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

 Test

Question 1

Find the equation of a line whose inclination is 60° and y-intercept is – 4.

Sol :

Angle of inclination = 60°

Slope = tan θ = tan 60° = √3

Equation of the line will be,

y = mx + c = √3x + ( – 4)

⇒ y – √3x – 4


Question 2

Write down the gradient and the intercept on the y-axis of the line 3y + 2x = 12.

Sol :

Slope of the line 3y + 2x = 12

⇒ 3y = 12 – 2x

⇒ 3y = -2x + 12

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

$\therefore$ Slope $=\frac{-2}{3}$ and y-intercept =4


Question 3

If the equation of a line is y – √3x + 1, find its inclination.

Sol :

In the line

y = √3 x + 1

Slope = √3

⇒ tan θ = √3

⇒ θ = 60° (∵ tan 60° = √3)


Question 4

If the line y = mx + c passes through the points (2, – 4) and ( – 3, 1), determine the values of m and c.

Sol :

The equation of line y = mx + c

∵ it passes through (2, – 4) and ( – 3, 1)

Now substituting the value of these points -4 = 2m + c …(i)

and 1 = -3m + c …(ii)

Subtracting we get,

$-5=5 \mathrm{~m}$
$ \Rightarrow m=\frac{-5}{5}=-1$

Substituting the value of m in (i)

-4=2(-1)+c 

$\Rightarrow-4=-2+c$

c=-4+2=-2

∴ m=-1, c=-2


Question 5

If the point (1, 4), (3, – 2) and (p, – 5) lie on a st. line, find the value of p.

Sol :

Let the points to be A (1, 4), B (3, -2) and C (p, -5) are collinear and let B (3, -2)

divides AC in the ratio of $m_{1}: m_{2}$

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

$\Rightarrow 3=\frac{m_{1} p+m_{2} \times 1}{m_{1}+m_{2}}$

$3 m_{1}+3 m_{2}=m_{1} p+m_{2}$

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

$\Rightarrow m_{1}(3-p)=-2 m_{2}$

$\Rightarrow \frac{m_{1}}{m_{2}}=\frac{-2}{3-p}$...(i)

and $-2=\frac{m_{1}(-5)+m_{2} \times 4}{m_{1}+m_{2}}$

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

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

$\Rightarrow 3 m_{1}=6 m_{2}$

$\Rightarrow \frac{m_{1}}{m_{2}}=\frac{6}{3}=2$...(ii)

From (i) and (ii)

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

$\Rightarrow-2=6-2 p$

$\Rightarrow 2 p=6+2=8 $

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


Question 6

Find the inclination of the line joining the points P (4, 0) and Q (7, 3).

Sol :

Slope of the line joining the points P (4, 0) and Q (7, 3)

$=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-0}{7-4}=\frac{3}{3}=1$

$\therefore \tan \theta=1 \Rightarrow \theta=45^{\circ}$ $\left(\because \tan 45^{\circ}=1\right)$

Hence inclination of line $=45^{\circ}$


Question 7

Find the equation of the line passing through the point of intersection of the lines 2x + y = 5 and x – 2y = 5 and having y-intercept equal to $-\frac{3}{7}$

Sol :

Equation of lines are

2x + y = 5 …(i)

x – 2y = 5 …(ii)

Multiply (i) by 2 and (ii) by 1, we get

4x + 2y = 10

x – 2y = 5

Adding we get,

5x=15
$ \Rightarrow x=\frac{15}{5}=3$

Substituting the values of x in (i)

$2 \times 3+y=5 \Rightarrow 6+y=5$

$\Rightarrow y=5-6=-1$

$\therefore$ Co-ordinates of point of intersection are(3,-1)

$\because$ the line passes through (3,-1)

$\therefore -1=m \times 3-\frac{3}{7}$ (y=m x+c)

$3 m=-1+\frac{3}{7}=\frac{-4}{7}$

$m=\frac{-4}{7 \times 3}=\frac{-4}{21}$

$\therefore$ Equation of line

$y=\frac{-4}{21} x-\frac{3}{7}$

$ \Rightarrow 21 y=-4 x-9$

$\Rightarrow 4 x+21 y+9=0$


Question 8

If point A is reflected in the y-axis, the co-ordinates of its image $A_1$, are (4, – 3),

(i) Find the co-ordinates of A

(ii) Find the co-ordinates of $A_2,A_3$ the images of the points A, $A_1$, Respectively under reflection in the line x = – 2

Sol :

(i) ∵ A is reflected in the y-axis and its image is A1 (4, -3)

Co-ordinates of A will be (-4, -3)



















(ii) $\because \mathrm{A}_{2}$ is the image of A(-4,-3) in the line

x=-2 which is parallel to y-axis

$\therefore \mathrm{AA}_{2}$ is perpendicular to the line x=-2

and this line bisects $\mathrm{AA}_{2}$ at M. 

