = -130 ……(iii)
For point (-2, 12),
Substituting x = -2 and y = 12 in (i), we get
4 + 144 – 4g + 24f + c = 0
⇒ -4g + 24f + c = -148 …..(iv)
By (ii) – (iii), we get
4g – 16f = 48
⇒ g – 4f = 12 …..(v)
By (iii) – (iv), we get
18g – 6f = 18
⇒ 3g – f = 3 ……(vi)
By 3 × (v) – (vi), we get
-11f = 33
⇒ f = -3
Substituting f = -3 in (vi), we get
3g – (-3) = 3
⇒ 3g + 3 = 3
⇒ g = 0
Substituting g = 0 and f = -3 in (ii), we get
18(0) + 2(-3) + c = – 82
⇒ -6 + c = -82
⇒ c = -76
Equation of the circle becomes
x² + y² + 2(0)x + 2(-3)y + (-76) = 0
⇒ x² + y² – 6y – 76 = 0 ……(vii)
Now for the point (6, 10),
Substituting x = 6 and y = 10 in L.H.S. of (vii), we get
L.H.S = 6² + 10² – 6(10) – 76
= 36 + 100 – 60 – 76
= 0
= R.H.S.
∴ Point (6,10) satisfies equation (vii).
∴ the given points are concyciic.

Question 6.
The line 2x – y + 6 = 0 meets the circle x² + y² + 10x + 9 = 0 at A and B. Find the equation of circle with AB as diameter. Solution:
2x – y + 6 = 0
⇒ y = 2x + 6
Substituting y = 2x + 6 in x² + y² + 10x + 9 = 0, we get
⇒ x² + (2x + 6)² + 10x + 9 = 0
⇒ x² + 4x² + 24x + 36 + 10x + 9 = 0
⇒ 5x² + 34x + 45 = 0
⇒ 5x² + 25x + 9x + 45 = 0
⇒ (5x + 9) (x + 5) = 0
⇒ 5x = -9 or x = -5
⇒ x = \(\frac{-9}{5}\) or x = -5
When x = \(\frac{-9}{5}\),
y = 2 × \(\frac{-9}{5}\) + 6
= \(\frac{-18}{5}\) + 6
= \(\frac{-18+30}{5}\)
= \(\frac{12}{5}\)
∴ Point of intersection is A\(\left(\frac{-9}{5}, \frac{12}{5}\right)\)
When x = -5,
y = -10 + 6 = -4
∴ Point of intersection in B (-5, -4).
By diameter form, equation of circle with AB as diameter is
(x + \(\frac{9}{5}\)) (x + 5) + (y – \(\frac{12}{5}\)) (y + 4) = 0
⇒ (5x + 9) (x + 5) + (5y – 12) (y + 4) = 0
⇒ 5x² + 25x + 9x + 45 + 5y² + 20y – 12y – 48 = 0
⇒ 5x² + 5y² + 34x + 8y – 3 = 0

Question 7.
Show that x = -1 is a tangent to circle x² + y² – 4x – 2y – 4 = 0 at (-1, 1).
Solution:
Given equation of circle is x² + y² – 4x – 2y – 4 = 0.
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -4, 2f = -2, c = -4
⇒ g = -2, f = -1, c = -4
The equation of a tangent to the circle
x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
the equation of the tangent at (-1, 1) is
⇒ x(-1) + y(1) – 2(x – 1) – 1(y + 1) – 4 = 0
⇒ -3x – 3 = 0
⇒ -x – 1 = 0
⇒ x = -1
∴ x = -1 is the tangent to the given circle at (-1, 1).

Question 8.
Find the equation of tangent to the circle x² + y² = 64 at the point P(\(\frac{2 \pi}{3}\)).
Solution:
Given equation of circle is x² + y² = 64
Comparing this equation with x² + y² = r², we get r = 8
The equation of a tangent to the circle x² + y² = r² at P(θ) is x cos θ + y sin θ = r
∴ the equation of the tangent at P(\(\frac{2 \pi}{3}\)) is
⇒ x cos \(\frac{2 \pi}{3}\) + y sin \(\frac{2 \pi}{3}\) = 9
⇒ \(x\left(\frac{-1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=8\)
⇒ -x + √3y = 16
⇒ x – √3y + 16 = 0

Question 9.
Find the equation of locus of the point of intersection of perpendicular tangents drawn to the circle x = 5 cos θ and y = 5 sin θ.
Solution:
The locus of the point of intersection of perpendicular tangents is the director circle of the given circle.
x = 5 cos θ and y = 5 sin θ
⇒ x² + y² = 25 cos² θ + 25 sin² θ
⇒ x² + y² = 25 (cos² θ + sin² θ)
⇒ x² + y² = 25(1) = 25
The equation of the director circle of the circle x² + y² = a² is x² + y² = 2a².
Here, a = 5
∴ the required equation is
x² + y² = 2(5)² = 2(25)
∴ x² + y² = 50

Question 10.
Find the equation of the circle concentric with x² + y² – 4x + 6y = 1 and having radius 4 units.
Solution:
Given equation of circle is
x² + y² – 4x + 6y = 1 i.e., x² + y² – 4x + 6y – 1 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -4, 2f = 6
⇒ g = -2, f = 3
Centre of the circle = (-g, -f) = (2, -3)
Given circle is concentric with the required circle.
∴ They have same centre.
∴ Centre of the required circle = (2, -3)
The equation of a circle with centre at (h, k) and radius r is (x – h)² + (y – k)² = r²
Here, h = 2, k = -3 and r = 4
∴ the required equation of the circle is
(x – 2)² + [y – (-3)]² = 4²
⇒ (x – 2)² + (y + 3)² = 16
⇒ x² – 4x + 4 + y² + 6y + 9 – 16 = 0
⇒ x² + y² – 4x + 6y – 3 = 0

Question 11.
Find the lengths of the intercepts made on the co-ordinate axes, by the circles.
(i) x² + y² – 8x + y – 20 = 0
(ii) x² + y² – 5x + 13y – 14 = 0
Solution:
To find x-intercept made by the circle x² + y² + 2gx + 2fy + c = 0,
substitute y = 0 and get a quadratic equation in x, whose roots are, say, x₁ and x₂.

These values represent the abscissae of ends A and B of the x-intercept.
Length of x-intercept = |AB| = |x₂ – x₁|
Similarly, substituting x = 0, we get a quadratic equation in y whose roots, say, y₁ and y₂ are ordinates of the ends C and D of the y-intercept.
Length of y-intercept = |CD| = |y₂ – y₁|
(i) Given equation of the circle is
x² + y² – 8x + y – 20 = 0 ……(i)
Substituting y = 0 in (i), we get
x² – 8x – 20 = 0 ……(ii)
Let AB represent the x-intercept, where
A = (x₁, 0), B = (x₂, 0)
Then from (ii),
x₁ + x₂ = 8 and x₁x₂ = -20
(x₁ – x₂)² = (x₁ + x₂)² – 4x₁x₂
= (8)² – 4(-20)
= 64 + 80
= 144
∴ |x₁ – x₂| = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}}\) = √144 = 12
∴ Length of x – intercept =12 units
Substituting x = 0 in (i), we get
y² + y – 20 = 0 …..(iii)
Let CD represent the y – intercept,
where C = (0, y₁) and D = (0, y₂)
Then from (iii),
y₁ + y₂ = -1 and y₁y₂ = -20
(y₁ – y₂)² = (y₁ + y₂)² – 4y₁y₂
= (-1)² – 4(-20)
= 1 + 80
= 81
∴ |y₁ – y₂| = \(\sqrt{\left(y_{1}-y_{2}\right)^{2}}\) = √81 = 9
∴ Length of y – intercept = 9 units.

Alternate Method:
Given equation of the circle is x² + y² – 8x + y – 20 = 0 ……(i)
x-intercept:
Substituting y = 0 in (i), we get
x² – 8x – 20 = 0
⇒ (x – 10)(x + 2) = 0
⇒ x = 10 or x = -2
length of x-intercept = |10 – (-2)| = 12 units
y-intercept:
Substituting x = 0 in (i), we get
y² + y – 20 = 0
⇒ (y + 5)(y – 4) = 0
⇒ y = -5 or y = 4
length of y-intercept = |-5 – 4| = 9 units

(ii) Given equation of the circle is
x² + y² – 5x + 13y – 14 = 0
Substituting y = 0 in (i), we get
x² – 5x – 14 = 0 ……(ii)
Let AB represent the x-intercept, where
A = (x₁, 0), B = (x₂, 0)
Then from (ii),
x₁ + x₂ = 5 and x₁x₂ = -14
(x₁ – x₂)² = (x₁ + x₂)² – 4x₁x₂
= (5)² – 4(-14)
= 25 + 56
= 81
∴ |x₁ – x₂| = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}}\) = √81 = 9
∴ Length of x-intercept = 9 units
Substituting x = 0 in (i), we get
y² + 13y – 14 = 0 ……(iii)
Let CD represent they-intercept,
where C = (0, y₁), D = (0, y₂).
Then from (iii),
y₁ + y₂ = -13 and y₁y₂ = -14
(y₁ – y₂)² = (y₁ + y₂)² – 4y₁y₂
= (-13)² – 4(-14)
= 169 + 56
= 225
∴ |y₁ – y₂| = \(\sqrt{\left(y_{1}-y_{2}\right)^{2}}\) = √225 = 15
∴ Length ofy-intercept = 15 units

Question 12.
Show that the circles touch each other externally. Find their point of contact and the equation of their common tangent.
(i) x² + y² – 4x + 10y + 20 = 0
x² + y² + 8x – 6y – 24 = 0
(ii) x² + y² – 4x – 10y + 19 = 0
x² + y² + 2x + 8y – 23 = 0
Solution:
(i) Given equation of the first circle is x² + y² – 4x + 10y + 20 = 0
Here, g = -2, f = 5, c = 20
Centre of the first circle is C₁ = (2, -5)
Radius of the first circle is
r₁ = \(\sqrt{(-2)^{2}+5^{2}-20}\)
= \(\sqrt{4+25-20}\)
= √9
= 3
Given equation of the second circle is x² + y² + 8x – 6y – 24 = 0
Here, g = 4, f = -3, c = -24
Centre of the second circle is C₂ = (-4, 3)
Radius of the second circle is
r₂ = \(\sqrt{4^{2}+(-3)^{2}+24}\)
= \(\sqrt{16+9+24}\)
= √49
= 7
By distance formula,
C₁C₂ = \(\sqrt{(-4-2)^{2}+[3-(-5)]^{2}}\)
= \(\sqrt{36+64}\)
= √1oo
= 10
r₁ + r₂ = 3 + 7 = 10
Since, C₁C₂ = r₁ + r₂
∴ the given circles touch each other externally.

Let P(x, y) be the point of contact.
∴ P divides C₁C₂ internally in the ratio r₁ : r₂ i.e. 3 : 7.
∴ By internal division,

Equation of common tangent is
(x² + y² – 4x + 10y + 20) – (x² + y² + 8x – 6y – 24) = 0
⇒ -4x + 10y + 20 – 8x + 6y + 24 = 0
⇒ -12x + 16y + 44 = 0
⇒ 3x – 4y – 11 = 0

(ii) Given equation of the first circle is x² + y² – 4x – 10y + 19 = 0
Here, g = -2, f = -5, c = 19
Centre of the first circle is C₁ = (2, 5)
Radius of the first circle is
r₁ = \(\sqrt{(-2)^{2}+(-5)^{2}-19}\)
= \(\sqrt{4+25-19}\)
= √10
Given equation of the second circle is x² + y² + 2x + 8y – 23 = 0
Here, g = 1, f = 4, c = -23
Centre of the second circle is C₂ = (-1, -4)
Radius of the second circle is
r₂ = \(\sqrt{(-1)^{2}+4^{2}+23}\)
= \(\sqrt{1+16+23}\)
= √40
= 2√10
By distance formula,
C₁C₂ = \(\sqrt{(-1-2)^{2}+(-4-5)^{2}}\)
= \(\sqrt{9+81}\)
= √90
= 3√10
r₁ + r₂ = √10 + 2√10 = 3√10
Since, C₁C₂ = r₁ + r₂
the given circles touch each other externally.
r₁ : r₂ = √10 : 2√10 = 1 : 2
Let P(x, y) be the point of contact.

∴ P divides C₁ C₂ internally in the ratio r₁ : r₂ i.e. 1 : 2
∴ By internal division,

Point of contact = (1, 2)
Equation of common tangent is
(x² + y² – 4x – 10y + 19) – (x² + y² + 2x + 8y – 23) = 0
⇒ -4x – 10y + 19 – 2x – 8y + 23 = 0
⇒ -6x – 18y + 42 = 0
⇒ x + 3y – 7 = 0

Question 13.
Show that the circles touch each other internally. Find their point of contact and the equation of their common tangent.
(i) x² + y² – 4x – 4y – 28 = 0,
x² + y² – 4x – 12 = 0
(ii) x² + y² + 4x – 12y + 4 = 0,
x² + y² – 2x – 4y + 4 = 0
Solution:
(i) Given equation of the first circle is x² + y² – 4x – 4y – 28 = 0
Here, g = -2, f = -2, c = -28
Centre of the first circle is C₁ = (2, 2)
Radius of the first circle is
r₁ = \(\sqrt{(-2)^{2}+(-2)^{2}+28}\)
= \(\sqrt{4+4+28}\)
= √36
= 6
Given equation of the second circle is x² + y² – 4x – 12 = 0
Here, g = -2, f = 0, c = -12
Centre of the second circle is C₂ = (2, 0)
Radius of the second circle is
r₂ = \(\sqrt{(-2)^{2}+0^{2}+12}\)
= \(\sqrt{4+12}\)
= √16
= 4
By distance formula,
C₁C₂ = \(\sqrt{(2-2)^{2}+(0-2)^{2}}\)
= √4
= 2
|r₁ – r₂| = 6 – 4 = 2
Since, C₁C₂ = |r₁ – r₂|
∴ the given circles touch each other internally.
Equation of common tangent is
(x² + y² – 4x – 4y – 28) – (x² + y² – 4x – 12) = 0
⇒ -4x – 4y – 28 + 4x + 12 = 0
⇒ -4y – 16 = 0
⇒ y + 4 = 0
⇒ y = -4
Substituting y = -4 in x² + y² – 4x – 12 = 0, we get
⇒ x² + (-4)² – 4x – 12 = 0
⇒ x² + 16 – 4x – 12 = 0
⇒ x² – 4x + 4 = 0 .
⇒ (x – 2)² = 0
⇒ x = 2
∴ Point of contact is (2, -4) and equation of common tangent is y + 4 = 0.

(ii) Given equation of the first circle is x² + y² + 4x – 12y + 4 = 0
Here, g = 2, f = -6, c = 4
Centre of the first circle is C₁ = (-2, 6)
Radius of the first circle is
r₁ = \(\sqrt{2^{2}+(-6)^{2}-4}\)
= \(\sqrt{4+36-4}\)
= √36
= 6
Given equation of the second circle is x² + y² – 2x – 4y + 4 = 0
Here, g = -1, f = -2, c = 4
Centre of the second circle is C₂ = (1, 2)
Radius of the second circle is
r₂ = \(\sqrt{(-1)^{2}+(-2)^{2}-4}\)
= \(\sqrt{1+4-4}\)
= √1
= 1
By distance formula,
C₁C₂ = \(\sqrt{[1-(-2)]^{2}+(2-6)^{2}}\)
= \(\sqrt{9+16}\)
= √25
= 5
|r₁ – r₂| = 6 – 1 = 5
Since, C₁C₂ = |r₁ – r₂|
the given circles touch each other internally.
Equation of common tangent is
(x² + y² + 4x – 12y + 4) – (x² + y² – 2x – 4y + 4) = 0
⇒ 4x – 12y + 4 + 2x + 4y – 4 = 0
⇒ 6x – 8y = 0
⇒ 3x – 4y = 0
⇒ y = \(\frac{3 x}{4}\)
Substituting y = \(\frac{3 x}{4}\) in x² + y² – 2x – 4y + 4 = 0, we get

∴ Point of contact is \(\left(\frac{8}{5}, \frac{6}{5}\right)\) and equation of common tangent is 3x – 4y = 0.

