ML Aggarwal Solution Class 10 Chapter 12 Equation of a Straight Line Exercise 12.2

 Exercise 12.2

Question 1

State which one of the following is true : The straight lines y = 3x – 5 and 2y = 4x + 7 are

(i) parallel

(ii) perpendicular

(iii) neither parallel nor perpendicular.

Sol :

Slope of line y = 3x – 5 = 3

and slope of line 2y = 4x + 7

$\Rightarrow y=2 x+\frac{7}{2}=2$

∴ Slope of both the lines are neither equal nor their product is – 1.

∴ These line are neither parallel nor perpendicular.

Question 2

If 6x + 5y – 7 = 0 and 2px + 5y + 1 = 0 are parallel lines, find the value of p.

Sol :

In equation

6x + 5 y – 7 = 0

⇒ 5y = -6x + 7

$\Rightarrow  y=-\frac{6}{5} x+\frac{2}{5}$

$\therefore$ Slope $(m)=-\frac{6}{5}$...(i)

Again in equation 2px+5y+1=0

$\Rightarrow 5 y=-2 p x-1$

$\Rightarrow y=\frac{-2}{5} p x-\frac{1}{5}$

$\therefore$ slope $(m)=-\frac{2}{5} p$...(ii)

∵lines are parallel

$m_{1}=m_{2}$

From (i) and (ii)

$-\frac{6}{5}=-\frac{2 p}{5}$

$\Rightarrow p=\frac{-6}{5} \times\left(\frac{-5}{2}\right)=3$

Question 3

Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel. Find the relation connecting a and b. (1991)

Sol :

In equation 2x – by + 5 = 0

⇒ – by = – 2x – 5

$\Rightarrow y=\frac{2}{b}+\frac{5}{b}$

slope $(m)=\frac{2}{b}$

and in equation $a x+3 y=2$

$\Rightarrow 3 y=-a x+2$

$\Rightarrow y=-\frac{a}{3} x+\frac{2}{3}$

$\therefore$ slope $\left(m_{2}\right)=-\frac{a}{3}$

∵Lines are parallel

$\therefore m_{1}=m_{2} \Rightarrow \frac{2}{b}=-\frac{a}{3}$

$\Rightarrow-a b=6 \Rightarrow a b=-6$

Question 4

Given that the line $\frac{y}{2}=x-p$ and the line ax + 5 = 3y are parallel, find the value of a. (1992)

Sol :

In equation y = x – p

⇒ y = 2x – 2p

Slope ($m_1$) = 2

In equation ax + 5 = 3y

$\Rightarrow y=\frac{a}{3} x+\frac{5}{3}$

Slope $\left(m_{2}\right)=\frac{a}{3}$

$\because$ Lines are paralle

$\therefore m_{1}=m_{2}$

$\frac{a}{3}=2 \Rightarrow a=6$

Question 5

If the lines y = 3x + 7 and 2y + px = 3 perpendicular to each other, find the value of p. (2006)

Sol :

Gradient $m_1$ of the line y = 3x + 7 is 3

2y + px = 3

$\Rightarrow y=\frac{-p x}{2}+\frac{3}{2}$

Gradient $m^{2}$ of this line is $-\frac{p}{2}$

Since, the given lines are perpendicular to each other.

$\therefore m_{1} \times m_{2}=-1$

$\Rightarrow 3 \times\left(-\frac{p}{2}\right)=-1$

$p=\frac{2}{3}$

Question 6

Find the value of k for which the lines kx – 5y + 4 = 0 and 4x – 2y + 5 = 0 are perpendicular to each other. (2003)

Sol :

Given

In equation, kx – 5y + 4 = 0

$\Rightarrow 5 y=k x+4$

$\Rightarrow y=\frac{k}{5}+\frac{4}{5}$

$\therefore$ Slope $\left(m_{1}\right)=\frac{k}{5}$

and in equation, 4x-2y+5=0

$\Rightarrow 2 y=4 x+5$

$\Rightarrow y=2 x+\frac{5}{2}$

$\because$ Lines are perpendicular to each other

$\therefore m_{1} m_{2}=-1$

$\frac{k}{5} \times 2=-1$

$\Rightarrow k=\frac{-1 \times 5}{2}=\frac{-5}{2}$

Question 7

If the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other, find the relation connecting a and b.

Sol :

Given

In the equation 3x + by + 5 = 0

by=-3x-5

$\Rightarrow y=-\frac{3}{b} x-\frac{5}{b}$

Slope $\left(m_{1}\right)=\frac{-3}{b}$

and in the equation ax-5y+7=0

$\Rightarrow 5 y=a x+7$

$\Rightarrow y=\frac{a}{5}+\frac{7}{5}$

$\therefore$ Slope $\left(m_{2}\right)=\frac{a}{5}$

$\because$ Lines are perpendicular to each other

$\therefore m_{1} m_{2}=-1 \Rightarrow \frac{-3}{b} \times \frac{a}{5}=-1$

$\Rightarrow \frac{-3 a}{5 b}=-1$

$\Rightarrow-3 a=-5 b$

$\Rightarrow 3 a=5 b$

Question 8

Is the line through ( – 2, 3) and (4, 1) perpendicular to the line 3x = y + 1 ?