Let co-ordinates of $\mathrm{A}_{2}$ be $(\alpha, \beta)$

$\therefore \alpha=0 $ and $ \beta=-3$

$\therefore$ Co-ordinates of $A_{2}$ will be (0,-3)

Again x=-2 is perpendicular bisector of

$A_{1}, A_{3}$ intersecting it at M

$\therefore M$ is mid point of $\mathrm{A}_{1} \mathrm{~A}_{3}$

$\therefore \quad A_{3} M=M A_{1}$

$\therefore$ Co-ordinates of $A_{3}$ will be (-8,-3)


Question 9

If the lines $\frac{x}{3}+\frac{y}{4}=7$ and 3x + ky = 11 are perpendicular to each other, find the value of k.

Sol :
Given Equation of lines are

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

$\Rightarrow 4 x+3 y=84$

$\Rightarrow 3 y=-4 x+84$

$ \Rightarrow y=\frac{-4}{3} x+28$...(i)

and 3x+ky=11 

$\Rightarrow k y=-3 x+11$

$\Rightarrow y=\frac{-3}{k} x+\frac{11}{k}$...(ii)

Let slope of line (i) be $m_{1}$ and of (ii) be $m_{2}$

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

$\because$ These lines are perpendicular to each other

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

$\Rightarrow \frac{-4}{3} \times\left(-\frac{3}{k}\right)=-1$

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

$ \Rightarrow \frac{-4}{3} \times\left(-\frac{3}{k}\right)=-1$

$\Rightarrow \frac{4}{k}=-1$

$ \Rightarrow-k=4 \Rightarrow k=-4$


Question 10

Write down the equation of a line parallel to x – 2y + 8 = 0 and passing through the point (1, 2).

Sol :

The equation of the line is x – 2y + 8 = 0

⇒ 2y = x + 8

$\Rightarrow  y=\frac{1}{2} x+4$

$\therefore$ Slope of the line $=\frac{1}{2}$

$\therefore$ Slope of the line parallel to the given line passing through $(1,2)=\frac{1}{2}$

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

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

$\Rightarrow 2 y-4=x-1$

$\Rightarrow  x-2 y-1+4=0$

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


Question 11

Write down the equation of the line passing through ( – 3, 2) and perpendicular to the line 3y = 5 – x.

Sol :

Equations of the line is

3y = 5 – x ⇒ 3y = -x + 5

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

$\therefore$ Slope of the line $=-\frac{1}{3}$

and slope of the line perpendicular to it and passing through (-3,2) will be=3

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

$\therefore$ Equation of the line will be

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

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

$\Rightarrow y-2=3 x+9$

$ \Rightarrow 3 x-y+9+2=0$

$\Rightarrow 3 x-y+11=0$ Ans.


Question 12

Find the equation of the line perpendicular to the line joining the points A (1, 2) and B (6, 7) and passing through the point which divides the line segment AB in the ratio 3 : 2.

Sol :
Let slope of the line joining the points A (1, 2) and B (6, 7) be $m_{1}$

$\therefore m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{7-2}{6-1}=\frac{5}{5}=1$

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

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

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

$\therefore m_{2}=-1$

Let the point P(x, y) divides the line AB in the ratio of 3: 2

$\therefore \quad x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{3 \times 6+2 \times 1}{3+2}$

$=\frac{18+2}{5}=\frac{20}{5}=4$


and $y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{3 \times 7+2 \times 2}{3+2}$

$=\frac{21+4}{5}=\frac{25}{5}=5$

$\therefore$ Co-ordinates of $\mathrm{P}$ will be (4,5)

Now equation of the line passing through P and having slope -1

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

$ \Rightarrow y-5=-1(x-4)$

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

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

$\Rightarrow x+y-9=0$


Question 13

The points A (7, 3) and C (0, – 4) are two opposite vertices of a rhombus ABCD. Find the equation of the diagonal BD.

Sol :

Slope of line AC $(m_1)$











$=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-4-3}{0-7}=\frac{-7}{-7}=1$

∵Diagonals of a rhombus bisect each other at right angles

$\therefore \mathrm{BD}$ is perpendicular to AC

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

and co-ordinates of O, the mid-point of AC will be

$\left(\frac{7+0}{2}, \frac{3-4}{2}\right)$ or $\left(\frac{7}{2}, \frac{-1}{2}\right)$

$\therefore$ Equation of BD will be

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

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

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

$ \Rightarrow 2 y+1=-2 x+7$

$\Rightarrow 2 x+2 y+1-7=0$

$\Rightarrow  2 x+2 y-6=0$

$\Rightarrow x+y-3=0$ (Dividing by 2)


Question 14

A straight line passes through P (2, 1) and cuts the axes in points A, B. If BP : PA = 3 : 1, find:













(i) the co-ordinates of A and B
(ii) the equation of the line AB
Sol :
A lies on x-axis and B lies on y-axis
Let co-ordinates of A be (x, 0) and B be (0, y)
and P (2, 1) divides BA in the ratio 3 : 1.

