Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Locus and Straight Line Ex 5.4 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Ex 5.4

Question 1.

Find the slope, x-intercept, y-intercept of each of the following lines.

(a) 2x + 3y – 6 = 0

(b) x + 2y = 0

Solution:

(a) 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

∴ Slope of the line = \(\frac{-a}{b}=\frac{-2}{3}\)

x-intercept = \(\frac{-\mathrm{c}}{\mathrm{a}}=\frac{-(-6)}{2}\) = 3

y-intercept = \(\frac{-\mathrm{c}}{\mathrm{b}}=\frac{-(-6)}{3}\) = 2

(b) 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

∴ Slope of the line = \(\frac{-a}{b}=\frac{-1}{2}\)

x-intercept = \(\frac{-c}{a}=\frac{-0}{1}\) = 0

y-intercept = \(\frac{-\mathrm{c}}{\mathrm{b}}=\frac{-0}{2}\) = 0

Question 2.

Write each of the following equations in ax + by + c = 0 form.

(a) y = 2x – 4

(b) y = 4

(c) \(\frac{x}{2}+\frac{y}{4}=1\)

(d) \(\frac{x}{3}=\frac{y}{2}\)

Solution:

(a) y = 2x – 4

∴ 2x – y – 4 = 0 is the equation in ax + by + c = 0 form.

(b) y = 4

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

(c) \(\frac{x}{2}+\frac{y}{4}=1\)

∴ \(\frac{2 x+y}{4}=1\)

∴ 2x + y = 4

∴ 2x + y – 4 = 0 is the equation in ax + by + c = 0 form.

(d) \(\frac{x}{3}=\frac{y}{2}\)

∴ 2x = 3y

∴ 2x – 3y + 0 = 0 is the equation in ax + by + c = 0 form.

Question 3.

Show that the lines x – 2y – 7 = 0 and 2x – 4y + 5 = 0 are parallel to each other.

Solution:

Let m_{1} be the slope of the line x – 2y – 7 = 0.

∴ m_{1} = \(\frac{-1}{-2}=\frac{1}{2}\)

Let m_{2} be the slope of the line 2x – 4y + 5 = 0.

∴ m_{2} = \(\frac{-2}{-4}=\frac{1}{2}\)

Since, m_{1} = m_{2}

∴ The given lines are parallel to each other.

Question 4.

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

∴ \(\frac{3 x}{\mathrm{p}}+\frac{4 y}{\mathrm{p}}=1\)

∴ \(\frac{x}{\frac{p}{3}}+\frac{y}{\frac{p}{4}}=1\)

The equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\), with 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

∴ \(\frac{1}{2}\) × OA × OB = 24

∴ \(\frac{1}{2}\) × \(\frac{p}{3}\) × \(\frac{p}{4}\) = 24

∴ p^{2} = 576

∴ p = ±24

Question 5.

Find the co-ordinates of the circumcentre of the triangle whose vertices are A(-2, 3), B(6, -1), C(4, 3).

Solution:

Here, A(-2, 3), B(6, -1), C(4, 3) are the vertices of ∆ABC.

Let F be the circumcentre of ∆ABC.

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.

∴ D = \(\left(\frac{6+4}{2}, \frac{-1+3}{2}\right)\) = (5, 1)

and E = \(\left(\frac{-2+4}{2}, \frac{3+3}{2}\right)\) = (1, 3)

Now, slope of BC = \(\frac{-1-3}{6-4}\) = -2

∴ slope of FD = \(\frac{1}{2}\) …..[∵ FD ⊥ BC]

Since, FD passes through (5, 1) and has slope \(\frac{1}{2}\)

∴ Equation of FD is y – 1 = \(\frac{1}{2}\)(x – 5)

∴ 2(y – 1) = x – 5

∴ x – 2y – 3 = 0 ……(i)

Since, both the points A and C have same y co-ordinates i.e. 3

∴ the points A and C lie on the liney = 3.