Does the line 3x = y + 1 bisect the join of ( – 2, 3) and (4, 1). (1993)

Sol :

Slope of the line passing through the points

(-2,3) and $(4,1)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$=\frac{1-3}{4+2}=\frac{-2}{6}=\frac{-1}{3}$

Slope of line 3x=y+1

$\Rightarrow y=3 x-1=3$

$\because m_{1} \times m_{2}=-\frac{1}{3} \times 3=-1$

$\therefore$ These lines are perpendicular to each other

(ii) Co-ordinates of mid point of line joining the points (-2,3) and (4,1) will be

$\left(\frac{-2+4}{2}, \frac{3+1}{2}\right)$ or $\left(\frac{2}{2}, \frac{4}{2}\right)$ or (1,2)

If mid-point (1,2) lies on the line 3x=y+1 then it will satisfy it

Now substituting the value of x and y is

$3 x=y+1 \Rightarrow 3(1)=2+1$

$\Rightarrow 3=3$ which is true

Hence the line 3x = y + 1 bisects the line joining the points (-2, 3), (4, 1).

Question 9

The line through A ( – 2, 3) and B (4, b) is perpendicular to the line 2x – 4y = 5. Find the value of b.

Sol :

Gradient ($m_1$) of the line passing through the

points A (-2, 3) and B (4, b)

$=\frac{b-3}{4+2}=\frac{b-3}{6}$

Gradient $\left(m_{2}\right)$ of the line 2x-4y=5

or $y=\frac{x}{2}-\frac{5}{2}$ is $\frac{1}{2}$

Since, the lines are perpendicular to each other,

$\therefore \quad m_{1} \times m_{2}=-1$

$\frac{b-3}{6} \times \frac{1}{2}=-1 \Rightarrow \frac{b-3}{12}=-1$

$\Rightarrow b-3=-12 \Rightarrow b=-9$

Question 10

If the lines 3x + y = 4, x – ay + 7 = 0 and bx + 2y + 5 = 0 form three consecutive sides of a rectangle, find the value of a and b.

Sol :

In the line 3x + y = 4 …(i)

⇒ y = – 3x + 4

Slope (m1) = – 3

In the line x – ay + 7 = 0…..(ii)

$\Rightarrow a y=x+7$

$\Rightarrow y=\frac{1}{a} x+\frac{7}{a}$

Slope $\left(m_{2}\right)=\frac{1}{a}$

and in the line bx+2y+5=0...(iii)

$\Rightarrow 2 y=-b x-5$

$\Rightarrow y=-\frac{b}{2} x-\frac{5}{2}$

$\therefore$ Slope $\left(m_{3}\right)=-\frac{b}{2}$

$\because$ These are the consecutive three sides of a rectangle.

$\therefore$(i) and (ii) are perpendicular to each other

$\therefore m_{1} m_{2}=-1 \Rightarrow-3 \times \frac{1}{a}=-1$

$\Rightarrow -3=-a \Rightarrow a=3$

and (i) and (iii) are parallel to each other

$\therefore m_{1}=m_{3} \Rightarrow-3=\frac{-b}{2}$

⇒ -b = -6 ⇒ b = 6

Hence a = 3, b = 6

Question 11

Find the equation of a line, which has the y-intercept 4, and is parallel to the line 2x – 3y – 7 = 0. Find the coordinates of the point where it cuts the x-axis. (1998)

Solution:

In the given line 2x – 3y – 7 = 0

⇒ 3y = 2x – 7

$\Rightarrow y=\frac{2}{3} x-\frac{7}{3}$

Hence slope $\left(m_{1}\right)=\frac{2}{3}$

$\therefore$ Equation of the line parallel to the given line will be

$y-y_{1}=m\left(x-x_{1}\right)$

$\because$ it passes through (0,4) then

$y-4=\frac{2}{3}(x-0)$

$\Rightarrow 3 y-12=2 x$

$\Rightarrow 2 x-3 y+12=0$...(i)

Now let it intersect x-axis at (x, y)

$\therefore  y=0$

Substituting the value of y in (i)

$2 x-3 \times 0+12=0$

$\Rightarrow 2 x=-12$

x=-6

Question 12

Find the equation of a straight line perpendicular to the line 2x + 5y + 7 = 0 and with y-intercept – 3 units.

Sol :

In the line 2x + 5y + 7 = 0

⇒ 5y = – 2x – 7

$\Rightarrow y=\frac{-2}{5} \times-\frac{7}{5}$

Here slope $\left(m_{1}\right)=-\frac{2}{5}$

Let the slope of the line perpendicular to the given line $=m_2$

$\therefore m_{1} m_{2}=-1$

$\Rightarrow-\frac{2}{5} m_{2}=-1$

$\therefore m_{2}=-1 \times \frac{-5}{2}=\frac{5}{2}$

$\therefore$ It makes y-intercept -3 units

$\therefore$ The point where it passes =(0,-3)

$\therefore$ Equations of the new line,

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-(-3)=\frac{5}{2}(x-0)$

$\Rightarrow y+3=\frac{5}{2} x$

$\Rightarrow 2 y+6=5 x$

$\Rightarrow 5 x-6 y-6=0$

Question 13

Find the equation of a st. line perpendicular to the line 3x – 4y + 12 = 0 and having same y-intercept as 2x – y + 5 = 0.