Question 14.
Find the length of the tangent segment drawn from the point (5, 3) to the circle x² + y² + 10x – 6y – 17 = 0.
Solution:
Given equation of circle is x² + y² + 10x – 6y – 17 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = 10, 2f = -6, c = -17
⇒ g = 5, f = -3, c = -17
Centre of circle = (-g, -f) = (-5, 3)

In right angled ∆ABC,
BC² = AB² + AC² …..[Pythagoras theorem]
⇒ (10)² = AB²+ (√51)²
⇒ AB² = 100 – 51 = √49
⇒ AB = 7
∴ Length of the tangent segment from (5, 3) is 7 units.

Alternate method:
Given equation of circle is x² + y² + 10x – 6y – 17 = 0
Here, g = 5, f = -3, c = -17
Length of the tangent segment to the circle x² + y² + 2gx + 2fy + c = 0 from the point (x₁, y₁) is \(\sqrt{x_{1}^{2}+y_{1}^{2}+2 g x_{1}+2 f y_{1}+c}\)
Length of the tangent segment from (5, 3)
= \(\sqrt{(5)^{2}+(3)^{2}+10(5)-6(3)-17}\)
= \(\sqrt{25+9+50-18-17}\)
= √49
= 7 units

Question 15.
Find the value of k, if the length of the tangent segment from the point (8, -3) to the circle x² + y² – 2x + ky – 23 = 0 is √10.
Solution:
Given equation of the circle is x² + y² – 2x + ky – 23 = 0
Here, g = -1, f = \(\frac{\mathrm{k}}{2}\), c = -23
Length of the tangent segment to the circle x² + y² + 2gx + 2fy + c = 0 from the point (x₁, y₁) is \(\sqrt{x_{1}^{2}+y_{1}^{2}+2 g x_{1}+2 f y_{1}+c}\)
Length of the tangent segment from (8, -3) = √10
⇒ \(\sqrt{8^{2}+(-3)^{2}-2(8)+k(-3)-23}=\sqrt{10}\)
⇒ 64 + 9 – 16 – 3k – 23 = 10 …..[Squaring both the sides]
⇒ 34 – 3k = 10
⇒ 3k = 24
⇒ k = 8

Question 16.
Find the equation of tangent to circle x² + y² – 6x – 4y = 0, at the point (6, 4) on it.
Solution:
Given equation of the circle is x² + y² – 6x – 4y = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -6, 2f = -4, c = 0
⇒ g = -3, f = -2, c = 0
The equation of a tangent to the circle x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
the equation of the tangent at (6, 4) is
x(6) + y(4) – 3(x + 6) – 2(y + 4) + 0 = 0
⇒ 6x + 4y – 3x – 18 – 2y – 8 = 0
⇒ 3x + 2y – 26 = 0

Alternate method:
Given equation of the circle is x² + y² – 6x – 4y = 0
x(x – 6) + y(y – 4) = 0, which is in diameter form where (0, 0) and (6, 4) are endpoints of diameter.

Slope of OP = \(\frac{4-0}{6-0}=\frac{2}{3}\)
Since, OP is perpendicular to the required tangent.
Slope of the required tangent = \(\frac{-3}{2}\)
the equation of the tangent at (6, 4) is
y – 4 = \(\frac{-3}{2}\) (x – 6)
⇒ 2(y – 4) = 3(x – 6)
⇒ 2y – 8 = -3x + 18
⇒ 3x + 2y – 26 = 0

Question 17.
Fihd the equation of tangent to circle x² + y² = 5, at the point (1, -2) on it.
Solution:
Given equation of the circle is x² + y² = 5
Comparing this equation with x² + y² = r², we get
r² = 5
The equation of a tangent to the circle x² + y² = r² at (x₁, y₁) is xx₁ + yy₁ = r²
the equation of the tangent at (1, -2) is
x(1) + y(-2) = 5
⇒ x – 2y = 5

Question 18.
Find the equation of tangent to circle x = 5 cos θ, y = 5 sin θ, at the point θ = \(\frac{\pi}{3}\) on it.
Solution:
The equation of a tangent to the circle x² + y² = r² at P(θ) is x cos θ + y sin θ = r
Here, r = 5, θ = \(\frac{\pi}{3}\)
the equation of the tangent at P(\(\frac{\pi}{3}\)) is
x cos \(\frac{\pi}{3}\) + y sin \(\frac{\pi}{3}\) = 5
⇒ \(x\left(\frac{1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)=5\)
⇒ x + √3y = 10

Question 19.
Show that 2x + y + 6 = 0 is a tangent to x² + y² + 2x – 2y – 3 = 0. Find its point of contact.
Solution:
Given equation of circle is
x² + y² + 2x – 2y – 3 = 0 ….(i)
Given equation of line is 2x + y + 6 = 0
y = -6 – 2x ……(ii)
Substituting y = -6 – 2x in (i), we get
x + (-6 – 2x)² + 2x – 2(-6 – 2x) – 3 = 0
⇒ x² + 36 + 24x + 4x² + 2x + 12 + 4x – 3 = 0
⇒ 5x² + 30x + 45 = 0
⇒ x² + 6x + 9 = 0
⇒ (x + 3)² = 0
⇒ x = -3
Since, the roots are equal.
∴ 2x + y + 6 = 0 is a tangent to x² + y² + 2x – 2y – 3 = 0
Substituting x = -3 in (ii), we get
y = -6 – 2(-3) = -6 + 6 = 0
Point of contact = (-3, 0)

Question 20.
If the tangent at (3, -4) to the circle x² + y² = 25 touches the circle x² + y² + 8x – 4y + c = 0, find c.
Solution:
The equation of a tangent to the circle
x² + y² = r² at (x₁, y₁) is xx₁ + yy₁ = r²
Equation of the tangent at (3, -4) is
x(3) + y(-4) = 25
⇒ 3x – 4y – 25 = 0 ……(i)
Given equation of circle is x² + y² + 8x – 4y + c = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = 8, 2f = -4
⇒ g = 4, f = -2
∴ C = (-4, 2) and r = \(\sqrt{4^{2}+(-2)^{2}-c}=\sqrt{20-c}\)
Since line (i) is a tangent to this circle also, the perpendicular distance from C(-4, 2) to line (i) is equal to radius r.

Question 21.
Find the equations of the tangents to the circle x² + y² = 16 with slope -2.
Solution:
Given equation of the circle is x² + y² = 16
Comparing this equation with x² + y² = a², we get
a² = 16
Equations of the tangents to the circle x² + y² = a² with slope m are
\(y=m x \pm \sqrt{a^{2}\left(1+m^{2}\right)}\)
Here, m = -2, a² = 16
the required equations of the tangents are
y = \(-2 x \pm \sqrt{16\left[1+(-2)^{2}\right]}\)
⇒ y = \(-2 x \pm \sqrt{16(5)}\)
⇒ y = -2x ± 4√5
⇒ 2x + y ± 4√5 = 0

Question 22.
Find the equations of the tangents to the circle x² + y² = 4 which are parallel to 3x + 2y + 1 = 0.
Solution:
Given equation of the circle is x² + y² = 4
Comparing this equation with x² + y² = a², we get
a² = 4
Given equation of the line is 3x + 2y + 1 = 0
Slope of this line = \(\frac{-3}{2}\)
Since, the required tangents are parallel to the given line.
Slope of required tangents (m) = \(\frac{-3}{2}\)
Equations of the tangents to the circle x² + y² = a² with slope m are
y = mx ± \(\sqrt{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}\)
the required equations of the tangents are

Question 23.
Find the equations of the tangents to the circle x² + y² = 36 which are perpendicular to the line 5x + y = 2.
Solution:
Given equation of the circle is x² + y² = 36
Comparing this equaiton with x² + y² = a², we get
a² = 36
Given equation of line is 5x + y = 2
Slope of this line = -5
Since, the required tangents are perpendicular to the given line.
Slope of required tangents (m) = \(\frac{1}{5}\)
Equations of the tangents to the circle x² + y² = a² with slope m are
y = mx ± \(\sqrt{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}\)
the required equations of the tangents are

Question 24.
Find the equations of the tangents to the circle x² + y² – 2x + 8y – 23 = 0 having slope 3.
Solution:
Let the equation of the tangent with slope 3 be y = 3x + c.
3x – y + c = 0 ……(i)
Given equation of circle is x² + y² – 2x + 8y – 23 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -2, 2f = 8, c = -23
g = -1, f = 4, c = -23
The centre of the circle is C(1, -4)
and its radius = \(\sqrt{1+16+23}\)
= √40
= 2√10
Since line (i) is a tangent to this circle the perpendicular distance from C(1, -4) to line (i) is equal to radius r.
\(\left|\frac{3(1)+4+c}{\sqrt{9+1}}\right|\) = 2√10
⇒ \(\left|\frac{7+c}{\sqrt{10}}\right|\) = 2√10
⇒ (7 + c) = ± 20
⇒ 7 + c = 20 or 7 + c = -20
⇒ c = 13 or c = – 27
∴ Equations of the tangents are 3x – y + 13 = 0 and 3x – y – 21 = 0

Question 25.
Find the equation of the locus of a point, the tangents from which to the circle x² + y² = 9 are at right angles.
Solution:
Given equation of the circle is x² + y² = 9
Comparing this equation with x² + y² = a², we get
a² = 9
The locus of the point of intersection of perpendicular tangents is the director circle of the given circle.
The equation of the director circle of the circle x² + y² = a² is x² + y² = 2a².
the required equation is
x² + y² = 2(9)
x² + y² = 18

Alternate method:
Given equation of the circle is x² + y² = 9
Comparing this equation with x² + y² = a², we get a² = 9
Let P(x₁, y₁) be a point on the required locus.
Equations of the tangents to the circle x² + y² = a² with slope m are
y = mx ± \(\sqrt{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}\)
∴ Equations of the tangents are
y = mx ± \(\sqrt{9\left(\mathrm{~m}^{2}+1\right)}\)
⇒ y = mx ± 3\(\sqrt{1+m^{2}}\)
Since, these tangents pass through (x₁, y₁).
y₁ = mx₁ ± 3\(\sqrt{1+m^{2}}\)
⇒ y₁ – mx₁ = ± 3\(\sqrt{1+m^{2}}\)
⇒ (y₁ – mx₁)² = 9(1 + m²) ……[Squaring both the sides]
⇒ \(y_{1}^{2}-2 m x_{1} y_{1}+m^{2} x_{1}^{2}=9+9 m^{2}\)
⇒ \(\left(x_{1}^{2}-9\right) \mathrm{m}^{2}-2 \mathrm{~m} x_{1} y_{1}+\left(y_{1}^{2}-9\right)=0\)
This is a quadratic equation which has two roots m₁ and m₂.
m₁m₂ = \(\frac{y_{1}^{2}-9}{x_{1}^{2}-9}\)
Since, the tangents are at right angles.
m₁m₂ = -1
⇒ \(\frac{y_{1}^{2}-9}{x_{1}^{2}-9}=-1\)
⇒ \(y_{1}^{2}-9=9-x_{1}^{2}\)
⇒ \(x_{1}^{2}+y_{1}^{2}=18\)
Equation of the locus of point P is x² + y² = 18.

Question 26.
Tangents to the circle x² + y² = a² with inclinations, θ₁ and θ₂ intersect in P. Find the locus of P such that
(i) tan θ₁ + tan θ₂ = 0
(ii) cot θ₁ + cot θ₂ = 5
(iii) cot θ₁ . cot θ₂ = c
Solution:
Let P(x₁, y₁) be a point on the required locus.
Equations of the tangents to the circle x² + y² = a² with slope m are
y = mx ± \(\sqrt{\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)}\)
Since, these tangents pass through (x₁, y₁).

Class 11 Maharashtra State Board Maths Solution 

• Exercise 6.1 Class 11 Maths 1
• Exercise 6.2 Class 11 Maths 1
• Exercise 6.3 Class 11 Maths 1
• Miscellaneous Exercise 6 Class 11 Maths 1
• Exercise 7.1 Class 11 Maths 1
• Exercise 7.2 Class 11 Maths 1
• Exercise 7.3 Class 11 Maths 1
• Miscellaneous Exercise 7 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Ex 6.3 Questions and Answers.

11th Maths Part 1 Circle Exercise 6.3 Questions And Answers Maharashtra Board

Question 1.
Write the parametric equations of the circles:
(i) x² + y² = 9
(ii) x² + y² + 2x – 4y – 4 = 0
(iii) (x – 3)² + (y + 4)² = 25
Solution:
(i) Given equation of the circle is
x² + y² = 9
⇒ x² + y² = 3²
Comparing this equation with x² + y² = r², we get r = 3
The parametric equations of the circle in terms of θ are
x = r cos θ and y = r sin θ
⇒ x = 3 cos θ and y = 3 sin θ

(ii) Given equation of the circle is
x² + y² + 2x – 4y – 4 = 0
⇒ x² + 2x + y² – 4y – 4 = 0
⇒ x² + 2x + 1 – 1 + y² – 4y + 4 – 4 – 4 = 0
⇒ (x² + 2x + 1 ) + (y² – 4y + 4) – 9 = 0
⇒ (x + 1)² + (y – 2)² = 9
⇒ (x + 1)² + (y – 2)² = 3²
Comparing this equation with (x – h)² + (y – k)² = r², we get
h = -1, k = 2 and r = 3
The parametric equations of the circle in terms of θ are
x = h + r cos θ and y = k + r sin θ
⇒ x = -1 + 3 cos θ and y = 2 + 3 sin θ

(iii) Given equation of the circle is
(x – 3)² + (y + 4)² = 25
⇒ (x – 3)² + (y + 4)² = 5²
Comparing this equation with (x – h)² + (y – k)² = r², we get
h = 3, k = -4 and r = 5
The parametric equations of the circle in terms of θ are
x = h + r cos θ and y = k + r sin θ
⇒ x = 3 + 5 cos θ and y = -4 + 5 sin θ

Question 2.
Find the parametric representation of the circle 3x² + 3y² – 4x + 6y – 4 = 0.
Solution:
Given equation of the circle is 3x² + 3y² – 4x + 6y – 4 = 0
Dividing throughout by 3, we get

Comparing this equation with (x – h)² + (y – k)² = r², we get
h = \(\frac{2}{3}\), k = -1 and r = \(\frac{5}{3}\)
The parametric representation of the circle in terms of θ are
x = h + r cos θ and y = k + r sin θ
⇒ x = \(\frac{2}{3}\) + \(\frac{5}{3}\) cos θ and y = -1 + \(\frac{5}{3}\) sin θ

Question 3.
Find the equation of a tangent to the circle x² + y² – 3x + 2y = 0 at the origin.
Solution:
Given equation of the circle is x² + y² – 3x + 2y = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -3, 2f = 2, c = 0
⇒ g = \(-\frac{3}{2}\), f = 1, c = 0
The equation of a tangent to the circle
x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is xx₁ +yy₁ + g(x + x₁) + f(y + y₁) + c = 0
The equation of the tangent at (0, 0) is
x(0) + y(0) + (\(-\frac{3}{2}\)) (x + 0) + 1(y + 0) + 0 = 0
⇒ \(-\frac{3}{2}\)x + y = 0
⇒ 3x – 2y = 0

Question 4.
Show that the line 7x – 3y – 1 = 0 touches the circle x² + y² + 5x – 7y + 4 = 0 at point (1, 2).
Solution:
Given equation of the circle is x² + y² + 5x – 7y + 4 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = 5, 2f = -7, c = 4
⇒ g = \(\frac{5}{2}\), f = \(\frac{-7}{2}\), c = 4
The equation of a tangent to the circle x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
The equation of the tangent at (1, 2) is

7x – 3y – 1 = 0, which is same as the given line.
The line 7x – 3y – 1 = 0 touches the given circle at (1, 2).

