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

## Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.3

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, (x_{1}y_{1}) = (2,0) and (x_{1},y_{2}) = (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-x_{1})

∴ 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-y_{1} = m(x -x_{1})

∴ 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-y_{1} = m(x -x_{1})

∴ 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 – y_{1} = m(x – x_{1})

∴ 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 – y_{1} = m(x – x_{1})

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