Sol :

In the given line 3x – 4y + 12 = 0

⇒ 4y = 3x + 12

$\Rightarrow y=\frac{3}{4} x+3$

Here slope $\left(m_{1}\right)=\frac{3}{4}$

Let the slope of the line perpendicular to the given line be $=m_{2}$

$\therefore m_{1} m_{2}=-1 \Rightarrow \frac{3}{4} m_{2}=-1$

$m_{2}=-\frac{4}{3}$

y-intercept in the equation

2 x-y+5=0

$\Rightarrow 2 \times 0-y+5=0$

$\Rightarrow y=5$

$\therefore$ The equation of the line passing through (0,5) will be

y-y_{1}=m\left(x-x_{1}\right) $

$\Rightarrow y-5=\frac{-4}{3}(x-0)

$\Rightarrow 3 y-15=-4 x$

$\Rightarrow 4 x+3 y-15=0$

Question 14

Find the equation of the line which is parallel to 3x – 2y = – 4 and passes through the point (0, 3). (1990)

Sol :

In the given line 3x – 2y = – 4

⇒ 2y = 3x + 4

$\Rightarrow y=\frac{3}{2} x+2$

Here slope $\left(m_{1}\right)=\frac{3}{2}$

$\therefore$ Slope of the line parallel to the given line

$=\frac{3}{2}$ and passes through (0,3)

$\therefore$ Equation of the line will be

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-3=\frac{3}{2}(x-0)$

$\Rightarrow  2 y-6=3 x$

$\Rightarrow 3 x-2 y+6=0$

Question 15

Find the equation of the line passing through (0, 4) and parallel to the line 3x + 5y + 15 = 0. (1999)

Sol :

In the given equation 3x + 5y + 15 = 0

⇒ 5y = – 3x – 15

$\Rightarrow y=\frac{-3}{5} x-3$

How slope $\left(m_{1}\right)=-\frac{3}{5}$

$\therefore$ Slope of the line parallel to the given line

$=-\frac{3}{5}$ and passes through the point (0,4)

$\therefore$ Equation of the line will be

$y-y_{1}=m\left(x-x_{1}\right) \Rightarrow y-4=-\frac{3}{5}(x-0)$

5y-20=-3 x 

$\Rightarrow 3 x+5 y-20=0$

Question 16

The equation of a line is y = 3x – 5. Write down the slope of this line and the intercept made by it on the y-axis. Hence or otherwise, write down the equation of a line which is parallel to the line and which passes through the point (0, 5).

Sol :

In the given line y = 3x – 5

Here slope ($m_1$) = 3

Substituting x = 0, then y = – 5

y-intercept = – 5

The slope of the line parallel to the given line

will be 3 and passes through the point (0, 5).

Equation of the line will be

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-5=3(x-0) \Rightarrow y-5=3 x$

$\Rightarrow 3 x-y+5=0 \Rightarrow y=3 x+5$

Question 17

Write down the equation of the line perpendicular to 3x + 8y = 12 and passing through the point ( – 1, – 2).

Sol :

In the given line 3x + 8y = 12

⇒ 8y = -3x + 12

$\Rightarrow y=\frac{-3}{8} x+\frac{12}{8}$

Here slope $\left(m_{1}\right)=\frac{-3}{8}$

Let the slope of the line perpendicular to the given line be $=m_{2}$

$\therefore m_{1} m_{2}=-1 \Rightarrow \frac{-3}{8} \times m_{2}=-1$

$m_{2}=\frac{8}{3}$

$\therefore$ Equation of the line where slope is $\frac{8}{3}$ and passes through the point (-1,-2) will be

$y-y_{1}=m\left(x-x_{1}\right)$

$y-(-2)=\frac{8}{3}[x-(-1)]$

$\Rightarrow y+2=\frac{8}{3}(x+1)$

$\Rightarrow 3 y+6=8 x+8$

$\Rightarrow 8 x-3 y+8-6=0$

$\Rightarrow 8 x-3 y+2=0$

Question 18

(i) The line 4x – 3y + 12 = 0 meets the x-axis at A. Write down the co-ordinates of A.

(ii) Determine the equation of the line passing through A and perpendicular to 4x – 3y + 12 = 0. (1993)

Sol :

(i) In the line 4x – 3y + 12 = 0 …(i)

3y = 4x + 12

$\Rightarrow y=\frac{4}{3} x+4$

Here slope $\left(m_{1}\right)=\frac{4}{3}$

Let the slope of the line perpendicular to the given line be $=m_{2}$

$\therefore m_{1} m_{2}=-1$

$\Rightarrow \frac{4}{3} \times m_{2}=-1$

$\Rightarrow m_{2}=-\frac{3}{4}$

Let the point on x-axis be A(x, 0)

$\therefore$ Substituting the value of y in (i)

$4 x-3 \times 0+12=0$

$\Rightarrow 4 x+12=0$

$\Rightarrow 4 x=-12$

$\Rightarrow x=-3$

$\therefore$ Co-ordinates of A will be (-3,0)

(ii) Equation of the line perpendicular to the given line passing through A will be.

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-0=\frac{-3}{4}(x+3)$

$\Rightarrow 4 y=-3 x-9$

$\Rightarrow 3 x+4 y+9=0$

Question 19

Find the equation of the line that is parallel to 2x + 5y – 7 = 0 and passes through the mid-point of the line segment joining the points (2, 7) and ( – 4, 1).

Sol :

The given line 2x + 5y – 7 = 0

5y = -2x + 7

$\Rightarrow y=\frac{-2}{5} x+\frac{7}{5}$

Co-ordinates of the mid point joining the points (2,7) and (-4,1) will be

$=\left(\frac{2-4}{2}, \frac{7+1}{2}\right)$ or $\left(\frac{-2}{2}, \frac{8}{2}\right)$ or (-1,4)

$\therefore$ Equation of the line will be,

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-4=\frac{-2}{5}(x+1)$

$\Rightarrow 5 y-20=-2 x-2$

$\Rightarrow 2 x+5 y-20+2=0$

$\Rightarrow 2 x+5 y-18=0$

Question 20

Find the equation of the line that is perpendicular to 3x + 2y – 8 = 0 and passes through the mid-point of the line segment joining the points (5, – 2), (2, 2).