Question 5.
Find the equation of tangent to the circle x² + y² – 4x + 3y + 2 = 0 at the point (4, -2).
Solution:
Given equation of the circle is x² + y² – 4x + 3y + 2 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -4, 2f = 3, c = 2
g = -2, f = \(\frac{3}{2}\), c = 2
The equation of a tangent to the circle x² + y² + 2gx + 2fy + c = 0 at (x₁, y₁) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
The equation of the tangent at (4, -2) is
x(4) + y(-2) – 2(x + 4) + \(\frac{3}{2}\)(y – 2) + 2 = 0
⇒ 4x – 2y – 2x – 8 + \(\frac{3}{2}\) y – 3 + 2 = 0
⇒ 2x – \(\frac{1}{2}\)y – 9 = 0
⇒ 4x – y – 18 = 0

Class 11 Maharashtra State Board Maths Solution 

• Exercise 6.1 Class 11 Maths 1
• Exercise 6.2 Class 11 Maths 1
• Exercise 6.3 Class 11 Maths 1
• Miscellaneous Exercise 6 Class 11 Maths 1
• Exercise 7.1 Class 11 Maths 1
• Exercise 7.2 Class 11 Maths 1
• Exercise 7.3 Class 11 Maths 1
• Miscellaneous Exercise 7 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Ex 6.2 Questions and Answers.

11th Maths Part 1 Circle Exercise 6.2 Questions And Answers Maharashtra Board

Question 1.
Find the centre and radius of each of the following circles:
(i) x² + y² – 2x + 4y – 4 = 0
(ii) x² + y² – 6x – 8y – 24 = 0
(iii) 4x² + 4y² – 24x – 8y – 24 = 0
Solution:
(i) Given equation of the circle is x² + y² – 2x + 4y – 4 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -2, 2f = 4 and c = -4
⇒ g = -1, f = 2 and c = -4
Centre of the circle = (-g, -f) = (1, -2)
and radius of the circle

(ii) Given equation of the circle is x² + y² – 6x – 8y – 24 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -6, 2f = -8 and c = -24
⇒ g = -3, f = -4 and c = -24
Centre of the circle = (-g, -f) = (3, 4)
and radius of the circle

(iii) Given equation of the circle is 4x² + 4y² – 24x – 8y – 24 = 0
Dividing throughout by 4, we get x² + y² – 6x – 2y – 6 = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = -6, 2f = -2 and c = -6
⇒ g = -3, f = -1 and c = -6
Centre of the circle = (-g, -f) = (3, 1)
and radius of the circle

Question 2.
Show that the equation 3x² + 3y² + 12x + 18y – 11 = 0 represents a circle.
Solution:
Given equation is 3x² + 3y² + 12x + 18y – 11 = 0
Dividing throughout by 3, we get
x² + y² + 4x + 6y – \(\frac{11}{3}\) = 0
Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get
2g = 4, 2f = 6, c = \(\frac{-11}{3}\)
⇒ g = 2, f = 3, c = \(\frac{-11}{3}\)
Now, g² + f² – c = (2)² + (3)² – (\(\frac{-11}{3}\))
= 4 + 9 + \(\frac{11}{3}\)
= \(\frac{50}{3}\) > 0
∴ The given equation represents a circle.

Question 3.
Find the equation of the circle passing through the points (5, 7), (6, 6), and (2, -2).
Solution:
Let C(h, k) be the centre of the required circle.
Since the required circle passes through points A(5, 7), B(6, 6), and D(2, -2),
CA = CB = CD = radius

Consider, CA = CD
By distance formula,
\(\sqrt{(\mathrm{h}-5)^{2}+(\mathrm{k}-7)^{2}}=\sqrt{(\mathrm{h}-2)^{2}+[\mathrm{k}-(-2)]^{2}}\)
Squaring both the sides, we get
⇒ (h – 5)² + (k – 7)² = (h – 2)² + (k + 2)²
⇒ h² – 10h + 25 + k² – 14k + 49 = h² – 4h + 4 + k² + 4k + 4
⇒ -10h – 14k + 74 = -4h + 4k + 8
⇒ 6h + 18k – 66 = 0
⇒ h + 3k – 11 = 0 …..(i)
Consider, CB = CD
By distance formula,
\(\sqrt{(h-6)^{2}+(k-6)^{2}}=\sqrt{(h-2)^{2}+[k-(-2)]^{2}}\)
Squaring both the sides, we get
⇒ (h – 6)² + (k – 6)² = (h – 2)² + (k + 2)²
⇒ h² – 12h + 36 + k² – 12k + 36 = h² – 4h + 4 + k² + 4k + 4
⇒ -12h – 12k + 72 = -4h + 4k + 8
⇒ 8h + 16k – 64 = 0
⇒ h + 2k – 8 = 0 ……(ii)
By (i) – (ii), we get k = 3
Substituting k = 3 in (i), we get
h + 3(3) – 11 = 0
⇒ h + 9 – 11 = 0
⇒ h = 2
Centre of the circle is C(2, 3).
radius (r) = CD
= \(\sqrt{(2-2)^{2}+(3+2)^{2}}\)
= \(\sqrt{0+5^{2}}\)
= √25
= 5
The equation of a circle with centre at (h, k) and radius r is given by (x – h)²+ (y – k)² = r²
Here, h = 2, k = 3
The required equation of the circle is
(x – 2)² + (y – 3)² = 5²
⇒ x² – 4x + 4 + y² – 6y + 9 = 25
⇒ x² + y² – 4x – 6y + 4 + 9 – 25 = 0
⇒ x² + y² – 4x – 6y – 12 = 0

Question 4.
Show that the points (3, -2), (1, 0), (-1, -2) and (1, -4) are concyclic.
Solution:
Let the equation of the circle passing through the points (3, -2), (1, 0) and (-1, -2) be
x² + y² + 2gx + 2fy + c = 0 …..(i)
For point (3, -2),
Substituting x = 3 and y = -2 in (i), we get
9 + 4 + 6g – 4f + c = 0
⇒ 6g – 4f + c = -13 ….(ii)
For point (1, 0),
Substituting x = 1 andy = 0 in (i), we get
1 + 0 + 2g + 0 + c = 0
⇒ 2g + c = -1 ……(iii)
For point (-1, -2),
Substituting x = -1 and y = -2, we get
1 + 4 – 2g – 4f + c = 0
⇒ 2g + 4f – c = 5 …….(iv)
Adding (ii) and (iv), we get
8g = -8
⇒ g = -1
Substituting g = -1 in (iii), we get
-2 + c = -1
⇒ c = 1
Substituting g = -1 and c = 1 in (iv), we get
-2 + 4f – 1 = 5
⇒ 4f = 8
⇒ f = 2
Substituting g = -1, f = 2 and c = 1 in (i), we get
x² + y² – 2x + 4y + 1 = 0 ……….(v)
If (1, -4) satisfies equation (v), the four points are concyclic.
Substituting x = 1, y = -4 in L.H.S of (v), we get
L.H.S. = (1)² + (-4)² – 2(1) + 4(-4) + 1
= 1 + 16 – 2 – 16 + 1
= 0
= R.H.S.
Point (1, -4) satisfies equation (v).
∴ The given points are concyclic.

Class 11 Maharashtra State Board Maths Solution 

• Exercise 6.1 Class 11 Maths 1
• Exercise 6.2 Class 11 Maths 1
• Exercise 6.3 Class 11 Maths 1
• Miscellaneous Exercise 6 Class 11 Maths 1
• Exercise 7.1 Class 11 Maths 1
• Exercise 7.2 Class 11 Maths 1
• Exercise 7.3 Class 11 Maths 1
• Miscellaneous Exercise 7 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Ex 6.1 Questions and Answers.

11th Maths Part 1 Circle Exercise 6.1 Questions And Answers Maharashtra Board

Question 1.
Find the equation of a circle with
(i) centre at origin and radius 4.
(ii) centre at (-3, -2) and radius 6.
(iii) centre at (2, -3) and radius 5.
(iv) centre at (-3, -3) passing through point (-3, -6).
Solution:
(i) The equation of a circle with centre at origin and radius ‘r’ is given by
x² + y² = r²
Here, r = 4
∴ The required equation of the circle is x² + y² = 4² i.e., x² + y² = 16.

(ii) The equation of a circle with centre at (h, k) and radius ‘r’ is given by
(x – h)² + (y – k)² = r²
Here, h = -3, k = -2 and r = 6
∴ The required equation of the circle is
[x – (-3)]² + [y – (-2)]² = 6²
⇒ (x + 3)² + (y + 2)² = 36
⇒ x² + 6x + 9 + y² + 4y + 4 – 36 = 0
⇒ x² + y² + 6x + 4y – 23 = 0

(iii) The equation of a circle with centre at (h, k) and radius ‘r’ is given by
(x – h)² + (y – k)² = r²
Here, h = 2, k = -3 and r = 5
The required equation of the circle is
(x – 2)² + [y – (-3)]² = 5²
⇒ (x – 2)² + (y + 3)² = 25
⇒ x² – 4x + 4 + y² + 6y + 9 – 25 = 0
⇒ x² + y² – 4x + 6y – 12 = 0

(iv) Centre of the circle is C (-3, -3) and it passes through the point P (-3, -6).

The equation of a circle with centre at (h, k) and radius ‘r’ is given by
(x – h)² + (y – k)² = r²
Here, h = -3, k = -3, r = 3
The required equation of the circle is
[x – (-3)]² + [y – (-3)]² = 3²
⇒ (x + 3)² + (y + 3)² = 9
⇒ x² + 6x + 9 + y² + 6y + 9 – 9 = 0
⇒ x² + y² + 6x + 6y + 9 = 0

Check:
If the point (-3, -6) satisfies x² + y² + 6x + 6y + 9 = 0, then our answer is correct.
L.H.S. = x² + y² + 6x + 6y + 9
= (-3)² + (-6)² + 6(-3) – 6(-6) + 9
= 9 + 36 – 18 – 36 + 9
= 0
= R.H.S.
Thus, our answer is correct.

Question 2.
Find the centre and radius of the following circles:
(i) x² + y² = 25
(ii) (x – 5)² + (y – 3)² = 20
(iii) \(\left(x-\frac{1}{2}\right)^{2}+\left(y+\frac{1}{3}\right)^{2}=\frac{1}{36}\)
Solution:
(i) Given equation of the circle is
x² + y² = 25
⇒ x² + y² = (5)²
Comparing this equation with x² + y² = r², we get r = 5
Centre of the circle is (0, 0) and radius of the circle is 5.

(ii) Given equation of the circle is
(x – 5)² + (y – 3)² = 20
⇒ (x – 5)² + (y – 3)² = (√20)²
Comparing this equation with (x – h)² + (y – k)² = r², we get
h = 5, k = 3 and r = √20 = 2√5
Centre of the circle = (h, k) = (5, 3)
and radius of the circle = 2√5.

(iii) Given the equation of the circle is

Comparing this equation with (x – h)² + (y – k)² = r², we get
h = \(\frac{1}{2}\), k = \(\frac{-1}{3}\) and r = \(\frac{1}{6}\)
Centre of the circle = (h, k) = (\(\frac{1}{2}\), \(\frac{-1}{3}\)) and radius of the circle = \(\frac{1}{6}\)

Question 3.
Find the equation of the circle with centre
(i) at (a, b) and touching the Y-axis.
(ii) at (-2, 3) and touching the X-axis.
(iii) on the X-axis and passing through the origin having radius 4.
(iv) at (3, 1) and touching the line 8x – 15y + 25 = 0.
Solution:
(i) Since the circle is touching the Y-axis, the radius of the circle is X-co-ordinate of the centre.

∴ r = a
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = a, k = b
The required equation of the circle is
⇒ (x – a)² + (y – b)² = a²
⇒ x² – 2ax + a² + y² – 2by + b² = a²
⇒ x² + y² – 2ax – 2by + b² = 0

(ii) Since the circle is touching the X-axis, the radius of the circle is the Y co-ordinate of the centre.

∴ r = 3
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = -2, k = 3
The required equation of the circle is
⇒ (x + 2)² + (y – 3)² = 3²
⇒ x² + 4x + 4 + y² – 6y + 9 = 9
⇒ x² + y² + 4x – 6y + 4 = 0

(iii) Let the co-ordinates of the centre of the required circle be C (h, 0).
Since the circle passes through the origin i.e., O(0, 0)
OC = radius
⇒ \(\sqrt{(h-0)^{2}+(0-0)^{2}}=4\)
⇒ h² = 16
⇒ h = ±4

the co-ordinates of the centre are (4, 0) or (-4, 0).
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = ± 4, k = 0, r = 4
The required equation of the circle is
⇒ (x – 4)² + (y – 0)² = 4² or (x + 4)² + (y – 0)² = 4²
⇒ x² – 8x + 16 + y² = 16 or x² + 8x + 16 + y² = 16
⇒ x² + y² – 8x = 0 or x² + y² + 8x = 0

(iv) Centre of the circle is C (3, 1).
Let the circle touch the line 8x – 15y + 25 = 0 at point M.

CM = radius (r)
CM = Length of perpendicular from centre C(3, 1) on the line 8x – 15y + 25 = 0

The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = 3, k = 1 and r = 2
The required equation of the circle is
⇒ (x – 3)² + (y – 1)² = 2²
⇒ x² – 6x + 9 + y² – 2y + 1 = 4
⇒ x² + y² – 6x – 2y + 10 – 4 = 0
⇒ x² + y² – 6x – 2y + 6 = 0

Question 4.
Find the equation of the circle, if the equations of two diameters are 2x + y = 6 and 3x + 2y = 4 and radius is 9.
Solution:

Given equations of diameters are 2x + y = 6 and 3x + 2y = 4.
Let C (h, k) be the centre of the required circle.
Since point of intersection of diameters is the centre of the circle,
x = h, y = k
Equations of diameters become
2h + k = 6 …..(i)
and 3h + 2k = 4 ……..(ii)
By (ii) – 2 × (i), we get
-h = -8
⇒ h = 8
Substituting h = 8 in (i), we get
2(8) + k = 6
⇒ k = 6 – 16
⇒ k = -10
Centre of the circle is C (8, -10) and radius, r = 9
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)² + (y – k)² = r²
Here, h = 8, k = -10
The required equation of the circle is
⇒ (x – 8)² + (y + 10)² = 9²
⇒ x² – 16x + 64 + y² + 20y + 100 = 81
⇒ x² + y² – 16x + 20y + 100 + 64 – 81 = 0
⇒ x² + y² – 16x + 20y + 83 = 0

Question 5.
If y = 2x is a chord of the circle x² + y² – 10x = 0, find the equation of the circle with this chord as diameter.
Solution:

y = 2x is the chord of the given circle.
It satisfies the equation of a given circle.
Substituting y = 2x in x² + y² – 10x = 0, we get
⇒ x² + (2x)² – 10x = 0
⇒ x² + 4x² – 10x = 0
⇒ 5x² – 10x = 0
⇒ 5x(x – 2) = 0
⇒ x = 0 or x = 2
When x = 0, y = 2x = 2(0) = 0
∴ A = (0, 0)
When x = 2, y = 2x = 2 (2) = 4
∴ B = (2, 4)
End points of chord AB are A(0, 0) and B(2, 4).
Chord AB is the diameter of the required circle.
The equation of a circle having (x₁, y₁) and (x₂, y₂) as end points of diameter is given by
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
Here, x₁ = 0, y₁ = 0, x₂ = 2, y₂ = 4
The required equation of the circle is
⇒ (x – 0) (x – 2) + (y – 0) (y – 4 ) = 0
⇒ x² – 2x + y² – 4y = 0
⇒ x² + y² – 2x – 4y = 0

Question 6.
Find the equation of a circle with a radius of 4 units and touch both the co-ordinate axes having centre in the third quadrant. Solution:
The radius of the circle = 4 units
Since the circle touches both the co-ordinate axes and its centre is in the third quadrant,
the centre of the circle is C(-4, -4).