$\therefore x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}} \Rightarrow 2=\frac{3 \times x+1 \times 0}{3+1}$

$\Rightarrow \frac{3 x}{4}=2 $

$\Rightarrow 3 x=8 \Rightarrow x=\frac{8}{3}$

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

$ \Rightarrow 1=\frac{3 \times 0+1 \times y}{3+1}$

$\Rightarrow  \frac{y}{4}=1 $

$\Rightarrow y=4$

$\therefore$ Co-ordinates of A will be $\left(\frac{8}{3}, 0\right)$ and of B  will be (0,4)

Slope of the line $A B=\frac{y_{2}+y_{1}}{x_{2}-x_{1}}$

$=\frac{4-0}{0-\frac{8}{3}}=\frac{4}{-\frac{8}{3}}$

$=-4 \times \frac{3}{8}=-\frac{3}{2}$

$\therefore$ Equation of AB will be

$y-y_{1}=m\left(x-x_{1}\right) \Rightarrow y-1=\frac{-3}{2}(x-2)$

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

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

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


Question 15

A straight line makes on the co-ordinates axes positive intercepts whose sum is 7. If the line passes through the point ( – 3, 8), find its equation.

Sol :

Let the line make intercept a and b with the

x-axis and y-axis respectively then the line passes through











A(a, 0) and B(0, b)
but a+b=7
b=7-a

Now slope of the line

$=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{b-0}{0-a}=\frac{-b}{a}$

$\therefore$ Equation of the line 

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

$\Rightarrow y-0=\frac{-b}{a}(x-a)$...(i)

∵the line passes through the point (-3,8)

$\therefore 8-0=\frac{-b}{a}(-3-a)=-\frac{7-a}{a}(-3-a)$

$\Rightarrow  8 a=(7-a)(3+a)$

$\Rightarrow  8 a=21+7 a-3 a+a^{2} $

$\Rightarrow  a^{2}+8 a-7 a+3 a-21=0 $

$ \Rightarrow  a^{2}+4 a-21=0 $

$ \Rightarrow  a^{2}+7 a-3 a-21=0 $

$ \Rightarrow  a(a+7)-3(a+7)=0 $

$ \Rightarrow  (a+7)(a-3)=0 $

Either a+7=0, then a=-7, which is not possible as it is not positive

or a-3=0 , then a=3

and b=7-3=4

∴Equation of the line

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

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

$\Rightarrow 4 x+3 y=12$


Question 16

If the coordinates of the vertex A of a square ABCD are (3, – 2) and the quation of diagonal BD is 3 x – 7 y + 6 = 0, find the equation of the diagonal AC. Also find the co-ordinates of the centre of the square.

Sol :

Co-ordinates of A are (3, -2).







Diagonals AC and BD of the square ABCD
bisect each other at right angle at O.
∴ O is the mid-point of AC and BD Equation

Equation of BD is 3x-7y+6=0

$\Rightarrow 7 y=3 x+6 $

$\Rightarrow y=\frac{3}{7} x+\frac{6}{7}$

$\therefore$ Slope of $B D=\frac{3}{7}$

and slope of $\mathrm{AC}=-\frac{7}{3}$ $\left(\because m_{1} m_{2}=-1\right)$

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

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

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

$\Rightarrow  3 y+6=-7 x+21$

$\Rightarrow  7 x+3 y+6-21=0$

$\Rightarrow  7 x+3 y-15=0$

Now we will find the co-ordinates of O, the points of intersection of AC and BD We will solve the equations,

3x-7y=-6...(i)

7x+3y=15...(ii)

Multiplying (i) by 3 and (ii) by 7, we get,

9x-21 y=-18...(iii)

49x+21 y=105...(iv)

Adding , we get

58x=87 

$\Rightarrow x=\frac{87}{58}=\frac{3}{2}$

Substituting the value of x in (iii)

$9\left(\frac{3}{2}\right)-21 y=-18$

$ \Rightarrow \frac{27}{2}-21 y=-18$

$21 y=\frac{27}{2}+18=\frac{27+36}{2}=\frac{63}{2}$

$y=\frac{63}{2 \times 21}=\frac{3}{2}$

$\therefore$ Co-ordinates of O will be $\left(\frac{3}{2}, \frac{3}{2}\right)$

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