Sol :

In the given line 3x + 2y – 8=0

⇒ 2y = – 3x + 8

$\Rightarrow y=\frac{-3}{2} x+4$

Here slope $\left(m_{1}\right)=\frac{-3}{2}$

Co-ordinates of the mid-point of the line segment joining the points (5,-2) and (2,2) will be

$\left(\frac{5+2}{2}, \frac{-2+2}{2}\right)$ or $\left(\frac{7}{2}, 0\right)$

and let the slope of the line perpendicular to the given line be $=m_{2}$

$\therefore m_{1} m_{2}=-1$

$\Rightarrow \frac{-3}{2} m_{2}=-1$

$\Rightarrow m_{2}=\frac{2}{3}$

$\therefore$ Equations of the line perpendicular to the given line and passing through $\left(\frac{7}{2}, 0\right)$ will be

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-0=\frac{2}{3}\left(x-\frac{7}{2}\right)$

$\Rightarrow 3 y=2 x-7$

$\Rightarrow 2 x-3 y-7=0$

Question 21

Find the equation of a straight line passing through the intersection of 2x + 5y – 4 = 0 with x-axis and parallel to the line 3x – 7y + 8 = 0.

Solution:

Let the point of intersection of the line 2x + 5y – 4 = 0 and x-axis be (x, 0)

Substituting the value of y in the equation

2x + 5 × 0 – 4 = 0

⇒ 2x – 4 = 0

⇒ 2x = 4

$\Rightarrow x=\frac{4}{2}=2$

Coordinates of the points of intersection will be (2, 0)

Now in the line 3x-7y+8=0

$\Rightarrow 7 y=3 x+8$

$\Rightarrow y=\frac{3}{7} x+\frac{8}{7}$

Slope $\left(m_{1}\right)=\frac{3}{7}$

and the slope of the line parallel to the above line will be $=\frac{3}{7}$

$\therefore$ Equation of the line will be

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-0=\frac{3}{7}(x-2)$

7y=3x-6

$\Rightarrow 3 x-7 y-6=0$

Question 22

The equation of a line is 3x + 4y – 7 = 0. Find

(i) the slope of the line. .

(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0. (2010)

Sol :

(i) Equation of the line is 3x + 4y – 1 = 0

⇒ 4y = 7 – 3x

$\Rightarrow y=\frac{-3}{4} x+\frac{7}{4}$

Comparing it with y=mx+c

$m=\frac{-3}{4}$

$\therefore$ Slope of the line $=\frac{-3}{4}$

(ii) Slope of the line perpendicular to the given line will be $m_{1}=-\frac{1}{m}=-\left(\frac{-4}{3}\right)=\frac{4}{3}$

Now x-y+2=0..(i)

3x+y-10=0...(ii)

Adding we get

4x-8=0

$\Rightarrow 4 x=8$

$\Rightarrow x=\frac{8}{4}=2$

From (i),

2-y+2=0

$\Rightarrow 4-y=0$

$\Rightarrow y=4$

$\therefore$ Point of intersection of the two lines is (2,4)

Now equation of the line perpendicular to the given line passing through (2,4) will be

$y-y_{1}=m_{1}\left(x-x_{1}\right)$

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

$\Rightarrow 3 y-12=4 x-8$

$\Rightarrow 4 x-3 y-8+12=0$

$\Rightarrow 4 x-3 y+4=0$

Question 23

Find the equation of the line perpendicular from the point (1, – 2) on the line 4x – 3y – 5 = 0. Also find the co-ordinates of the foot of perpendicular.

Sol :

In the equation 4x – 3y – 5 = 0,

⇒ 3y = 4x – 5

$\Rightarrow y=\frac{4}{3} x-\frac{5}{3}$

Slope $\left(m_{1}\right)=\frac{4}{3}$

Let the slope of the perpendicular $=m_{2}$

$m_{1} \times m_{2}=-1 \Rightarrow \frac{4}{3} \times m_{2}=-1$

$\therefore m_{2}=-\frac{3}{4}$

$\therefore$ Equation of the perpendicular where slope is $-\frac{3}{4}$ and drawn through the point (1,-2)

$y-y_{1}=m\left(x-x_{1}\right)$

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

$\Rightarrow 4 y+8=-3 x+3$

$\Rightarrow 3 x+4 y+8-3=0$

$\Rightarrow 3 x+4 y+5=0$

For finding the co-ordinates of the foot of the perpendicular we have to solve the equation

4 x-3 y-5=0 ..(i)

and 3 x+4 y+5=0..(ii)

Multiplying (i) by 4 and (ii) by 3, we get

16 x-12 y-20=0

9 x+12 y+15=0

25x-5=0 

$\Rightarrow 25 x=5$

$\therefore x=\frac{5}{25}=\frac{1}{5}$

Substituting the value of x in (i)

$4 \times\left(\frac{1}{5}\right)-3 y-5=0$

$\Rightarrow \frac{4}{5}-3 y-5=0$

$\Rightarrow 3 y=\frac{4}{5}-5=\frac{4-25}{5}=\frac{-21}{5}$

$\Rightarrow y=\frac{-21}{5 \times 3}=\frac{-7}{5}$

$\therefore$ Co-ordinates are $\left(\frac{1}{5}, \frac{-7}{5}\right)$

Question 24

Prove that the line through (0, 0) and (2, 3) is parallel to the line through (2, – 2) and (6, 4).