The equation of a circle with centre at (h, k) and radius r is given by (x – h)² + (y – k)² = r²
Here, h = -4, k = -4, r = 4
the required equation of the circle is
⇒ [x – (-4)]² + [y – (-4)]² = 4²
⇒ (x + 4)² + (y + 4)² = 16
⇒ x² + 8x + 16 + y² + 8y + 16 – 16 = 0
⇒ x² + y² + 8x + 8y + 16 = 0

Question 7.
Find the equation of the circle passing through the origin and having intercepts 4 and -5 on the co-ordinate axes.
Solution:
Let the circle intersect X-axis at point A and intersect Y-axis at point B.
the co-ordinates of point A are (4, 0) and the co-ordinates of point B are (0, -5).

Since ∠AOB is a right angle,
AB represents the diameter of the circle.
The equation of a circle having (x₁, y₁) and (x₂, y₂) as end points of diameter is given by
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
Here, x₁ = 4, y₁ = 0, x₂ = 0, y₂ = -5
The required equation of the circle is
⇒ (x – 4) (x – 0) + (y – 0) [y – (-5)] = 0
⇒ x(x – 4) + y(y + 5) = 0
⇒ x² – 4x + y² + 5y = 0
⇒ x² + y² – 4x + 5y = 0

Question 8.
Find the equation of a circle passing through the points (1, -4), (5, 2) and having its centre on line x – 2y + 9 = 0.
Solution:

Let C(h, k) be the centre of the required circle which lies on the line x – 2y + 9 = 0.
Equation of line becomes
h – 2k + 9 = 0 …..(i)
Also, the required circle passes through points A(1, -4) and B(5, 2).
CA = CB = radius
CA = CB
By distance formula,
\(\sqrt{(\mathrm{h}-1)^{2}+[\mathrm{k}-(-4)]^{2}}=\sqrt{(\mathrm{h}-5)^{2}+(\mathrm{k}-2)^{2}}\)
Squaring both the sides, we get
⇒ (h – 1)² + (k + 4)² = (h – 5)² + (k – 2)²
⇒ h² – 2h + 1 + k² + 8k + 16 = h² – 10h + 25 + k² – 4k + 4
⇒ -2h + 8k + 17 = -10h – 4k + 29
⇒ 8h + 12k – 12 = 0
⇒ 2h + 3k – 3 = 0 ……(ii)
By (ii) – (i) × 2, we get
7k = 21
⇒ k = 3
Substituting k = 3 in (i), we get
h – 2(3) + 9 = 0
⇒ h – 6 + 9 = 0
⇒ h = -3
Centre of the circle is C(-3, 3).
radius (r) = CA

The equation of a circle with centre at (h, k) and radius r is given by (x – h)² + (y – k)² = r²
Here, h = -3, k = 3, r = √65
The required equation of the circle is
⇒ [x – (-3)]² + (y – 3)² = (√65)²
⇒ (x + 3)² + (y – 3)² = 65
⇒ x² + 6x + 9 + y² – 6y + 9 – 65 = 0
⇒ x² + y² + 6x – 6y – 47 = 0

Class 11 Maharashtra State Board Maths Solution 

• Exercise 6.1 Class 11 Maths 1
• Exercise 6.2 Class 11 Maths 1
• Exercise 6.3 Class 11 Maths 1
• Miscellaneous Exercise 6 Class 11 Maths 1
• Exercise 7.1 Class 11 Maths 1
• Exercise 7.2 Class 11 Maths 1
• Exercise 7.3 Class 11 Maths 1
• Miscellaneous Exercise 7 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Miscellaneous Exercise 5 Questions and Answers.

11th Maths Part 1 Straight Line Miscellaneous Exercise 5 Questions And Answers Maharashtra Board

(I) Select the correct option from the given alternatives.

Question 1.
If A is (5, -3) and B is a point on the X-axis such that the slope of line AB is -2, then B ≡
(a) (7, 2)
(b) (\(\frac{7}{2}\), 0)
(c) (0, \(\frac{7}{2}\))
(d) (\(\frac{2}{7}\), 0)
Answer:
(b) (\(\frac{7}{2}\), 0)
Hint:
Let B(x, 0) be the point on X-axis.
We have A = (5, -3)
slope of AB = -2
⇒ \(\frac{0-(-3)}{x-5}\) = -2
⇒ 3 = -2(x – 5)
⇒ 3 = -2x + 10
⇒ x = \(\frac{7}{2}\)
Co-ordinates of point B = (\(\frac{7}{2}\), 0)

Question 2.
If the point (1, 1) lies on the line passing through the points (a, 0) and (0, b), then \(\frac{1}{a}+\frac{1}{b}=\)
(a) -1
(b) 0
(c) 1
(d) \(\frac{1}{a b}\)
Answer:
(c) 1
Hint:
Line passes through (a, 0), (0, b).
x-intercept = a, y-intercept = b
∴ Equation of line is \(\frac{x}{a}+\frac{y}{b}=1\) …….(i)
Since line (i) passes through (1, 1), (1, 1) satisfies (i)
∴ \(\frac{1}{a}+\frac{1}{b}=1\)

Question 3.
If A(1, -2), B(-2, 3) and C(2, -5) are the vertices of ΔABC, then the equation of median BE is
(a) 7x + 13y + 47 = 0
(b) 13x + 7y + 5 = 0
(c) 7x – 13y + 5 = 0
(d) 13x – 7y – 5 = 0
Answer:
(b) 13x + 7y + 5 = 0
Hint:

Question 4.
The equation of the line through (1, 2), which makes equal intercepts on the axes, is
(a) x + y = 1
(b) x + y = 2
(c) x + y = 4
(d) x + y = 3
Answer:
(d) x + y = 3
Hint:
Let the equation of required line be
\(\frac{x}{a}+\frac{y}{b}=1\) ……..(i)
Since the line makes equal intercepts on the axes, a = b
\(\frac{x}{a}+\frac{y}{a}=1\)
∴ x + y = a ……(ii)
But, equation (ii) passes through (1, 2).
1 + 2 = a
∴ a = 3
Substituting a = 3 in equation (ii), we get
x + y = 3

Question 5.
If the line kx + 4y = 6 passes through the point of intersection of the two lines 2x + 3y = 4 and 3x + 4y = 5, then k =
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2
Hint:
Given two lines are
2x + 3y = 4 ……(i)
3x + 4y = 5 …….(ii)
Multiplying (i) by 3 and (ii) by 2 and then subtracting, we get
y = 2
Substituting y = 2 in (i), we get
x = -1
∴ Point of intersection of lines (i) and (ii) is (-1, 2).
Given that the line kx + 4y = 6 passes through (-1, 2).
k(-1) + 4(2) = 6
∴ k = 2

Question 6.
The equation of a line, having inclination 120° with positive direction of X-axis, which is at a distance of 3 units from the origin is
(a) √3x ± y + 6 = 0
(b) √3x + y ± 6 = 0
(c) x + y = 6
(d) x + y = -6
Answer:
(b) √3x + y ± 6 = 0
Hint:

Here, α = 30° and p = 3 units
Equation of line with inclination a and distance from origin as p is
x cos α + y sin α = p
∴ x cos 30° + y sin 30° = ±3
∴ \(\frac{\sqrt{3} x}{2}+\frac{y}{2}=\pm 3\)
∴ √3x + y ± 6 = 0

Question 7.
A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is
(a) \(\frac{1}{3}\)
(b) \(\frac{2}{3}\)
(c) 1
(d) \(\frac{4}{3}\)
Answer:
(d) \(\frac{4}{3}\)
Hint:
Slope of line 3x + y = 3 is -3
∴ Slope of line perpendicular to given line = \(\frac{1}{3}\)
Equation of required line passing through (2, 2) and having slope \(\frac{1}{3}\) is
y – 2 = \(\frac{1}{3}\)(x – 2)
3y – 6 = x – 2
∴ x – 3y + 4 = 0
∴ y-intercept = \(\frac{-4}{-3}=\frac{4}{3}\)

Question 8.
The angle between the line √3x – y – 2 = 0 and x – √3y + 1 = 0 is
(a) 15°
(b) 30°
(c) 45°
(d) 60°
Answer:
(b) 30°
Hint:

Question 9.
If kx + 2y – 1 = 0 and 6x – 4y + 2 = 0 are identical lines, then determine k.
(a) -3
(b) \(-\frac{1}{3}\)
(c) \(\frac{1}{3}\)
(d) 3
Answer:
(a) -3
Hint:
Lines kx + 2y – 1 = 0 and 6x – 4y + 2 = 0 are identical.
∴ \(\frac{k}{6}=\frac{2}{-4}=\frac{-1}{2}\)
∴ k = -3

Question 10.
Distance between the two parallel lines y = 2x + 7 and y = 2x + 5 is
(a) \(\frac{\sqrt{2}}{\sqrt{5}}\)
(b) \(\frac{1}{\sqrt{5}}\)
(c) \(\frac{\sqrt{5}}{2}\)
(d) \(\frac{2}{\sqrt{5}}\)
Answer:
(d) \(\frac{2}{\sqrt{5}}\)
Hint:
Here, c₁ = 7, c₂ = 5, a = 2 and b = -1
Distance between parallel lines

II. Answer the following questions.

Question 1.
Find the value of k:
(a) if the slope of the line passing through the points P(3, 4), Q(5, k) is 9.
(b) the points A(1, 3), B(4, 1), C(3, k) are collinear.
(c) the point P(1, k) lies on the line passing through the points A(2, 2) and B(3, 3).
Solution:
(a) Given, P(3, 4), Q(5, k) and
Slope of PQ = 9
\(\frac{\mathrm{k}-4}{5-3}\) = 9
\(\frac{\mathrm{k}-4}{2}\) = 9
k – 4 = 18
k = 22

(b) Given, points A(1, 3), B(4, 1) and C(3, k) are collinear.
Slope of AB = Slope of BC
\(\frac{1-3}{4-1}=\frac{k-1}{3-4}\)
\(\frac{-2}{3}=\frac{\mathrm{k}-1}{-1}\)
2 = 3k – 3
k = \(\frac{5}{3}\)

(c) Given, point P(1, k) lies on the line joining A(2, 2) and B(3, 3).
Slope of AB = Slope of BP
\(\frac{3-2}{3-2}=\frac{3-k}{3-1}\)
1 = \(\frac{3-\mathrm{k}}{2}\)
2 = 3 – k
k = 1

Question 2.
Reduce the equation 6x + 3y + 8 = 0 into slope-intercept form. Hence, find its slope.
Solution:
Given equation is 6x + 3y + 8 = 0, which can be written as
3y = – 6x – 8
y = \(\frac{-6 x}{3}-\frac{8}{3}\)
y = -2x – \(\frac{8}{3}\)
This is of the form y = mx + c with m = -2
y = -2x – \(\frac{8}{3}\) is in slope-intercept form with slope = -2

Question 3.
Find the distance of the origin from the line x = -2.
Solution:
Given equation of line is x = -2
This equation represents a line parallel to Y-axis and at a distance of 2 units to the left of Y-axis.
∴ Distance of the origin from the line is 2 units.

Question 4.
Does point A(2, 3) lie on the line 3x + 2y – 6 = 0? Give reason.
Solution:
Given equation is 3x + 2y – 6 = 0.
Substituting x = 2 and y = 3 in L.H.S. of given equation, we get
L.H.S. = 3x + 2y – 6
= 3(2) + 2(3) – 6
= 6
≠ R.H.S.
∴ Point A does not lie on the given line.

Question 5.
Which of the following lines passes through the origin?
(a) x = 2
(b) y = 3
(c) y = x + 2
(d) 2x – y = 0
Answer:
(d) 2x – y = 0
Hint:
Any line passing through origin is of the form y = mx or ax + by = 0.
Here in the given option, 2x – y = 0 is in the form ax + by = 0.
∴ Option (d) is the correct answer.

Question 6.
Obtain the equation of the line which is:
(a) parallel to the X-axis and 3 units below it.
(b) parallel to the Y-axis and 2 units to the left of it.
(c) parallel to the X-axis and making an intercept of 5 on the Y-axis.
(d) parallel to the Y-axis and making an intercept of 3 on the X-axis.
Solution:
(a) Equation of a line parallel to X-axis is y = k.
Since the line is at a distance of 3 units below X-axis, k = -3
∴ The equation of the required line is y = -3.

(b) Equation of a line parallel to Y-axis is x = h.
Since the line is at a distance of 2 units to the left of Y-axis, h = -2
∴ The equation of the required line is x = -2.

(c) Equation of a line parallel to X-axis with y-intercept ‘k’ is y = k.
Here, y-intercept = 5
∴ The equation of the required line is y = 5.

(d) Equation of a line parallel to Y-axis with x-intercept ‘h’ is x = h.
Here, x-intercept = 3
∴ The equation of the required line is x = 3.

Question 7.
Obtain the equation of the line containing the point:
(i) (2, 3) and parallel to the X-axis.
(ii) (2, 4) and perpendicular to the Y-axis.
Solution:
(i) Equation of a line parallel to X-axis is of the form y = k.
Since the line passes through (2, 3), k = 3
∴ The equation of the required line is y = 3.

(ii) Equation of a line perpendicular to Y-axis
i.e., parallel to X-axis, is of the form y = k.
Since the line passes through (2, 4), k = 4
∴ The equation of the required line is y = 4.

Question 8.
Find the equation of the line:
(a) having slope 5 and containing point A(-1, 2).
(b) containing the point T(7, 3) and having inclination 90°.
(c) through the origin which bisects the portion of the line 3x + 2y = 2 intercepted between the co-ordinate axes.
Solution:
(a) Given, slope(m) = 5 and the line passes through A(-1, 2).
Equation of the line in slope point form is y – y1 = m(x – x1)
The equation of the required line is
y – 2 = 5(x + 1)
y – 2 = 5x + 5
∴ 5x – y + 7 = 0

(b) Given, Inclination of line = θ = 90°
the required line is parallel to Y-axis.
Equation of a line parallel to Y-axis is of the form x = h.
Since the line passes through (7, 3), h = 7
∴ The equation of the required line is x = 7.

(c) Given equation of the line is 3x + 2y = 2.

\(\frac{3 x}{2}+\frac{2 y}{2}=1\)
\(\frac{x}{\frac{2}{3}}+\frac{y}{1}=1\)
This equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\), with a = \(\frac{2}{3}\), b= 1.
The line 3x + 2y = 2 intersects the X-axis at A(\(\frac{2}{3}\), 0) and Y-axis at B(0, 1).
Required line is passing through the midpoint of AB.
Midpoint of AB = \(\left(\frac{\frac{2}{3}+0}{2}, \frac{0+1}{2}\right)=\left(\frac{1}{3}, \frac{1}{2}\right)\)
∴ Required line passes through (0, 0) and \(\left(\frac{1}{3}, \frac{1}{2}\right)\).
Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the required line is
\(\frac{y-0}{\frac{1}{2}-0}=\frac{x-0}{\frac{1}{3}-0}\)
2y = 3x
∴ 3x – 2y = 0

Question 9.
Find the equation of the line passing through the points S(2, 1) and T(2, 3).
Solution:
The required line passes through the points S(2, 1) and T(2, 3).
Since both the given points have same x co-ordinates i.e. 2
the given points lie on a line parallel to Y-axis.
∴ The equation of the required line is x = 2.