Since, 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 6.

Find the equation of the line whose x-intercept is 3 and which is perpendicular to the line 3x – y + 23 = 0.

Solution:

Slope of the line 3x – y + 23 = 0 is 3.

∴ slope of the required line which is 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 7.

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, x_{1} = -2, y_{1} = 3

Question 8.

Find the distance between parallel lines 9x + 6y – 1 = 0 and 9x + 6y – 32 = 0.

Solution:

Equations of the given parallel lines are 9x + 6y – 7 = 0 and 9x + 6y – 32 = 0.

Here, a = 9, b = 6, C_{1} = -7 and C_{2} = -32

∴ Distance between the parallel lines

Question 9.

Find the equation of the line passing through the point of intersection of lines x + y – 2 = 0 and 2x – 3y + 4 = 0 and making intercept 3 on the X-axis.

Solution:

Given equations of lines are

x + y – 2 = 0 ……(i)

and 2x – 3y – 4 = 0 ……(ii)

Multiplying equation (i) by 3, we get

3x – 3y – 6 = 0 …..(iii)

Adding equation (ii) and (iii), we get

5x – 2 = 0

∴ x = \(\frac{2}{5}\)

Substituting x = \(\frac{2}{5}\) in equation (i), we get

\(\frac{2}{5}\) + y – 2 = 0

∴ y = 2 – \(\frac{2}{5}\) = \(\frac{8}{5}\)

∴ The required line passes through point (\(\frac{2}{5}\), \(\frac{8}{5}\)).

Also, the line makes intercept of 3 on X-axis

∴ it also passes through point (3, 0).

∴ required equation of line passing through points (\(\frac{2}{5}\), \(\frac{8}{5}\)) and (3, 0) is

∴ 13(5y – 8) = -8(5x – 2)

∴ 65y – 104 = -40x + 16

∴ 40x + 65y – 120 = 0

∴ 8x + 13y – 24 = 0 which is the equation of the required line.

Question 10.

D(-1, 8), E(4, -2), F(-5, -3) are midpoints of sides BC, CA and AB of ∆ABC. Find

(i) equations of sides of ∆ABC.

(ii) co-ordinates of the circumcentre of ∆ABC.

Solution:

(i) Let A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) be the vertices of ∆ABC.

Given, points D, E and F are midpoints of sides BC, CA and AB respectively of ∆ABC.

For x-coordinates:

Adding (i), (iii) and (v), we get

2x_{1} + 2x_{2} + 2x_{3} = -4

∴ x_{1} + x_{2} + x_{3} = -2 …..(vii)

Solving (i) and (vii), we get x_{1} = 0

Solving (iii) and (vii), we get x_{2} = -10

Solving (v) and (vii), we get x_{3} = 8

For y-coordinates:

Adding (ii), (iv) and (vi), we get

2y_{1} + 2y_{2} + 2y_{3} = 6

∴ y_{1} + y_{2} + y_{3} = 3 …..(viii)

Solving (ii) and (viii), we get y_{1} = -13

Solving (iv) and (viii), we get y_{2} = 7

Solving (vi) and (viii), we get y_{3} = 9

∴ Vertices of ∆ABC are A(0, -13), B(-10, 7), C(8, 9)

(ii) Here, A(0, -13), B(-10, 7), C(8, 9) are the vertices of ∆ABC.

Let F be the circumcentre of ∆ABC.

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.

∴ D = \(\left(\frac{-10+8}{2} \cdot \frac{7+9}{2}\right)\) = (-1, 8)

and E = \(\left(\frac{0+8}{2}, \frac{-13+9}{2}\right)\) = (4, -2)

Now, slope of BC = \(\frac{7-9}{-10-8}=\frac{1}{9}\)

∴ 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 …..(i)

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 \(\frac{-4}{11}\)

∴ 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 of y 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 = \(\left(\frac{-1}{19}, \frac{-10}{19}\right)\)