Sol :

Given that

Slope of the line through (0, 0) and (2, 3)

$\left(m_{1}\right)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-0}{2-0}=\frac{3}{2}$

and slope of the line through (2,-2) and (6,4)

$\left(m_{2}\right)=\frac{4+2}{6-2}=\frac{6}{4}=\frac{3}{2}$

$\therefore m_{1}=m_{2}=\frac{3}{2}$

$\therefore$ The lines are parallel to each other

Question 25

Prove that the line through,( – 2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (4, 24).

Sol :

Given that

Slope of the line through (-2, 6) and (4, 8)

$\left(m_{1}\right)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{8-6}{4+2}=\frac{2}{6}=\frac{1}{3}$

and slope of the line through (8,12) and (4,24)

$\left(m_{2}\right)=\frac{24-12}{4-8}=\frac{12}{-4}=-3$

$\because m_{1} \times m_{2}=\frac{1}{3} \times(-3)=-1$

∴ Lines are perpendicular to each other.

Question 26

Show that the triangle formed by the points A (1, 3), B (3, – 1) and C ( – 5, – 5) is a right angled triangle by using slopes.

Sol :

Slope $(m_1)$ of line by joining the points

$A(1,3), B(3,-1)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\therefore m_{1}=\frac{-1-3}{3-1}=\frac{-4}{2}=-2$

Slope $\left(m_{2}\right)$ of the line joinig the points B

(3,-1) and $\mathrm{C}(-5,-5)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\Rightarrow m_{2}=\frac{-5+1}{-5-3}=\frac{-4}{-8}=\frac{1}{2}$

$\therefore m_{1} \times m_{2}=-2 \times \frac{1}{2}=-1$

$\therefore$ Lines AB and BC are perpendicular to each other.

Hence  ΔABC  is a right angled triangle. Ans.

Question 27

Find the equation of the line through the point ( – 1, 3) and parallel to the line joining the points (0, – 2) and (4, 5).

Sol :

Slope of the line joining the points (0, -2) and (4, 5)$=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$=\frac{5+2}{4-0}=\frac{7}{4}$

Slope of the line parallel to it passing through (-1, 3) $=\frac{7}{4}$

and Equation of the line

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-3=\frac{7}{4}(x+1)$

$\Rightarrow 4 y-12=7 x+7$

$\Rightarrow  7 x-4 y+7+12=0$

$\Rightarrow  7 x-4 y+19=0$

Question 28

A ( – 1, 3), B (4, 2), C (3, – 2) are the vertices of a triangle.

(i) Find the coordinates of the centroid G of the triangle.

(ii) Find the equation of the line through G and parallel to AC

Sol :

Given, A (-1, 3), B (4, 2), C (3, -2)

(i) Coordinates of centroid G

$=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$

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

$=\left(\frac{6}{3}, \frac{3}{3}\right)=(2,1)$

So, the coordinates are (2,1)

(ii) Slope of $\mathrm{AC}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-3}{3-(-1)}=\frac{-5}{4}$

$\therefore$ Slope of the required line $(m)=\frac{-5}{4}$

Let the equation of the line through G, be

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-1=-\frac{5}{4}(x-2)$

$\Rightarrow 4 y-4=-5 x+10$

$\Rightarrow 5 x+4 y-14=0$ which is the required line.

Question 29

The line through P (5, 3) intersects y-axis at Q.

(i) Write the slope of the line.

(ii) Write the equation of the line.

(iii) Find the coordinates of Q.

Sol :

(i) Here θ = 45°

So, slope of the line = tan θ = tan 45° = 1

(ii) Equation of the line through P and Q is

y – 3 = 1(x – 5) ⇒ y – x + 2 = 0

(iii) Let the coordinates of Q be (0, y)

Then $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\Rightarrow 1=\frac{3-y}{5-0}$

$\Rightarrow 5=3-y \Rightarrow y=-2$

So, coordinates of Q are (0,-2)

Question 30

In the adjoining diagram, write down

(i) the co-ordinates of the points A, B and C.

(ii) the equation of the line through A parallel to BC. (2005)

Sol :

From the given figure, it is clear that

co-ordinates of A are (2, 3) of B are ( -1, 2)

and of C are (3, 0).

Now slope of $\mathrm{BC}(m)$

$=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{0-2}{3-(-1)}$

$=\frac{-2}{3+1}=\frac{-2}{4}=\frac{-1}{2}$

$\therefore$ Slope of line parallel to $\mathrm{BC}=\frac{-1}{2}$

$\because$ It passes through A(2,3)

$\therefore$ Its equations will be, $y-y_{1}=m\left(x-x_{1}\right)$

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

$\Rightarrow 2 y-6=-x+2$

$\Rightarrow x+2 y=2+6$

$\Rightarrow x+2 y=8$

Question 31

Find the equation of the line through (0, – 3) and perpendicular to the line joining the points ( – 3, 2) and (9, 1).

Sol :

The slope $(m_1)$ of the line joining the points (-3, 2) and (9, 1)

$=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{1-2}{9+3}=\frac{-1}{12}$

Let slope of the line perpendicular to the line $=m_{2}$

$\therefore m_{1} m_{2}=-1 \Rightarrow \frac{-1}{12} \times m_{2}=-1$

$\Rightarrow m_{2}=-1 \times \frac{(-12)}{1}=12$

$\therefore$ Equation of the line passing through

(0,-3) and of slope $m_{2}=12$

$y-y_{1}=m\left(x-x_{1}\right) \Rightarrow y+3=12(x-0)$

$\Rightarrow y+3=12 x$

$\Rightarrow 12 x-y-3=0$

Question 32

The vertices of a triangle are A (10, 4), B (4, – 9) and C ( – 2, – 1). Find the equation of the altitude through A. The perpendicular drawn from a vertex of a triangle to the opposite side is called altitude.