Question 10.
Find the distance of the origin from the line 12x + 5y + 78 = 0.
Solution:
Let p be the perpendicular distance of origin from the line 12x + 5y + 78 = 0.
Here, a = 12, b = 5, c = 78

Question 11.
Find the distance between the parallel lines 3x + 4y + 3 = 0 and 3x + 4y + 15 = 0.
Solution:
Equations of the given parallel lines are 3x + 4y + 3 = 0 and 3x + 4y + 15 = 0
Here, a = 3, b = 4, c₁ = 3 and c₂ = 15
∴ Distance between the parallel lines

Question 12.
Find the equation of the line which contains the point A(3, 5) and makes equal intercepts on the co-ordinates axes.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}=1\) …….(i)
This line passes through A(3, 5).
∴ \(\frac{3}{a}+\frac{5}{b}=1\) ……..(ii)
Since the required line makes equal intercepts on the co-ordinates axes,
a = b …….(iii)
Substituting the value of b in (ii), we get
\(\frac{3}{a}+\frac{5}{a}=1\)
∴ a = 8
∴ b = 8 …… [From (iii)]
Substituting the values of a and b in equation (i), the equation of the required line is
\(\frac{x}{8}+\frac{y}{8}=1\)
∴ x + y = 8

Case II: Line passing through origin.
Slope of line passing through origin and A(3, 5) is
m = \(\frac{5-0}{3-0}=\frac{5}{3}\)
∴ Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is
y = \(\frac{5}{3}\)x
∴ 5x – 3y = 0

Question 13.
The vertices of a triangle are A(1, 4), B(2, 3) and C(1, 6). Find equations of
(a) the sides
(b) the medians
(c) perpendicular bisectors of sides
(d) altitudes of ?ABC
Solution:
Vertices of ∆ABC are A(1, 4), B(2, 3) and C(1, 6)
(a) Equation of the line in two point form is \(\frac{y-y_{1}}{y_{2}-y_{1}}\) = \(\frac{x-x_{1}}{x_{2}-x_{1}}\)
Equation of side AB is
\(\frac{y-4}{3-4}=\frac{x-1}{2-1}\)
y – 4 = -1(x – 1)
y – 4 = -x + 1
x + y = 5
Equation of side BC is
\(\frac{y-3}{6-3}=\frac{x-2}{1-2}\)
-1(y – 3) = 3(x – 2)
-y + 3 = 3x – 6
∴ 3x + y = 9
Since both the points A and C have same x co-ordinates i.e. 1
the points A and C lie on a line parallel to Y-axis.
∴ The equation of side AC is x = 1.

(b) Let D, E and F be the midpoints of sides AC and AB respectively of ∆ABC.

(c) Slope of side BC = \(\left(\frac{6-3}{1-2}\right)=\left(\frac{3}{-1}\right)\) = -3
Slope of perpendicular bisector of BC is \(\frac{1}{3}\) and the line passes through \(\left(\frac{3}{2}, \frac{9}{2}\right)\).
Equation of the perpendicular bisector of side BC is
\(\left(y-\frac{9}{2}\right)=\frac{1}{3}\left(x-\frac{3}{2}\right)\)
3(2y – 9) = (2x – 3)
6y – 27 = 2x – 3
2x – 6y + 24 = 0
∴ x – 3y + 12 = 0
Since both the points A and C have same x co-ordinates i.e. 1
the points A and C lie on the line x = 1.
AC is parallel to Y-axis and therefore, perpendicular bisector of side AC is parallel to X-axis.
Since, the perpendicular bisector of side AC passes through E(1, 5).
The equation of perpendicular bisector of side AC is y = 5.
Slope of side AB = \(\left(\frac{3-4}{2-1}\right)\) = -1
Slope of perpendicular bisector of AB is 1 and the line passes through \(\left(\frac{3}{2}, \frac{7}{2}\right)\).
Equation of the perpendicular bisector of side AB is
\(\left(y-\frac{7}{2}\right)=1\left(x-\frac{3}{2}\right)\)
2y – 7 = 2x – 3
2x – 2y + 4 = 0
∴ x – y + 2 = 0

(d) Let AX, BY, and CZ be the altitudes through the vertices A, B and C respectively of ∆ABC.

Slope of BC = -3
Slope of AX = \(\frac{1}{3}\) ……[∵ AX ⊥ BC]
Since altitude AX passes through (1, 4) and has slope \(\frac{1}{3}\),
equation of altitude AX is
y – 4 = \(\frac{1}{3}\)(x – 1)
3y – 12 = x – 1
∴ x – 3y + 11 = 0
Since both the points A and C have same x co-ordinates i.e. 1
the points A and C lie on the line x = 1.
AC is parallel to Y-axis and therefore, altitude BY is parallel to X-axis.
Since the altitude BY passes through B(2, 3), the equation of altitude BY is y = 3.
Also, slope of AB = -1
Slope of CZ = 1
Since altitude CZ passes through (1, 6) and has slope 1,
equation of altitude CZ is
y – 6 = 1(x – 1)
∴ x – y + 5 = 0

Question 14.
Find the equation of the line which passes through the point of intersection of lines x + y – 3 = 0, 2x – y + 1 = 0 and which is parallel to X-axis.
Solution:
Let u ≡ x + y – 3 = 0 and v ≡ 2x – y + 1 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
(x + y – 3) + k(2x – y + 1) = 0 …..(i)
x + y – 3 + 2kx – ky + k = 0
x + 2kx + y – ky – 3 + k = 0
(1 + 2k)x + (1 – k)y – 3 + k = 0
But, this line is parallel to X-axis
Its slope = 0
⇒ \(\frac{-(1+2 k)}{1-k}=0\)
⇒ 1 + 2k = 0
⇒ k = \(\frac{-1}{2}\)
Substituting the value of k in (i), we get
(x + y – 3) + \(\frac{-1}{2}\) (2x – y + 1) = 0
⇒ 2(x + y – 3) – (2x – y + 1 ) = 0
⇒ 2x + 2y – 6 – 2x + y – 1 = 0
⇒ 3y – 7 = 0, which is the equation of the required line.

Question 15.
Find the equation of the line which passes through the point of intersection of lines x + y + 9 = 0, 2x + 3y + 1 = 0 and which makes x-intercept 1.
Solution:
Let u ≡ x + y + 9 = 0 and v ≡ 2x + 3y + 1 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
(x + y + 9) + k(2x + 3y + 1) = 0 ……(i)
⇒ x + y + 9 + 2kx + 3ky + k = 0
⇒ (1 + 2k)x + (1 + 3k)y + 9 + k = 0
But, x-intercept of this line is 1.
⇒ \(\frac{-(9+\mathrm{k})}{1+2 \mathrm{k}}\)
⇒ -9 – k = 1 + 2k
⇒ k = \(\frac{-10}{3}\)
Substituting the value of k in (i), we get
(x + y + 9) + (\(\frac{-10}{3}\)) (2x + 3y + 1) = 0
⇒ 3(x + y + 9) – 10(2x + 3y + 1) = 0
⇒ 3x + 3y + 27 – 20x – 30y – 10 = 0
⇒ -17x – 27y+ 17 = 0
⇒ 17x + 27y – 17 = 0, which is the equation of the required line.

Question 16.
Find the equation of the line through A(-2, 3) and perpendicular to the line through S(1, 2) and T(2, 5).
Solution:
Slope of ST = \(\frac{5-2}{2-1}\) = 3
Since the required line is perpendicular to ST,
slope of required line = \(\frac{-1}{3}\) and line passes through A(-2, 3)
Equation of the line in slope point form is y – y₁ = m(x – x₁)
The equation of the required line is
y – 3 = \(\frac{-1}{3}\)(x + 2)
⇒ 3(y – 3) = -(x + 2)
⇒ 3y – 9 = -x – 2
⇒ x + 3y = 7

Question 17.
Find the x-intercept of the line whose slope is 3 and which makes intercept 4 on the Y-axis.
Solution:
Equation of a line having slope ‘m’ and y-intercept ‘c’ is y = mx + c
Given, m = 3, c = 4
The equation of the line is y = 3x + 4
3x – y = -4
\(\frac{3 x}{(-4)}-\frac{y}{(-4)}=1\)
\(\frac{x}{\left(\frac{-4}{3}\right)}+\frac{y}{4}=1\)
This equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\), where
x-intercept = a
x-intercept = \(\frac{-4}{3}\)

Alternate Method:

Let θ be the inclination of the line.
Then tan θ = 3 …..[∵ slope = 3 (given)]
\(\frac{\mathrm{OB}}{\mathrm{OA}}=3\)
\(\frac{4}{\mathrm{OA}}=3\)
OA = \(\frac{4}{3}\)
x-intercept = –\(\frac{4}{3}\) as point A is to the left side of Y-axis.

Question 18.
Find the distance of P(-1, 1) from the line 12(x + 6) = 5(y – 2).
Solution:
Given equation of the line is
12(x + 6) = 5(y – 2)
12x + 72 = 5y – 10
12x – 5y + 82 = 0
Let p be the perpendicular distance of the point (-1, 1) from the line 12x – 5y + 82 = 0.

Question 19.
Line through A(h, 3) and B(4,1) intersect the line lx – 9y -19 = 0 at right angle. Find the value of h.
Solution:
Given, A(h, 3) and B(4, 1)
Slope of AB (m₁) = \(\frac{1-3}{4-h}\)
m₁ = \(\frac{2}{h-4}\)
Slope of line 7x – 9y – 19 = 0 is m₂ = \(\frac{7}{9}\)
Since line AB and line 7x – 9y – 19 = 0 are perpendicular to each other,
m₁ × m₂ = -1
\(\frac{2}{h-4} \times \frac{7}{9}=-1\)
14 = 9(4 – h)
14 = 36 – 9h
9h = 22
h = \(\frac{22}{9}\)

Question 20.
Two lines passing through M(2, 3) intersect each other at an angle of 45°. If slope of one line is 2, find the equation of the other line.
Solution:
Let m be the slope of the required line which make an angle of 45° with the other line.
Slope of one of the lines is 2.
tan 45° = \(\left|\frac{\mathrm{m}-2}{1+\mathrm{m}(2)}\right|\)
1 = \(\left|\frac{m-2}{1+2 m}\right|\)
\(\frac{m-2}{1+2 m}=\pm 1\)
\(\frac{\mathrm{m}-2}{1+2 \mathrm{~m}}\) = 1 or \(\frac{\mathrm{m}-2}{1+2 \mathrm{~m}}\) = -1
m – 2 = 1 + 2m or m – 2 = -1 – 2m
m = -3 or 3m = 1
m = -3 or m = \(\frac{1}{3}\)
Required line passes through M(2, 3)
When m = -3, equation of the line is
y – 3 = -3(x – 2)
y – 3 = -3x + 6
∴ 3x + y = 9
When m = \(\frac{1}{3}\), equation of the line is
y – 3 = \(\frac{1}{3}\)(x – 2)
3y – 9 = x – 2
∴ x – 3y + 7 = 0

Question 21.
Find the y-intercept of the line whose slope is 4 and which has x-intercept 5.
Solution:
Given, slope = 4, x-intercept = 5
Since the x-intercept of the line is 5, it passes through (5, 0).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
Equation of the required line is
y – 0 = 4(x – 5)
y = 4x – 20
4x – y = 20
\(\frac{4 x}{20}-\frac{y}{20}=1\)
\(\frac{x}{5}+\frac{y}{(-20)}=1\)
This equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\), where
x-intercept = b, y-intercept = -20

Question 22.
Find the equations of the diagonals of the rectangle whose sides are contained in the lines x = 8, x = 10, y = 11 and y = 12.
Solution:
Given, equations of sides of rectangle are x = 8, x = 10, y = 11 and y = 12

From the above diagram,
Vertices of rectangle are A(8, 11), B(10, 11), C(10, 12) and D(8, 12).
Equation of diagonal AC is
\(\frac{y-11}{12-11}=\frac{x-8}{10-8}\)
\(\frac{y-11}{1}=\frac{x-8}{2}\)
2y – 22 = x – 8
x – 2y + 14 = 0
Equation of diagonal BD is
\(\frac{y-11}{12-11}=\frac{x-10}{8-10}\)
\(\frac{y-11}{1}=\frac{x-10}{-2}\)
-2y + 22 = x – 10
x + 2y = 32

Question 23.
A(1, 4), B(2, 3) and C(1, 6) are vertices of AABC. Find the equation of the altitude through B and hence find the co-ordinates of the point where this altitude cuts the side AC of ∆ABC.
Solution:
Vertices of triangle are A(1, 4), B(2, 3) and C(1, 6).
Let BD be the altitude through the vertex B.

Since both the points A and C have same x co-ordinates i.e. 1
the given points lie on a line parallel to Y-axis.
The equation of the line AC is x = 1 …..(i)
AC is parallel to Y-axis and therefore, altitude BD is parallel to X-axis.
Since the altitude BD passes through B(2, 3), the equation of altitude BD is y = 3 ……(ii)
From (i) and (ii),
Point of intersection of AC and altitude BD is (1, 3).

Question 24.
The vertices of ∆PQR are P(2, 1), Q(-2, 3) and R(4, 5). Find the equation of the median through R.
Solution:

Let S be the midpoint of side PQ.
Then RS is the median through R.
S = \(\left(\frac{2-2}{2}, \frac{3+1}{2}\right)\) = (0, 2)
The median RS passes through the points R(4, 5) and S(0, 2).
∴ Equation of median RS is
\(\frac{y-5}{2-5}=\frac{x-4}{0-4}\)
⇒ \(\frac{y-5}{-3}=\frac{x-4}{-4}\)
⇒ 4(y – 5) = 3(x – 4)
⇒ 4y – 20 = 3x – 12
∴ 3x – 4y + 8 = 0

Question 25.
A line perpendicular to segment joining A(1, 0) and B(2, 3) divides it internally in the ratio 1 : 2. Find the equation of the line. Solution:
Given, A(1, 0), B(2, 3)
Slope of AB = \(\frac{3-0}{2-1}\) = 3
Required line is perpendicular to AB.
Slope of required line = \(\frac{-1}{3}\)
Let point C divide AB in the ratio 1 : 2.

Required line passes through \(\left(\frac{4}{3}, 1\right)\) and has slope = \(\frac{-1}{3}\)
Equation of the line in slope point form is y – y₁ = m(x – x₁)
The equation of the required line is
y – 1 = \(\frac{-1}{3}\left(x-\frac{4}{3}\right)\)
⇒ 3(y – 1) = \(-1\left(x-\frac{4}{3}\right)\)
⇒ 3y – 3 = -x + \(\frac{4}{3}\)
⇒ 9y – 9 = -3x + 4
⇒ 3x + 9y = 13

Question 26.
Find the co-ordinates of the foot of the perpendicular drawn from the point P(-1, 3) to the line 3x – 4y – 16 = 0.
Solution:

Let M be the foot of perpendicular drawn from P(-1, 3) to the line 3x – 4y – 16 = 0
Slope of the line 3x – 4y – 16 = 0 is \(\frac{-3}{-4}=\frac{3}{4}\)
Since PM ⊥ to line (i),
slope of PM = \(\frac{-4}{3}\)
Equation of PM is
y – 3 = \(\frac{-4}{3}\) (x + 1)
⇒ 3(y – 3) = -4(x + 1)
⇒ 3y – 9 = -4x – 4
∴ 4x + 3y – 5 = 0 ……(ii)
The foot of perpendicular i.e., point M, is the point of intersection of equation (i) and (ii).
By (i) × 3 + (ii) × 4, we get
25x = 68
x = \(\frac{68}{25}\)
Substituting x = \(\frac{68}{25}\) in (ii), we get

The co-ordinates of the foot of perpendicular M are \(\left(\frac{68}{25}, \frac{-49}{25}\right)\)

Question 27.
Find points on the X-axis whose distance from the line \(\frac{x}{3}+\frac{y}{4}=1\) is 4 units.
Solution:
The equation of line is \(\frac{x}{3}+\frac{y}{4}=1\)
i.e. 4x + 3y – 12 = 0 …..(i)
Let (h, 0) be a point on the X-axis.
The distance of this point from line (i) is 4.
⇒ \(\frac{|4 h+3(0)-12|}{\sqrt{4^{2}+3^{2}}}=4\)
⇒ \(\frac{|4 \mathrm{~h}-12|}{5}=4\)
⇒ |4h – 12| = 20
⇒ 4h – 12 = 20 or 4h – 12 = -20
⇒ 4h = 32 or 4h = -8
⇒ h = 8 or h = -2
∴ The required points are (8, 0) and (-2, 0).

