Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Locus and Straight Line Ex 5.3 Questions and Answers.

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

Question 1.

Write the equation of the line:

(a) parallel to the X-axis and at a distance of 5 units from it and above it.

(b) parallel to the Y-axis and at a distance of 5 units from it and to the left of it.

(c) parallel to the X-axis and at a distance of 4 units from the point (-2, 3).

Solution:

(a) Equation of a line parallel to the 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.

(b) 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.

(c) 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 = 3 + 4 = 7 or k = 3 – 4 = -1

∴ the equation of the required line is y = 7 or y = -1.

Question 2.

Obtain the equation of the line:

(a) parallel to the X-axis and making an intercept of 3 units on the Y-axis.

(b) parallel to the Y-axis and making an intercept of 4 units on the X-axis.

Solution:

(a) Equation of a line parallel to X-axis with y-intercept ‘k’ is y = k.

Here, y-intercept = 3

∴ the equation of the required line is y = 3.

(b) 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:

(a) A(2, -3) and parallel to the Y-axis.

(b) B(4, -3) and parallel to the X-axis.

Solution:

(a) Equation of a line parallel to the 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.

(b) Equation of a line parallel to the X-axis is of the form y = 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 passing through the points A(2, 0) and B(3, 4).

Solution:

The required line passes through the points A(2, 0) = (x_{1}, y_{1}) and B(3, 4) = (x_{2}, y_{2}) say.

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

∴ \(\frac{y-0}{4-0}=\frac{x-2}{3-2}\)

∴ \(\frac{y}{4}=\frac{x-2}{1}\)

∴ y = 4(x – 2)

∴ y = 4x – 8

∴ 4x – y – 8 = 0

Check:

If the points A(2, 0) and B(3, 4) satisfy 4x – y – 8 = 0, then our answer is correct.

For point A(2, 0),

L.H.S. = 4x – y – 8

= 4(2) – 0 – 8

= 8 – 8

= 0

= R.H.S.

For point B(3, 4),

L.H.S. = 4x – y – 8

= 4(3) – 4 – 8

= 12 – 12

= 0

= R.H.S.

Thus, our answer is correct.

Question 5.

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 a 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 passing through A and B 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 6.

The vertices of a triangle are A(3, 4), B(2, 0), and C(-1, 6). Find the equations of

(a) side BC

(b) the median AD

(c) the midpoints of sides AB and BC.

Solution:

Vertices of ∆ABC are A(3, 4), B(2, 0) and C(-1, 6).

(a) Equation of a 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

\(\frac{y-0}{6-0}=\frac{x-2}{-1-2}\) ……[B = (x_{1}, y_{1}) = (2, 0), C = (x_{2}, y_{2}) = (-1, 6)]

∴ \(\frac{y}{6}=\frac{x-2}{-3}\)

∴ y = -2(x – 2)

∴ 2x + y – 4 = 0

(b) 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

\(\frac{y-4}{3-4}=\frac{x-3}{\frac{1}{2}-3}\)

∴ \(\frac{y-4}{-1}=\frac{x-3}{-\frac{5}{2}}\)

∴ \(\frac{1}{2}\) (y – 4) = x – 3

∴ 5y – 20 = 2x – 6

∴ 2x – 5y + 14 = 0

(c) Let D and E be the midpoints of side AB and side BC respectively.

∴ D = \(\left(\frac{3+2}{2}, \frac{4+0}{2}\right)=\left(\frac{5}{2}, 2\right)\) and

E = \(\left(\frac{2-1}{2}, \frac{0+6}{2}\right)=\left(\frac{1}{2}, 3\right)\)

the equation of the line DE is

\(\frac{y-2}{3-2}=\frac{x-\frac{5}{2}}{\frac{1}{2}-\frac{5}{2}}\)

∴ \(\frac{y-2}{1}=\frac{2 x-5}{-4}\)

∴ -4(y – 2) = 2x – 5

∴ -4y + 8 = 2x – 5

∴ 2x + 4y – 13 = 0

Question 7.

Find the x and y-intercepts of the following lines:

(a) \(\frac{x}{3}+\frac{y}{2}=1\)

(b) \(\frac{3 x}{2}+\frac{2 y}{3}=1\)

(c) 2x – 3y + 12 = 0

Solution:

(a) Given equation of the line is \(\frac{x}{3}+\frac{y}{2}=1\)

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

(b) 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 = \(\frac{3}{2}\)

(c) 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 \(\frac{x}{a}+\frac{y}{b}=1\),

where x-intercept = a, y-intercept = b

∴ x-intercept = -6 and y-intercept = 4

Question 8.

Find the equations of a line containing the point A(3, 4) and make equal intercepts on the co-ordinate axes.

Solution:

Let the equation of the line be

\(\frac{x}{a}+\frac{y}{b}=1\) …..(i)

Since, the required line make equal intercepts on the co-ordinate axes.

∴ a = b

∴ (i) reduces to x + y = a …..(ii)

Since the line passes through A(3, 4).

∴ 3 + 4 = a

i.e. a = 7

Substituting a = 7 in (ii) to get

x + y = 7

Question 9.

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 BC = \(\frac{-3-(-1)}{-4-6}=\frac{-2}{-10}=\frac{1}{5}\)

∴ slope of AD = -5 ……..[∵ AD ⊥ BC]

Since, altitude AD passes through (2, 5) and has slope -5.

∴ the 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 \(\frac{-3}{4}\).

∴ the equation of the altitude BE is

y – (-1) = \(\frac{-3}{4}\)(x – 6)

∴ 4(y + 1) = -3(x – 6)

∴ 3x + 4y – 14 = 0

Also, slope of AB = \(\frac{-1-5}{6-2}=\frac{-6}{4}=\frac{-3}{2}\)

∴ slope of CF = \(\frac{2}{3}\) ………[∵ CF ⊥ AB]

Since, altitude CF passes through (-4, -3) and has slope \(\frac{2}{3}\).

∴ the equation of the altitude CF is

y – (-3) = \(\frac{2}{3}\) [x – (-4)]

∴ 3(y + 3) = 2(x + 4)

∴ 2x – 3y – 1 = 0