Sol :

Vertices of ∆ABC are A (10, 4), B (4, -9) and C( – 2, – 1)

Slope of the line BC $\left(m_{1}\right)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$=\frac{-1+9}{-2-4}=\frac{8}{-6}=\frac{-4}{3}$

Let the slope of the altitude from A(10,4) to $\mathrm{BC}=m_{2}$

$\therefore m_{1} m_{2}=-1 \Rightarrow=\frac{-4}{3} \times m_{2}=-1$

$\Rightarrow m_{2}=-1\left(-\frac{3}{4}\right)=\frac{3}{4}$

$\therefore$ Equation of the line will be,

$y-y_{1}=m\left(x-x_{1}\right) \Rightarrow y-4=\frac{3}{4}(x-10)$

$\Rightarrow 4 y-16=3 x-30$

$\Rightarrow 3 x-4 y+16-30=0$

$\Rightarrow 3 x-4 y-14=0$ 

Question 33

A (2, – 4), B (3, 3) and C ( – 1, 5) are the vertices of triangle ABC. Find the equation of :

(i) the median of the triangle through A

(ii) the altitude of the triangle through B

Sol :

(i) D is the mid-point of BC

Co-ordinates of D will

$\left(\frac{3-1}{2}, \frac{3+5}{2}\right)$ or $\left(\frac{2}{2}, \frac{8}{2}\right)$ or (1,4)

$\therefore$ Slope of median $\mathrm{AD}$

$(m)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{4+4}{1-2}=\frac{8}{-1}=-8$

Then equation of AD will be.

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-4=-8(x-1)$

$\Rightarrow y-4=-8 x+8$

$\Rightarrow 8 x+y-4-8=0$

$\Rightarrow 8 x+y-12=0$

(ii) BE is the altitude from B to AC

$\therefore$ Slope of $\mathrm{AC}\left(m_{1}\right)=\frac{y_{2}-y_{1}}{x_{2}-x_{\mathrm{1}}}=\frac{5+4}{-1-2}$

$=\frac{9}{-3}=-3$

Let slope of $\mathrm{BE}=m_{2}$

But $m_{1} m_{2}=-1 \Rightarrow-3 \times m_{2}=-1$

$m_{2}=\frac{-1}{-3}=\frac{1}{3}$

$\therefore$ Equation of BE will be,

$y-y_{1}=m\left(x-x_{1}\right)$

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

$\Rightarrow 3 y-9=x-3$

$\Rightarrow x-3 y-3+9=0$

$\Rightarrow x-3 y+6=0$

Question 34

Find the equation of the right bisector of the line segment joining the points (1, 2) and (5, – 6).

Sol :

Slope of the line joining the points (1, 2) and (5, -6)

$m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-6-2}{5-1}=\frac{-8}{4}=-2$

Let $m_{2}$ be the right bisector of the line

$\therefore m_{1} m_{2}=-1 \Rightarrow-2 \times m_{2}=-1$

$m_{2}=\frac{-1}{-2}=\frac{1}{2}$

mid point of the line segment joining (1,2) and (5,-6) will be

$\left(\frac{1+5}{2}, \frac{2-6}{2}\right)$ or $\left(\frac{6}{2}, \frac{-4}{2}\right)$ or (3,-2)

$\therefore$ Equation of line, the right bisector will be

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y+2=\frac{1}{2}(x-3)$

$\Rightarrow 2 y+4=x-3$

$\Rightarrow x-2 y-3-4=0$

$\Rightarrow x-2 y-7=0$ Ans.

Question 35

Points A and B have coordinates (7, – 3) and (1, 9) respectively. Find

(i) the slope of AB.

(ii) the equation of the perpendicular bisector of the line segment AB.

(iii) the value of ‘p’ if ( – 2, p) lies on it.

Sol :

Coordinates of A are (7, -3), of B = (1, 9)

(i) ∴ slope (m)

$=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{9-(-3)}{1-7}$

$=\frac{9+3}{1-7}=\frac{12}{-6}=-2$

(ii) Let PQ is the perpendicular bisector of AB intersecting it at M

$\therefore$ Co-ordinates of M will be

$=\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}$

$=\frac{7+1}{2}, \frac{-3+9}{2}=\frac{8}{2}, \frac{6}{2}$

or (4,3)

$\therefore$ Slope of $\mathrm{PQ}=\frac{1}{2}\left(m_{1}, m_{2}=-1\right)$

$\therefore$ Equation of $\mathrm{PQ}=y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-3=\frac{1}{2}(x-4) \Rightarrow 2 y-6=x-4$

$\Rightarrow x-2 y+6-4=0 \Rightarrow x-2 y+2=0$

(iii) $\therefore$ Point $(-2, p)$ lies on it

$\therefore-2-2 p+2=0$

$\Rightarrow-2 p+0=0 \Rightarrow-2 p=0$

$\therefore p=0$

Question 36

The points B (1, 3) and D (6, 8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC.