Question 28.
The perpendicular from the origin to a line meets it at (-2, 9). Find the equation of the line.
Solution:

Slope of ON = \(\frac{9-0}{-2-0}=\frac{-9}{2}\)
Since line AB ⊥ ON,
slope of the line AB perpendicular to ON is \(\frac{2}{9}\) and it passes through point N(-2, 9).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
Equation of line AB is
y – 9 = \(\frac{2}{9}\)(x + 2)
⇒ 9(y – 9) = 2(x + 2)
⇒ 9y – 81 = 2x + 4
⇒ 2x – 9y + 85 = 0

Question 29.
P(a, b) is the midpoint of a line segment intercepted between the axes. Show that the equation of the line is \(\frac{x}{a}+\frac{y}{b}=2\).
Solution:
Let the intercepts of a line AB be x₁ and y₁ on the X and Y-axes respectively.
A ≡ (x₁, 0), B = (0, y₁)
P(a, b) is the midpoint of a line segment AB intercepted between the axes.

Question 30.
Find the distance of the line 4x – y = 0 from the point P(4, 1) measured along the line making an angle of 135° with the positive X-axis.
Solution:

Let a line L make angle 135° with positive X-axis.
Required distance = PQ, where PQ || line L
Slope of PQ = tan 135°
= tan (180° – 45°)
= -tan 45°
= -1
Equation of PQ is
y – 1 = (-1)(x – 4)
y – 1 = -x + 4
x + y = 5 …..(i)
To get point Q we solve the equation 4x – y = 0 with (i)
Substituting y = 4x in (i), we get
5x = 5
x = 1
Substituting x = 1 in (i), we get
1 + y = 5
y = 4
∴ Q = (1, 4)
PQ = \(\sqrt{(4-1)^{2}+(1-4)^{2}}\)
= \(\sqrt{9+9}\)
= 3√2

Question 31.
Show that there are two lines which pass through A(3, 4) and the sum of whose intercepts is zero.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}=1\) ……(1)
This line passes through (3, 4)
\(\frac{3}{a}+\frac{4}{b}=1\) …..(ii)
Since the sum of the intercepts of the line is zero,
a + b = 0
a = -b ……(iii)
Substituting the value of a in (ii), we get
\(\frac{3}{-b}+\frac{4}{b}=1\)
\(\frac{1}{b}\) = 1
b = 1
a = -1 ……[From (iii)]
Substituting the values of a and b in (i),
the equation of the required line is
\(\frac{x}{-1}+\frac{y}{1}=1\)
x – y = -1
∴ x – y + 1 = 0

Case II: Line passing through origin.
Slope of line passing through origin and A(3, 4) is
m = \(\frac{4-0}{3-0}=\frac{4}{3}\)
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
The equation of the required line is
y = \(\frac{4}{3}\)x
∴ 4x – 3y = 0
∴ There are two lines which pass through A(3, 4) and the sum of whose intercepts is zero.

Question 32.
Show that there is only one line which passes through B(5, 5) and the sum of whose intercepts is zero.
Solution:
When line is passing through origin, the sum of intercepts made by the line is zero.
Slope of line passing through origin and B(5, 5) is
m = \(\frac{5-0}{5-0}\) = 1
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
The equation of the required line is y = x
∴ x – y = 0
∴ There is only one line which passes through B(5, 5) and the sum of whose intercepts is zero.

Class 11 Maharashtra State Board Maths Solution 

• Exercise 5.1 Class 11 Maths 1
• Exercise 5.2 Class 11 Maths 1
• Exercise 5.3 Class 11 Maths 1
• Exercise 5.4 Class 11 Maths 1
• Miscellaneous Exercise 5 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.4 Questions and Answers.

11th Maths Part 1 Straight Line Exercise 5.4 Questions And Answers Maharashtra Board

Question 1.
Find the slope, x-intercept, y-intercept of each of the following lines, i. 2x + 3y-6 = 0 ii. 3x-y-9 = 0 iii. x + 2y = 0
Solution:
i. Given equation of the line is 2x + 3y – 6 = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 2, b = 3, c = -6

ii. Given equation of the line is 3x – y – 9 = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 3, b = – 1, c = – 9

iii. Given equation of the line is x + 2y = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 1, b = 2, c = 0

Question 2.
Write each of the following equations in ax + by + c = 0 form.
i. y = 2x – 4
ii. y = 4
iii. \(\frac{x}{2}+\frac{y}{4}=1\)
iv. \(\frac{x}{3}-\frac{y}{2}=0\)
i. y = 2x – 4
∴ 2x – y – 4 = 0 is the equation in ax + by + c = 0 form.

ii. y = 4
∴ 0x + 1y – 4 = 0 is the equation in ax + by + c = 0 form.

iii. \(\frac{x}{2}+\frac{y}{4}=1\)
∴ \(\frac{2 x+y}{4}\)
∴ 2x + y – 4 = 0 is the equation in ax + by + c = 0 form.

iv. \(\frac{x}{3}-\frac{y}{2}=0\)
∴ 2x – 3y = 0
∴ 2x – 3y + 0 = 0 is the equation in ax + by + c = 0 form.
[Note: Answer given in the textbook is ‘2x – 3y – 6 = 0’. However, as per our calculation it is ‘2x-3y + 0 = 0’.]

Question 3.
Show that the lines x – 2y – 7 = 0 and 2x – 4y + 15 = 0 are parallel to each other.
Solution:
Let m₁ be the slope of the line x – 2y – 7 = 0.
∴ m₁ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-1}{-2}=\frac{1}{2}\)
Let m₂ be the slope of the line 2x – 4y + 15 = 0.
∴ m₂ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-2}{-4}=\frac{1}{2}\)
Since m₁ = m₂
the given lines are parallel to each other.

Question 4.
Show that the lines x – 2y – 7 = 0 and 2x + y + 1 = 0 are perpendicular to each other. Find their point of intersection.
Solution:
Let m₁ be the slope of the line x – 2y – 7 = 0.
∴ m₁ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-1}{-2}=\frac{1}{2}\)
Let m₂ be the slope of the line 2x + y + 1 = 0.
∴ m₂ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-2}{1}=-2\)
Since m₁ x m₁ = \(\frac{1}{2}\) x (- 2) = -1,
the given lines are perpendicular to each other. Consider,
x – 2y – 7 = 0 …(i)
2x + y + 1 =0 …(ii)
Multiplying equation (ii) by 2, we get
4x + 2y + 2 = 0 …(iii)
Adding equations (i) and (iii), we get
5x – 5 = 0
∴ x = 1
Substituting x = 1 in equation (ii), we get
2 + y + 1 = 0
∴ y = – 3
∴ The point of intersection of the given lines is (1,-3).

Question 5.
If the line 3x + 4y = p makes a triangle of area 24 square units with the co-ordinate axes, then find the value of p.
Solution:
Let the line 3x + 4y = p cuts the X and Y axes at points A and B respectively.
3x + 4y = p

This equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\),
where a = \(\frac{p}{3}\) and b = \(\frac{p}{4}\)
∴ A (a, 0) ≡ (\(\frac{p}{3}\), 0) and B ≡ (0, b) = (0, \(\frac{p}{4}\))
∴ OA = \(\frac{p}{3}\) and OB = \(\frac{p}{4}\)
Given, A (∆OAB) = 24 sq. units
∴ \(\left|\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\right|=24\)
∴ \(\left|\frac{1}{2} \times \frac{\mathrm{p}}{3} \times \frac{\mathrm{p}}{4}\right|=24\)
∴ p² = 576
∴ p = ± 24

Question 6.
Find the co-ordinates of the foot of the perpendicular drawn from the point A(- 2,3) to the line 3x-y -1 = 0.
Solution:
Let M be the foot of perpendicular drawn from
point A(- 2,3) to the line
3x-y- 1 = 0 …(i)
Slope of the line 3x-y – 1 = 0 is \(\frac{-3}{-1}\) =3.
Since AM ⊥ to line (i),
slope of AM = \(\frac{-1}{3}\)
∴ Equation of AM is
y – 3 = \(\frac{-1}{3}\)(x + 2)

∴ 3(y – 3) = – 1(x + 2)
∴ 3y – 9 = -x – 2
∴ x + 3y – 7 = 0 …………(ii)
The foot of perpendicular i.e., point M, is the point of intersection of equations (i) and (ii).
By (i) x 3 + (ii), we get 10x -10 = 0
∴ x = 1
Substituting x = 1 in (ii), we get
1 + 3y – 7 = 0
∴ 3y = 6
∴ y = 2
∴ The co-ordinates of the foot of the perpendicular Mare (1,2).

Question 7.
Find the co-ordinates of the circumcentre of the triangle whose vertices are A(- 2, 3), B(6, -1), C(4,3),e
Solution:
Here, A(-2, 3), B(6, -1), C(4, 3) are the vertices of ∆ABC.
Let F be the circumcentre of AABC.
Let FD and FE be the perpendicular bisectors of the sides BC and AC respectively.
∴ D and E are the midpoints of side BC and AC respectively.

Since FD passes through (5, 1) and has slope 1/2 equation of FD is
y – 1 = \(\frac{1}{2}\)(x-5)
∴ 2 (y – 1) = x – 5
∴ 2y – 2 = x – 5
∴ x – 2y – 3 = 0 …(i)
Since both the points A and C have same y co-ordinates i.e. 3,
the given points lie on the line y = 3.
Since the equation FE passes through E(1, 3),
the equation of FE is x = 1. .. .(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value of x in (i), we get
1 – 2y -3 = 0
∴ y = -1
∴ Co-ordinates of circumcentre F ≡ (1, – 1).

Question 8.
Find the co-ordinates of the orthocentre of the triangle whose vertices are A(3, – 2), B(7,6), C (-1,2).
Solution:
Let O be the orthocentre of ∆ABC.
Let AD and BE be the altitudes on the sides BC and AC respectively.

Slope of side BC = \(\frac{2-6}{-1-7}=\frac{-4}{-8}=\frac{1}{2}\)
∴ Slope of AD = – 2 [∵ AD ⊥ BC]
∴ Equation of line AD is
y – (-2) = (- 2) (x – 3)
∴ y + 2 = -2x + 6
∴ 2x + y -4 = 0 …(i)
Slope of side AC = \(\frac{-2-2}{3-(-1)}=\frac{-4}{4}\) = -1
∴ Slope of BE = 1 …[ ∵ BE ⊥ AC]
∴ Equation of line BE is
y – 6 = 1(x – 7)
∴ y – 6 = x – 1
∴ x = y + 1 …(ii)
Substituting x = y + 1 in (i), we get
2(y + 1) + y – 4 = 0
∴ 2y + 2 + y – 4 = 0
∴ 3y – 2 = 0
∴ y = \(\frac{2}{3} in (ii), we get
Substituting y = [latex]\frac{2}{3}\) in (ii), we get
x = \(\frac{2}{3}+1=\frac{5}{3}\)
∴ Co-ordinates of orthocentre, O = \(\left(\frac{5}{3}, \frac{2}{3}\right)\)

Question 9.
Show that the lines 3 – 4y + 5 = 0, lx – 8y + 5 = 0 and 4JC + 5y – 45 = 0 are concurrent. Find their point of concurrence.
Solution:
The number of lines intersecting at a point are called concurrent lines and their point of intersection is called the point of concurrence. Equations of the given lines are
3x – 4y + 5 = 0 …(i)
7x-8y + 5 = 0 …(ii)
4x + 5y – 45 = 0 …(iii)
By (i) x 2 – (ii), we get
– x + 5 = 0
∴ x = 5
Substituting x = 5 in (i), we get
3(5) – 4y + 5 = 0
∴ -4y = – 20
∴ y = 5
∴ The point of intersection of lines (i) and (ii) is given by (5, 5).
Substituting x = 5 and y = 5 in L.H.S. of (iii), we get
L.H.S. = 4(5) + 5(5) – 45
= 20 + 25 – 45
= 0
= R.H.S.
∴ Line (iii) also passes through (5, 5).
Hence, the given three lines are concurrent and the point of concurrence is (5, 5).

Question 10.
Find the equation of the line whose x-intercept is 3 and which ¡s perpendicular to the line 3x – y + 23 = 0.
Solution:
Slope of the line 3x – y + 23 = 0 is 3.
∴ Slope of the required line perpendicular to
3x – y + 23 = 0 is \(\frac{-1}{3}\)
Since the x-intercept of the required line is 3, it passes through (3, 0).
∴ The equation of the required line is ‘
y – 0 = \(\frac{-1}{3}\)(x – 3)
∴ 3y = x + 3
∴ x + 3y = 3

Question 11.
Find the distance of the origin from the line 7x + 24y – 50 = 0.
Solution:
Let p be the perpendicular distance of origin
fromtheline7x + 24y – 50 = 0
Here, a = 7, b = 24, c = -50

Question 12.
Find the distance of the point A(- 2, 3) from the line 12x – 5y – 13 = 0.
Solution:
Let p be the perpendicular distance of the point A(- 2, 3) from the line 12x – 5y – 13 = 0
Here, a = 12, b = – 5, c = – 13, x₁ = -2, y₁ = 3

Question 13.
Find the distance between parallel lines 4x – 3y + 5 = 0 and 4xr – 3y + 7 = 0.
Solution:
Equations of the given parallel lines are 4x – 3y + 5 = 0 and 4x – 3y + 1 = 0
Here, a = 4, b = – 3, c₁ = 5 and c₂ = 7
∴ Distance between the parallel lines

Question 14.
Find the distance between the parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.
Solution:
Equations of the given parallel lines are 3x + 2y + 6 = 0 and
9x + 6y – 1 = 0 i.e., 3x + 2y – \(\frac{7}{3}\) =0
Here, a = 3, b = 2, c₁ = 6 and c₂ = \(\frac{-7}{3}\)
∴ Distance between the parallel lines

Question 15.
Find the points on the line x + y – 4 = 0 which are at a unit distance from the line 4JC + 3y = 10.
Solution:
Let P(x₁, y₁) be a point on the line x + y – 4 = o.
∴ x₁ + y₁ – 4 = 0
∴ y₁ = 4 – x₁ …(i)
Also, distance of P from the line 4x + 3y- 10 = 0 is 1

∴ 5 = | x₁ + 2 |
∴ x₁ + 2 = ± 5
∴ x₁ + 2 = 5 or x₁ + 2 = – 5
∴ x₁ = 3 or x₁ = – 7
From (i), when x₁ = 3, y₁ = 1
and when x₁ = -7, y₁ = 11
∴ The required points are (3, 1) and (-7, 11).
[Note: The question has been modified]

Question 16.
Find the equation of the line parallel to the X-axis and passing through the point of intersection of lines x + y – 2 = 0 and 4x + 3y = 10.
Solution:
Let u = x + y – 2 = 0 and v = 4x + 3y – 10 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
∴ (x + y – 2) + k(4x + 3y – 10) = 0 …(i)
∴ x + y – 2 + 4kx + 3ky – 10k = 0
∴ x + 4kx + y + 3ky – 2 – 10k = 0
∴ (1+ 4k)x + (1 + 3k)y – 2 – 10k = 0
But, this line is parallel to X-axis.
∴ Its slope = 0
∴ \(\frac{-(1+4 k)}{1+3 k}\) = 0
∴ 1 + 4k = 0
∴ k = \(\frac{-1}{4}\)
Substituting the value of k in (i), we get
(x + y – 2) + (4x + 3y – 10) = 0
∴ 4(x +y – 2) – (4x + 3y -10 ) = 0
∴ 4x + 4y – 8 – 4x – 3y + 10 = 0
∴ y + 2 = 0, which is the equation of the required line.
[Note: Answer given in the textbook is 5y – 8= 0. However, as per our calculation it is y + 2 = 0.]

Question 17.
Find the equation of the line passing through the point of intersection of lines x + y – 2 = 0 and 2xr – 3y + 4 = 0 and making intercept 3 on the X-axis.
Solution:
Let u ≡ x + y – 2 = 0 and v ≡ 2x – 3y + 4 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
∴ (x +y – 2) + k(2x – 3y + 4) = 0 …(i)
But, x-intercept of line is 3.
∴ It passes through (3, 0).
Substituting x = 3 and y = 0 in (i), we get
(3 + 0 – 2) + k(6 – 0 + 4) = 0
∴ 1 + 10k = 0
k = \(\frac{-1}{10}\)
Substituting the value of k in (i), we get (x + y – 2) + \(\left(\frac{-1}{10}\right)\) (2x – 3y + 4) = 0
∴ 10(x + y – 2) – (2x – 3y + 4) = 0
∴ 10x + 10y -20 — 2x + 3y-4 = 0
∴ 8x + 13y – 24 = 0, which is the equation of the required line.

Question 18.
If A(4, 3), B(0, 0) and C(2, 3) are the vertices of ΔABC, then find the equation of bisector of angle BAC.
Solution:
Let the bisector of ∠ BAC meets BC at point D.
∴ Point D divides seg BC in the ratio l(AB) : l(AC)

∴ 18 (y – 3) = 6 (x – 4)
∴ 3(y – 3) = x – 4
∴ 3y – 9 = x – 4
∴ x – 3y + 5 = 0

Question 19.
D(- 1, 8), E(4, – 2), F(- 5, – 3) are midpoints of sides BC, CA and AB of AABC. Find
i. equations of sides of ΔABC.
ii. co-ordinates of the circumcentre of ΔABC.
Solution:
Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of ΔABC.
Given, points D, E and F are midpoints of sides BC, CA and AB respectively of ΔABC.

∴ x₁ + x₂ = -10 …………. (v)
and y₁ + y₂ = – 6 …………(vi)
For x-coordinates:
Adding (i), (iii) and (v), we get
2x₁ + 2x₂ + 2x₃ = – 4
∴ x₁ + x₂ + x₃ = -2 …………..(vii)
Solving (i) and (vii), we get x₁ = 0
Solving (iii) and (vii), we get x₂ = – 10
Solving (v) and (vii), we get x₃ = 8

For y-coordinates:
Adding (ii), (iv) and (vi), we get 2y₁ + 2y₂ + 2y₃ = 6
y₁ + y₂ + y₃ = 3 …….(viii)
Solving (ii) and (viii), we get y₁ = -13
Solving (iv) and (viii), we get y₂ = 7
Solving (vi) and (viii), we get y₃ = 9
∴ Vertices of AABC are A(0, – 13), B(- 10, 7), C(8, 9)

∴ 8(y + 13) = 22x
∴ 4(y + 13) = 11x
∴ 11x – 4y – 52 = 0

ii. Here, A(0, – 13), B(- 10, 7), C(8, 9) are the vertices of ΔABC.
Let F be the circumcentre of AABC.
Let FD and FE be perpendicular bisectors of the sides BC and AC respectively.
D and E are the midpoints of side BC and AC.

∴ Slope of FD = -9 … [ ∵ FD ⊥ BC]
Since FD passes through (-1, 8) and has slope -9, equation of FD is
y – 8 = -9 (x +1)
∴ y – 8 = -9x – 9
∴ y = -9x – 1
Also, slope of AC = \(\frac{-13-9}{0-8}=\frac{11}{4}\)
∴ Slope of FE = \(\frac{-4}{11}\) [ ∵ FE ⊥ AC]
Since FE passes through (4, -2) and has slope -4
\(\frac{-4}{11}\), equation of FE is
(y + 2) = \(\frac{-4}{11}\) (x – 4)
∴ 11(y + 2) = -4(x – 4)
∴ 11y + 22 = – 4x + 16
∴ 4x + 11y = -6 …………(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value ofy in (ii), we get
4x + 11(-9x- 1) = – 6
∴ 4x – 99x -11 = – 6
∴ -95x = 5
∴ x = \(\frac{-1}{19}\)
Substituting the value of x in (i), we get
y = -9(\(\frac{-1}{19}\)) – 1 = \(\frac{-10}{19}\)
∴ Co-ordinates of circumcentre F ≡ \(\left(\frac{-1}{19}, \frac{-10}{19}\right)\)

Question 20.
0(0, 0), A(6, 0) and B(0, 8) are vertices of a triangle. Find the co-ordinates of the incentre of ∆OAB.
Solution:
Let bisector of ∠O meet AB at point D and bisector of ∠A meet BO at point E
∴ Point D divides seg AB in the ratio l(OA): l(OB)
and point E divides seg BO in the ratio l(AB): l(AO)
Let I be the incentre of ∠OAB.

∴ Point D divides AB internally in 6 : 8
i.e. 3 :4

∴ y = x …(i)
Now, by distance formula,
l(AB) = \(\begin{aligned}
&=\sqrt{(6-0)^{2}+(0-8)^{2}} \\
&=\sqrt{36+64}=10
\end{aligned}\)
l(AO) = \(\sqrt{(6-0)^{2}+(0-0)^{2}}\) = 6
∴ Point E divides BO internally in 10 : 6 i.e. 5:3

∴ -2y = x – 6
∴ x + 2y = 6 …(ii)
To find co-ordinates of incentre, we have to solve equations (i) and (ii).
Substituting y = x in (ii), we get
x + 2x = 6
∴ x = 2
Substituting the value of x in (i), we get
y = 2
∴ Co-ordinates of incentre I ≡ (2, 2).

Alternate Method:
Let I be the incentre.
I lies in the 1st quadrant.
OPIR is a square having side length r.
Since OA = 6, OP = r,
PA = 6 – r
Since PA = AQ,
AQ = 6 – r …(i)
Since OB = 8, OR = r,
BR = 8 – r
∴ BR = BQ
∴ BQ = 8 – r …(ii)

AB = BQ + AQ
Also, AB = \(\begin{aligned}
&=\sqrt{\mathrm{OA}^{2}+\mathrm{OB}^{2}} \\
&=\sqrt{6^{2}+8^{2}} \\
&=\sqrt{100}=10
\end{aligned}\)
∴ BQ + AQ= 10
∴ (8 – r) + (6 – r) = 10
∴ 2r = 14- 10 = 4
∴ r = 2
∴ I = (2,2)

Class 11 Maharashtra State Board Maths Solution 

• Exercise 5.1 Class 11 Maths 1
• Exercise 5.2 Class 11 Maths 1
• Exercise 5.3 Class 11 Maths 1
• Exercise 5.4 Class 11 Maths 1
• Miscellaneous Exercise 5 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.3 Questions and Answers.

11th Maths Part 1 Straight Line Exercise 5.3 Questions And Answers Maharashtra Board

Question 1.
Write the equation of the line:
i. parallel to the X-axis and at a distance of 5 units from it and above it.
ii. parallel to the Y-axis and at a distance of 5 units from it and to the left of it.
iii. parallel to the X-axis and at a distance of 4 units from the point (- 2,3).
Solution:
i. Equation of a line parallel to X-axis is y = k. Since the line is at a distance of 5 units above the X-axis, k = 5
∴ The equation of the required line is y = 5.

ii. Equation of a line parallel to the Y-axis is x = h. Since the line is at a distance of 5 units to the left of the Y-axis, h = -5
∴ The equation of the required line is x = -5.
[Note: Answer given in the textbook is ‘y = -5
However, we found that ‘x = – 5’.]

iii. Equation of a line parallel to the X-axis is of the form y = k (k > 0 or k < 0).
Since the line is at a distance of 4 units from the point (- 2, 3),
k = 4 + 3 = 7 or k = 3- 4 = -1
∴ The equation of the required line is y = 1 or y = – 1.

Question 2.
Obtain the equation of the line:
i. parallel to the X-axis and making an intercept of 3 units on the Y-axis.
ii. parallel to the Y-axis and making an intercept of 4 units on the X-axis.
Solution:
i. Equation of a line parallel to X-axis with y-intercept ‘k’ isy = k.
Here, y-intercept = 3
∴ The equation of the required line is y = 3.

ii. Equation of a line parallel to Y-axis with x-intercept ‘h’ is x = h.
Here, x-intercept = 4
∴ The equation of the required line is x = 4.

Question 3.
Obtain the equation of the line containing the point:
i. A(2, – 3) and parallel to the Y-axis.
ii. B(4, – 3) and parallel to the X-axis.
Solution:
i. Equation of a line parallel to Y-axis is of the form x = h.
Since the line passes through A(2, – 3), h = 2
∴ The equation of the required line is x = 2.

ii. Equation of a line parallel to X-axis is of the formy = k.
Since the line passes through B(4, – 3), k = -3
∴ The equation of the required line is y = – 3.

Question 4.
Find the equation of the line:
i. passing through the points A(2, 0) and B(3,4)
ii. passing through the points P(2, 1) and Q(2,-1)
Solution:
i. The required line passes through the points A(2, 0) and B(3,4).
Equation of the line in two point form is \(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
Here, (x₁y₁) = (2,0) and (x₁,y₂) = (3,4)
∴ The equation of the required line is
∴ \(\frac{y-0}{4-0}=\frac{x-2}{3-2}\)
∴ \(\frac{y}{4}=\frac{x-2}{1}\)
∴ y = 4(x – 2)
∴ y = 4x – 8
∴ 4x – y – 8 = 0

ii. The required line passes through the points P(2, 1) and Q(2,-1).
Since both the given points have same
x co-ordinates i.e. 2,
the given points lie on the line x = 2.
∴ The equation of the required line is x = 2.

Question 5.
Find the equation of the line:
i. containing the origin and having inclination 60°.
ii. passing through the origin and parallel to AB, where A is (2,4) and B is (1,7).
iii. having slope 1/2 and containing the point (3, -2)
iv. containing the point A(3, 5) and having slope 2/3
v. containing the point A(4, 3) and having inclination 120°.
vi. passing through the origin and which bisects the portion of the line 3JC + y = 6 intercepted between the co-ordinate axes.
Solution:
i. Given, Inclination of line = θ = 60°
Slope of the line (m) = tan θ = tan 60°
= \(\sqrt{3}\)
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
.‘. The equation of the required line is y = \(\sqrt{3}\) x

ii. Given, A (2, 4) and B (1, 7)
Slope of AB = \(\frac{7-4}{1-2}\) = -3 1-2
Since the required line is parallel to line AB, slope of required line (m) = slope of AB
∴ m = – 3 and the required line passes through the origin.
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is y = – 3x

iii. Given, slope(m) = \(=\frac{1}{2}\) and the line passes through (3, – 2).
Equation of the line in slope point form is
y-y ₁= m(x-x₁)
∴ The equation of the required line is
[y-(- 2)]=\(\frac{1}{2}\)(x-3)
∴ 2(y + 2)=x – 3
∴ 2y + 4 = x – 3
∴ x – 2y – 7 = 0

iv. Given, slope(m) = \(\frac{2}{3}\) and the line passes through (3, 5).
Equation of the line in slope point form is y-y₁ = m(x -x₁)
∴ The equation of the required line is y – 5 = \(\frac{2}{3}\)(x-3)
∴ 3 (y – 5) = 2 (x – 3)
∴ 3y – 15 = 2x – 6
∴ 2x – 3y + 9 = 0

v. Given, Inclination of line = θ = 120°
Slope of the line (m) = tan θ = tan 120°
= tan (90° + 30°)
= – cot 30°
= – \(\sqrt{3}\)
and the line passes through A(4, 3).
Equation of the line in slope point form is y-y₁ = m(x -x₁)
∴ The equation of the required line is
y- 3 = –\(\sqrt{3}\)(x-4)
∴ y – 3 = –\(\sqrt{3}\) x + 4\(\sqrt{3}\)
∴ \(\sqrt{3}\)x + y – 3 -4\(\sqrt{3}\) = 0

vi.

Given equation of the line is 3x +y = 6.
∴ \(\frac{x}{2}+\frac{y}{6}=1\)
This equation is of the form \(\frac{x}{\mathrm{a}}+\frac{y}{\mathrm{~b}}\) = 1,
where a = 2, b = 6
∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2, 0) and B(0, 6) respectively. Required line is passing through the midpoint of AB.
∴ Midpoint of AB = ( \(\frac{2+0}{2}, \frac{0+6}{2}\) ) = (1,3)
∴ Required line passes through (0, 0) and (1,3).
Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the required line is
\(\frac{y-0}{3-0}=\frac{x-0}{1-0}\)
\(\frac{y}{3}=\frac{x}{1}\)
∴ y = 3x
∴ 3x – y = 0

Alternate Method:
Given equation of the line is 3x + y = 6 …(i)
Substitute y = 0 in (i) to get a point on X-axis.
∴ 3x + 0 = 6
∴ x = 2
Substitute x = 0 in (i) to get a point on Y-axis.
∴ 3(0) + 7 = 6
∴ y = 6
∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2,0) and B(0,6) respectively.
Let M be the midpoint of AB.
M = \(\left(\frac{2+0}{2}, \frac{0+6}{2}\right)\) = (1,3)
Slope of OM (m) = \(\frac{3-0}{1-0}\) = 3
Equation of OM is of the formy = mx.
∴ The equation of the required line is y = 3x
∴ 3x – y = 0

Question 6.
Line y = mx + c passes through the points A(2,1) and B(3,2). Determine m and c.
Solution:
Given, A(2, 1) and B(3,2)
Equation of the line in two point form is \(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the required line is
\(\frac{y-1}{2-1}=\frac{x-2}{3-2}\)
∴ \(\frac{y-1}{1}=\frac{x-2}{1}\)
∴ y – 1 = x – 2
∴ y = x – 1
Comparing this equation with y = mx + c, we get
m = 1 and c = – 1

Alternate Method:
Points A(2, 1) and B(3, 2) lie on the line y = mx + c.
∴ They must satisfy the equation.
∴ 2m + c = 1 …(i)
and 3m + c = 2 …(ii)
equation (ii) – equation (i) gives m = 1
Substituting m = 1 in (i), we get 2(1) + c = 1
∴ c = 1 – 2 = – 1

Question 7.
Find the equation of the line having inclination 135° and making x-intercept 7.
Solution:
Given, Inclination of line = 0 = 135°
∴ Slope of the line (m) = tan 0 = tan 135°
= tan (90° + 45°)
= – cot 45° = – 1 x-intercept of the required line is 7.
∴ The line passes through (7, 0).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
∴ The equation of the required line is y — 0 = – 1 (x – 7)
∴ y = -x + 7
∴ x + y – 7 = 0

Question 8.
The vertices of a triangle are A(3, 4), B(2, 0) and C(- 1, 6). Find the equations of the lines containing
i. side BC
ii. the median AD
iii. the midpoints of sides AB and BC.
Solution:
Vertices of AABC are A(3, 4), B(2, 0) and C(- 1, 6).
i. Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the side BC is
\(\frac{y-0}{6-0}=\frac{x-2}{-1-2}\)
\(\frac{y}{6}=\frac{x-2}{-3}\)
∴ – 3y = 6x – 12
∴ 6x + 3y – 12 = 0
∴ 2x + y – 4 = 0

ii. Let D be the midpoint of side BC.
Then, AD is the median through A.
∴ D = \(\left(\frac{2-1}{2}, \frac{0+6}{2}\right)=\left(\frac{1}{2}, 3\right)\)
The median AD passes through the points
A(3,4) and D( \(\frac{1}{2}\) , 3)