Sol :

Slope of $\mathrm{BD}\left(\mathrm{m}_{1}\right)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$=\frac{8-3}{6-1}=\frac{5}{5}=1$

Diagonal AC is perpendicular bisector of diagonal BD

$\therefore$ Slope of $\mathrm{AC}=-1\left(\because m_{1} m_{2}=-1\right)$

and co-ordinates of mid point of BD will be

$\left(\frac{1+6}{2}, \frac{3+8}{2}\right)$ or $\left(\frac{7}{2}, \frac{11}{2}\right)$

$\therefore$ Equation of $\mathrm{AC}$ 

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-\frac{11}{2}=-1\left(x-\frac{7}{2}\right)$

$\Rightarrow y-\frac{11}{2}=-x+\frac{7}{2}$

$\Rightarrow 2 y-11=-2 x+7$

$\Rightarrow 2 x+2 y-11-7=0$

$\Rightarrow 2 x+2 y-18=0$

or x+y-9=0

Question 37

ABCD is a rhombus. The co-ordinates of A and C are (3, 6) and ( – 1, 2) respectively. Write down the equation of BD. (2000)

Sol :

Co-ordinates of A (3, 6), C (-1, 2)

Slope of $\mathrm{AC}\left(\mathrm{m}_{1}\right)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$=\frac{2-6}{-1-3}=\frac{-4}{-4}=1$

$\therefore$ But line BD is the right bisector of AC.

$\therefore$ Slope of $\mathrm{BD}=-1\left(\because m_{1} m_{2}=-1\right)$

and co-ordinates of mid point of AC wil be

$\left(\frac{3-1}{2}, \frac{6+2}{2}\right)$ or $\left(\frac{2}{2}, \frac{8}{2}\right)$ or (1,4)

$\therefore$ Equation of BD will be, $y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-4=-1(x-1)$

$\Rightarrow y-4=-x+1$

$\Rightarrow x+y-4-1=0$

$\Rightarrow x+y-5=0$

Question 38

Find the equation of the line passing through the intersection of the lines 4x + 3y = 1 and 5x + 4y = 2 and

(i) parallel to the line x + 2y – 5 = 0   

(ii) perpendicular to the x-axis.

Sol :

4x + 3y = 1 …(i)

5x + 4y = 2 …(ii)

Multiplying (i) by 4 and (ii) by 3

$\begin{array}{c}16 x+12 y=4 \\15 x+12 y=6 \\-\quad-\quad- \\\hline x \quad=-2 \\\hline\end{array}$

Substituting the value of x in (i)

4(-2)+3y=1

$\Rightarrow-8+3 y=1$

$\Rightarrow \quad 3 y=1+8=9$

$\Rightarrow y=\frac{9}{3}=3$

$\therefore$ Point of intersection $=(-2,3)$

(i) In the line x+2y-5=0

$\Rightarrow 2 y=-x+5$

$\Rightarrow y=-\frac{1}{2} x+\frac{5}{2}$

$\because$ Slope $\left(m_{1}\right)=-\frac{1}{2}$

$\therefore$ Slope of its parallel line $=-\frac{1}{2}$

and equation of the parallel line

$y-y_{1}=m\left(x-{x}_{1}\right)$

$\Rightarrow \quad y-3=-\frac{1}{2}(x+2)$

$\Rightarrow 2 y-6=-x-2$

$\Rightarrow x+2 y-6+2=0$

$\Rightarrow x+2 y-4=0$

(ii) $\because$ Any line perpendicular to x-axis will be parallel to y-axis.

$\therefore$ Equation of the line will be

x=a i.e. $x=-2 \Rightarrow x+2=0$

Question 39

(i) Write down the co-ordinates of the point P that divides the line joining A ( – 4, 1) and B (17, 10) in the ratio 1 : 2.

(ii) Calculate the distance OP where 0 is the origin

(iii) In what ratio does the y-axis divide the line AB?

Sol :

(i) Co-ordinate A (-4, 1) and B (17, 10) P divides it in the ratio of 1 : 2

Let the co-ordinates of P will be (x, y)

$\therefore x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{1 \times 17+2 \times(-4)}{1+2}$

$=\frac{17-8}{3}=\frac{9}{3}=3$

$y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{1 \times 10+2 \times 1}{1+2}$

$=\frac{10+2}{3}=\frac{12}{3}=4$

$\therefore$ Co-ordinates of P wiil be (3,4)

(ii) O is the origin

$\therefore$ Distance between O and P

$=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\sqrt{(0-3)^{2}+(0-4)^{2}}=\sqrt{(-3)^{2}+(-4)^{2}}$

$=\sqrt{9+16}=\sqrt{25}=5$ units.

(iii) Let y-axis divides AB in the ratio of $m_{1},:$ $m_{2}$

$\therefore x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}$

$\Rightarrow 0=\frac{m_{1} \times 17+m_{2} \times(-4)}{m_{1}+m_{2}}$

$\Rightarrow 17 m_{1}-4 m_{2}=0$

$\Rightarrow 17 m_{1}=4 m_{2}$

$\Rightarrow \frac{m_{1}}{m_{2}}=\frac{4}{17}$

$\Rightarrow m_{1}: m_{2}=4: 17$

Question 40

Find the image of the point (1, 2) in the line x – 2y – 7 = 0

Sol :

Draw a perpendicular from the point P(1, 2) on the line, x – 2y – 7 = 0

Let P’ is the image of P and let its

co-ordinates sue (α, β) slope of line x – 2y – 7 = 0

$\Rightarrow 2 y=x-7$

$\Rightarrow y=\frac{1}{2} x-\frac{7}{2}$ is $\frac{1}{2}$

$\therefore$ Slope of PP'=-2  $\left(\because m_{1} m_{2}=-1\right)$

$\therefore$ Equation of PP'

$y-y_{1}=m\left(x-x_{1}\right)=y-2=-2(x-1)$

$\Rightarrow y-2=-2 x+2 \Rightarrow 2 x+y=2+2$

$\Rightarrow 2 x+y=4$

$\because P^{\prime}(\alpha, \beta)$ lies on it

$\therefore 2 \alpha+\beta=4$..(i)

$\because P^{\prime}$ is the image of P in the line

x-2y-7=0

$\therefore$ the line bisects PP' at M

or M is the mid-point of PP'

$\therefore$ Co-ordinates of M will be $\left(\frac{1+\alpha}{2}, \frac{2+\beta}{2}\right)$

$\therefore$ Substituting the value of x, y

$\frac{1+\alpha}{2}-2\left(\frac{2+\beta}{2}\right)-7=0$

$\Rightarrow \frac{1+\alpha}{2}-(2+\beta)-7=0$

$1+\alpha-4-2 \beta-14=0$

$\alpha-2 \beta=4+14-1=17$...(ii)

$\alpha=17+2 \beta$

Substituting the value of ⍺ in (i)

$2(17+2 \beta)+\beta=4$

$34+4 \beta+\beta=4$

$\Rightarrow 5 \beta=4-34=-30$

$\beta=-\frac{30}{5}=-6$

Substituting the value of $\beta$ in (i)

$2 \alpha-6=4 \Rightarrow 2 \alpha=4+6=10$

$\alpha=\frac{10}{2}=5$

$\therefore$ Co-ordinates of P' will be (5,-6)

Question 41

If the line x – 4y – 6 = 0 is the perpendicular bisector of the line segment PQ and the co-ordinates of P are (1, 3), find the co-ordinates of Q.

Sol :

Let the co-ordinates of Q be (α, β) and let the line x – 4y – 6 = 0 is the

perpendicular bisector of PQ and it intersects the line at M.

M is the mid point of PQ Now slope of line x – 4y – 6 = 0

$\Rightarrow 4 y=x-6$

$\Rightarrow y=\frac{1}{4} x-\frac{6}{4}$ is $\frac{1}{4}$

$\therefore$ Slope of $\mathrm{PQ}=-4$

$\left(\because m_{1} m_{2}=-1\right)$

and equation of line PQ

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-3=-4(x-1)$

$\Rightarrow y-3=-4 x+4$

$\Rightarrow 4 x+y-3-4=0$

$\Rightarrow 4 x+y-7=0$

$\Rightarrow 4 x+y \pm 7$

$\because \mathrm{Q}(\alpha, \beta)$ lies on it.

$\therefore 4 \alpha+B=7$...(i)

Now co-ordinates of M will be

$\left(\frac{1+\alpha}{2}, \frac{3+\beta}{2}\right)$

$\because \mathrm{M}$ lies on the line x-4y-6=0

$\therefore \frac{1+\alpha}{2}-4\left(\frac{3+\beta}{2}\right)-6=0$

$\Rightarrow 1+\alpha-4(3+\beta)-12=0$

$\Rightarrow 1+\alpha-12-4 \beta-12=0$

$\Rightarrow \alpha-4 \beta=24-1=23$..(ii)

Multiply (i) by 4 and (ii) by 1

$16 \alpha+4 \beta=28$

$\alpha-4 \beta=23$

Adding , we get

$17 \alpha =51 \quad \Rightarrow \quad \alpha=\frac{51}{17}=3$

Substituting the value of $\alpha$ in (i) 

$4 \times 3+\beta=7$

$\Rightarrow \beta=7-12=-5$

$\therefore$ Co-ordinates of Q will be (3,-5)

Question 42

OABC is a square, O is the origin and the points A and B are (3, 0) and (p, q). If OABC lies in the first quadrant, find the values of p and q. Also write down the equations of AB and BC.

Sol :

$O A=\sqrt{(3-0)^{2}+(0-0)^{2}}$

$=\sqrt{(3)^{2}+(0)^{2}}$

$=\sqrt{9+0}=\sqrt{9}=3$

$A B=\sqrt{(3-p)^{2}+(0-q)^{2}}$

$=\sqrt{(3-p)^{2}+q^{2}}$

$\because$ OA=AB (sides of a square)

$\therefore \sqrt{(3-p)^{2}+q^{2}}=3$

$(3-p)^{2}+q^{2}=9$ (Squaring both sides)

$9+p^{2}-6 p+q^{2}=9$

$\Rightarrow p^{2}+q^{2}-6 p=0$...(i)

$\mathrm{OB}=\sqrt{(p-0)^{2}+(q-0)^{2}}=\sqrt{p^{2}+q^{2}}$

But $\mathrm{OB}^{2}=\mathrm{OA}^{2}+\mathrm{AB}^{2}$

$\Rightarrow\left(\sqrt{p^{2}+q^{2}}\right)^{2}$

$=3^{2}+\left(\sqrt{(3-p)^{2}+q^{2}}\right)^{2}$

$\Rightarrow p^{2}+q^{2}=9+(3-p)^{2}+q^{2}$

$\Rightarrow p^{2}+q^{2}=9+9+p^{2}-6 p+q^{2}$

$6 p=18 \Rightarrow p=\frac{18}{6}=3$

Substituting the value of p in (i)

$(3)^{2}+q^{2}-6(3)=0$

$\Rightarrow 9+q^{2}-18=0$

$\Rightarrow q^{2}-9=0$

$\Rightarrow q^{2}=9$

$\Rightarrow q=3$

$\therefore  p=3, q=3$

$\because$ AB parallel to y-axis

$\therefore$ Equation AB will be x=3

$\Rightarrow x-3=0$

and equation of BC will be y=3 $(\because \mathrm{BC} \| x$ -axis $)$

⇒y-3=0