∴ The equation of the median AD is
\(\frac{y-4}{3-4}=\frac{x-3}{\frac{1}{2}-3}\)
\(\frac{y-4}{-1}=\frac{x-3}{-\frac{5}{2}}\)
\(\frac{5}{2}\)(y-4) = x – 3
∴ 5y – 20 = 2x – 6
∴ 2x – 5y + 14 = 0

iii. Let D and E be the midpoints of side AB and side BC respectively.
The equation of the line DE is

∴ -4(y-2) = 2x-5
∴ 2x + 4y – 13 = 0

Question 9.
Find the x and y-intercepts of the following lines:
i. \(\frac{x}{3}+\frac{y}{2}=1\)
ii. \(\frac{3 x}{2}+\frac{2 y}{3}=1\)
iii. 2x – 3y + 12 = 0
Solution:
i. Given equation of the line is latex]\frac{x}{3}+\frac{y}{2}=1[/latex]
This is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = 3, y-intercept = 2

ii. Given equation of the line is \(\frac{3 x}{2}+\frac{2 y}{3}\) = 1
∴ \(\frac{x}{\left(\frac{2}{3}\right)}+\frac{y}{\left(\frac{3}{2}\right)}\) = 1
This is of the form = \(\frac{x}{a}+\frac{y}{b}\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = \(\frac{2}{3}\) and y-intercept = \(\frac{3}{2}\)

iii. Given equation of the line is 2x – 3y + 12 = 0
∴ 2x – 3y = – 12
∴ \(\frac{2 x}{(-12)}-\frac{3 y}{(-12)}=1\)
∴ \(\frac{x}{-6}+\frac{y}{4}=1\)
This is of the form \(\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = – 6 and y-intercept = 4

Question 10.
Find equations of the line which contains the point A(l, 3) and the sum of whose intercepts on the co-ordinate axes is zero.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be
\(\frac{x}{a}+\frac{y}{b}=1\) ………..(i)
Since, the sum of the intercepts of the line is zero.
∴ a + b = 0
∴ b = – a
Substituting b = – a in (i), we get
\(\frac{x}{a}+\frac{y}{(-a)}=1\)
x – y = a .. .(ii)
Since, the line passes through A(1, 3).
∴ 1 – 3 = a
∴ a = – 2
Substituting the value of a in (ii), equation of the required line is
∴ x – y = – 2,
∴ x – y + 2 = 0

Case II: Line passing through origin.
Slope of line passing through origin and
A(1, 3) is m = \(\frac{3-0}{1-0}\) = 3
∴ Equation of the line having slope m and passing through origin (0, 0) is / = mx.
∴ The equation of the required line is y = 3x
∴ 3x – y = 0

Question 11.
Find equations of the line containing the point A(3, 4) and making equal intercepts on the co-ordinate axes.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}=1\) …………(i)
This line passes through A(3, 4).
∴ \(\frac{3}{a}+\frac{4}{b}=1\)……………..(ii)
Since, the required line make equal intercepts on the co-ordinate axes.
∴ a = b …(iii)
Substituting the value of b in (ii), we get
\(\frac{3}{a}+\frac{4}{a}=1\)
∴ \(\frac{7}{a}=1\)
∴ a = 7
∴ b = 7 …[From (iii)]
Substituting the values of a and b in (i), equation of the required line is
\(\frac{x}{7}+\frac{y}{7}=1\) = 1
∴ x + y = 7

Case II: Line passing through origin.
Slope of line passing through origin and A(3,4) is m = \(=\frac{4-0}{3-0}=\frac{4}{3}\)
∴ Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is 4
y = \(\frac{4}{3}\)x
∴ 4x – 3y = 0

Question 12.
Find the equations of the altitudes of the triangle whose vertices are A(2, 5), B(6, – 1 ) and C(- 4, – 3).
Solution:

A(2, 5), B(6, – 1), C(- 4, – 3) are the vertices of ∆ABC.
Let AD, BE and CF be the altitudes through the vertices A, B and C respectively of ∆ABC.
∴ Slope of AD = -5 …[∵AD ⊥ BC]
Since altitude AD passes through (2, 5) and has slope – 5,
equation of the altitude AD is y – 5 = -5 (x – 2)
∴ y – 5 = – 5x + 10
∴ 5x +y -15 = 0
Now, slope of AC = \(\frac{-3-5}{-4-2}=\frac{-8}{-6}=\frac{4}{3}\)
Slope of BE = \(\frac{-3}{4}\)
…[∵ BE ⊥ AC]
Since altitude BE passes through (6,-1) and has slope \(\frac{-3}{4}\),
equation of the altitude BE is
y-(-1) = \(\frac{-3}{4}\) (x – 6)
∴ 4 (y + 1) = – 3 (x – 6)
∴ 4y + 4 =-3x+ 18
∴ 3x + 4y – 14 = 0
Also, slope of AB = \(\frac{-1-5}{6-2}=\frac{-6}{4}=\frac{-3}{2}\)
∴ Slope of CF = \({2}{3}\) ….[∵ CF ⊥ AB]
Since altitude CF passes through (- 4, – 3) and has slope , \(\frac{2}{3}\)
equation of the altitude CF is
y-(-3) = \(\frac{2}{3}\)[x-(-4)]
∴ 3 (y + 3) = 2 (x + 4)
∴ 3y + 9 = 2x + 8
∴ 2x – 3y – 1 = 0

Question 13.
Find the equations of perpendicular bisectors of sides of the triangle whose vertices are P(-1, 8), Q(4, – 2) and R(- 5, – 3).
Solution:

Let A, B and C be the midpoints of sides PQ, QR and PR respectively of APQR.
A is the midpoint of side PQ.

Slope of perpendicular bisector of PQ is \(\frac{1}{2}\) and it passes through (\(\frac{3}{2}\)), 3).
Equation of the perpendicular bisector of side PQ is
y – 3 = \(\frac{1}{2}\)(x – \(\frac{3}{2}\))
y – 3 = (\(\frac{1}{2}\left(\frac{2 x-3}{2}\right)\))
∴ 4(y – 3) = 2x – 3
∴ 4y – 12 = 2x – 3
∴ 2x – 4y + 9 = 0
B is the midpoint of side QR
∴ B = \(\left(\frac{4-5}{2}, \frac{-2-3}{2}\right)=\left(\frac{-1}{2}, \frac{-5}{2}\right)\)
Slope of side QR = \(\frac{-3-(-2)}{-5-4}=\frac{-1}{-9}=\frac{1}{9}\)
∴ Slope of perpendicular bisector of QR is -9 and it passes through \(\left(-\frac{1}{2},-\frac{5}{2}\right)\)
∴ Equation of the perpendicular bisector of side QR is

∴ 2y + 5 = -18x – 9
∴ 18x + 2y + 14 = 0
∴ 9x + y + 7 = 0
C is the midpoint of side PR.

Equation of the perpendicular bisector of PR is \(y-\frac{5}{2}=-\frac{4}{11}(x+3)\)
∴ \(11\left(\frac{2 y-5}{2}\right)\) =-4(x + 3)
∴ 11(2y – 5) = – 8 (x + 3)
∴ 22y – 55 = – 8x – 24
∴ 8x + 22y -31 = 0

Question 14.
Find the co-ordinates of the orthocentre of the triangle whose vertices are A(2, – 2), B(l, 1) and C(-1,0).
Solution:
Let O be the orthocentre of AABC.
Let AM and BN be the altitudes of sides BC and AC respectively.
Now, slope of BC = \(\frac{0-1}{-1-1}=\frac{-1}{-2}=\frac{1}{2}\)
Slope of AM = -2 ,..[∵ AM ⊥ BC]
Since AM passes through (2, – 2) and has slope -2,
equation of the altitude AM is y – (- 2) = – 2 (x – 2)
∴ y + 2 = -2x + 4
∴ 2x + y – 2 = 0 …(i)

Also, slope of AC = \(\frac{0-(-2)}{-1-2}=\frac{2}{-3}\)
∴ Slope of BN = \(\frac{3}{2}\) …[∵ BN ⊥ AC]
Since BN passes through (1,1) and has slope \(\frac{3}{2}\), equation of the altitude BN is
y – 1 = \(\frac{3}{2}\)(x-1)
∴ 2y – 2 = 3x – 3
∴ 3x – 2y – 1 = 0 …(ii)
To find co-ordinates of orthocentre, we have to solve equations (i) and (ii).
By (i) x 2 + (ii), we get
7x – 5 = 0
∴ x = \(\frac{5}{7}\)
substituting x = \(\frac{5}{7}\) in eq (i), we get
2(\(\frac{5}{7}\)) + y – 2 = 0
∴ y = -2(\(\frac{5}{7}\)) + 2
∴ y = \(\frac{-10+14}{7}=\frac{4}{7}\)
∴ Coordinates of orthocentre O = \(\left(\frac{5}{7}, \frac{4}{7}\right)\)

Question 15.
N(3, – 4) is the foot of the perpendicular drawn from the origin to line L. Find the equation of line L.
Solution:

Slope of ON = \(\frac{-4-0}{3-0}=\frac{-4}{3}\)
Since line L ⊥ ON,
slope of the line L is \(\frac{3}{4}\) and it passes through point N(3, -4).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
Equation of line L is
y-(-4) = \(\frac{3}{4}\)(x-3)
∴ 4(y + 4) = 3(x – 3)
∴ 4y + 16 = 3x – 9
∴ 3x – 4y – 25 = 0

Class 11 Maharashtra State Board Maths Solution 

• Exercise 5.1 Class 11 Maths 1
• Exercise 5.2 Class 11 Maths 1
• Exercise 5.3 Class 11 Maths 1
• Exercise 5.4 Class 11 Maths 1
• Miscellaneous Exercise 5 Class 11 Maths 1

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.2 Questions and Answers.

11th Maths Part 1 Straight Line Exercise 5.2 Questions And Answers Maharashtra Board

Question 1.
Find the slope of each of the lines which passes through the following points:
i. A(2, -1), B(4,3)
ii. C(- 2,3), D(5, 7)
iii. E(2,3), F(2, – 1)
iv. G(7,1), H(- 3,1)
Solution:
i. Here, A = (2, -1) andB = (4, 3)
Slope of line AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-(-1)}{4-2}=\frac{4}{2}\) = 2

ii. Here, C = (-2, 3) and D = (5, 7)
Slope of line CD = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{7-3}{5-(-2)}=\frac{4}{7}\)

iii. Here, E s (2, 3) and F = (2, -1)
Slope of line EF = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-1-3}{2-2}=\frac{-4}{0}\), which ix not defined.

Alternate Method:
Points E and F have same x co-ordinates i.e. 2.
Points E and F lie on a line parallel to Y-axis.
∴ The slope of EF is not defined.

iv. Here, G = (7, 1) and H = (-3, 1)
Slope of line GH = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{1-1}{-3-7}\) = o

Alternate Method:
Points G and H have same y co-ordinate i.e. 1.
∴ Points G and H lie on a line parallel to the
X-axis.
∴ The slope of GH is 0.

Question 2.
If the x and x-intercepts of line L are 2 and 3 respectively, then find the slope of line L.
Solution:
Given, x-intercept of line L is 2 and y-intercept of line L is 3
∴ The line L intersects X-axis at (2, 0) and Y-axis at (0,3).
∴ The line L passes through (2, 0) and (0, 3).
Slope of line L = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-0}{0-2}=\frac{-3}{2}\)

Question 3.
Find the slope of the line whose inclination is 30°.
Solution:
Given, inclination (θ) = 30°
∴ Slope of the line = tanθ = tan30° = \(\frac{1}{\sqrt{3}}\)

Question 4.
Find the slope of the line whose inclination is \(\frac{\pi}{4}\)
Solution:
Given, inclination (0) = \(\frac{\pi}{4}\)
∴ Slope of the line = tan θ = tan\(\frac{\pi}{4}\) = 1

Question 5.
A line makes intercepts 3 and 3 on the co-ordinate axes. Find the inclination of the line.
Solution:
Given, x-intercept of line is 3 and y-intercept of line is 3
∴ The line intersects X-axis at (3, 0) and Y-axis at (0, 3).
∴ The line passes through (3, 0) and (0,3).
∴ Slope of line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-0}{0-3}\) = -1
But, slope of a line = tan θ
∴ tan θ = – 1
= – tan \(\frac{\pi}{4}\)
= tan(π-\(\frac{\pi}{4}\) ) …[v tan(π – θ) = -tan θ]
tan θ = tan \(\frac{3\pi}{4}\)
θ = \(\frac{3\pi}{4}\)
The inclination of the line is \(\frac{3\pi}{4}\).
[Note: Answer given in the textbook is ‘-1 However, as per our calculation it is \(\frac{3\pi}{4}\)]

Question 6.
Without using Pythagoras theorem, show that points A (4, 4), B (3, 5) and C (- 1, – 1) are the vertices of a right-angled triangle.
Solution:
Given, A(4,4), B(3, 5), C (-1, -1).
Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-4}{3-4}\) = – 1
Slope of BC = \(\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}\)
Slope of AC = \(\frac{-1-4}{-1-4}\) = 1
Slope of AB x slope of AC = – 1 x 1 = – 1
∴ side AB ⊥ side AC
∴ ∆ABC is a right angled triangle right angled at A.
∴ The given points are the vertices of a right angled triangle.

Question 7.
Find the slope of the line which makes angle of 45° with the positive direction of the Y-axis measured anticlockwise.
Solution:

Since the line makes an angle of 45° with positive direction of Y-axis in anticlockwise direction,
Inclination of the line (0) = (90° + 45°)
∴ Slope of the line = tan(90° + 45°)
= – cot 45°
= -1

Question 8.
Find the value of k for which the points P(k, -1), Q(2,1) and R(4,5) are collinear.
Solution:
Given, points P(k, – 1), Q (2, 1) and R(4, 5) are collinear.
∴ Slope of PQ = Slope of QR .
∴ \(\frac{1-(-1)}{2-k}=\frac{5-1}{4-2}\)
∴ \(\frac{2}{2-k}=\frac{4}{2}\)
∴ 4 = 4 (2 – k)
∴ 1 = 2 – k
∴ k = 2 – 1 = 1

Question 9.
Find the acute angle between the X-axis and the line joining the points A(3, -1) and B(4, – 2).
Solution:
Given, A (3, – 1) and B (4, – 2)
Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-(-1)}{4-3}\) = – 1
But, slope of a line = tan θ
∴ tan θ = – 1
= – tan 45°
= tan (180° -45°)
… [∵ tan (180° – θ) = -tan θ]
= tan 135°
∴ θ = 135°

Let α be the acute angle that line AB makes with X-axis.
Then, α + 0 = 180°
α = 180°- 135° = 45°
∴ The acute angle between the X-axis and the line joining the points A and B is 45°.

Question 10.
A line passes through points A(xi, y0 and B(h, k). If the slope of the line is m, then show that k – y₁ = m (h – x₁).
Solution:
Given, A(x₁, y₁), B(h, k) and
slope of line AB = m
Slope of line AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
∴ m = \(\frac{\mathrm{k}-y_{1}}{\mathrm{~h}-x_{1}}\)
∴ k – y₁ = m (h – x₁)

Question 11.
If the points A(h, 0), B(0, k) and C(a, b) lie on a line, then show that \(\frac{a}{h}+\frac{b}{k}\) = 1. ‘
Solution:
Given, A(h, 0), B(0, k) and C(a, b)
Since the points A, B and C lie on a line, they are collinear.
∴ Slope of AB = slope of BC

Class 11 Maharashtra State Board Maths Solution 

• Exercise 5.1 Class 11 Maths 1
• Exercise 5.2 Class 11 Maths 1
• Exercise 5.3 Class 11 Maths 1
• Exercise 5.4 Class 11 Maths 1
• Miscellaneous Exercise 5 Class 11 Maths 1