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

 MCQs

Choose the correct answer from the given four options (1 to 13) :

Question 1

The slope of a line parallel to y-axis is

(a) 0

(b) 1

(c) – 1

(d) not defined

Sol :

Slope of a line parallel to y-axis is not defined. (b)


Question 2

The slope of a line which makes an angle of 30° with the positive direction of x-axis is

(a) 1

(b) $\frac{1}{\sqrt{3}}$

(c) $\sqrt{3}$

$(d)-\frac{1}{\sqrt{3}}$

Sol :

Slope of a line which makes an angle of 30°

with positive direction of x-axis = tan 30°

$=\frac{1}{\sqrt{3}}$

Ans (b)


Question 3

The slope of the line passing through the points (0, – 4) and ( – 6, 2) is

(a) 0

(b) 1

(c) – 1

(d) 6

Sol :

Slope of the line passing through the points (0, -4) and (-6, 2)

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2+4}{-6-0}=\frac{6}{-6}=-1$

Ans (c)


Question 4

The slope of the line passing through the points (3, – 2) and ( – 7, – 2) is

(a) 0

(b) 1

(c) $-\frac{1}{10}$

(d) not defined

Sol :
Slope of the line passing through the points (3, -2) and (-7, -2)
$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2+2}{-7-3}=\frac{0}{-10}=0$
Ans (a)


Question 5

The slope of the fine passing through the points (3, – 2) and (3, – 4) is

(a) – 2

(b) 0

(c) 1

(d) not defined

Sol :

The slope of the line passing through (3, -2) and (3, -4)

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-4+2}{3-3}=\frac{-2}{0}$
Ans (d)


Question 6

The inclination of the line y = √3x – 5 is

(a) 30°

(b) 60°

(c) 45°

(d) 0°

Sol :

The inclination of the line y = √3x – 5 is

√3 = tan 60° = 60° (b)


Question 7

If the slope of the line passing through the points (2, 5) and (k, 3) is 2, then the value of k is

(a) -2

(b) -1

(c) 1

(d) 2

Sol :

Slope of the line passing through the points (2, 5) and (k, 3) is 2, then

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \Rightarrow 2=\frac{3-5}{k-2} \Rightarrow 2=\frac{-2}{k-2}$
$\Rightarrow 2 k-4=-2 \Rightarrow 2 k=4-2=2$
$\Rightarrow k=\frac{2}{2}=1$
Ans (c)

Question 8

The slope of a line parallel to the line passing through the points (0, 6) and (7, 3) is

(a) $-\frac{1}{5}$
(b) $\frac{1}{5}$

(c) -5

(d) 5

Sol :

Slope of the line parallel to the line passing through (0, 6) and (7, 3)

Slope of the line $=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-6}{7-0}=\frac{-3}{7}$

Ans (b)


Question 9

The slope of a line perpendicular to the line passing through the points (2, 5) and ( – 3, 6) is

(a) $-\frac{1}{5}$
(b) $\frac{1}{5}$
(c) -5
(d) 5
Sol :
Slope of the line joining the points (2, 5), (-3, 6)

$=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{6-5}{-3-2}=\frac{1}{-5}$

$\therefore$ Slope of the line perpendicular to this lie =5
$\left(\frac{-1}{5} \times 5=-1\right)$
Ans (d)

Question 10

The slope of a line parallel to the line 2x + 3y – 7 = 0 is
(a) $-\frac{2}{3}$
(b) $\frac{2}{3}$
(c) $-\frac{3}{2}$
(d) $\frac{3}{2}$
Sol :
The slope of a line parallel to the line 2x + 3y – 7 = 0 
slope of the line
3y=-2 x+7 
$\Rightarrow y=\frac{-2}{3}+\frac{7}{3}=\frac{-2}{3}$

Ans (a)

Question 11

The slope of a line perpendicular to the line 3x = 4y + 11 is

(a) $\frac{3}{4}$
(b) $-\frac{3}{4}$
(c) $\frac{4}{3}$
(d) $-\frac{4}{3}$

Sol :
slope of a line perpendicular to the line 3x = 4y + 11 is

$\Rightarrow 4 y=3 x-11 \Rightarrow y=\frac{3}{4} x-\frac{11}{4}=\frac{3}{4}$

$\therefore$ Slope of the line perpendicular to this line

$=\frac{-4}{3}$ $(\because m \times n=-1)$
Ans (d)

Question 12

If the lines 2x + 3y = 5 and kx – 6y = 7 are parallel, then the value of k is
(a) 4
(b) – 4
(c) $\frac{1}{4}$
(d) $-\frac{1}{4}$
Sol :
lines 2x + 3y = 5 and kx – 6y = 7 are parallel
Slope of 2x + 3y = 5 = Slope of kx – 6y = 7
⇒ 3y – 2x + 5

Slope of 2x+3 y=5 is $\frac{-2}{3}$

and Slope of  kx-6 y=7
$ \Rightarrow 6 y=k x-7$

$\Rightarrow y=\frac{k}{6} x-\frac{7}{6}$

$\therefore$ Slope $=\frac{k}{6}$

Since both lines are parallel

$\therefore \frac{-2}{3}=\frac{k}{6} \Rightarrow k=\frac{-2 \times 6}{3}=-4$

Ans (b)


Question 13

If the line 3x – 4y + 7 = 0 and 2x + ky + 5 = 0 are perpendicular to each other, then the value of k is

(a) $\frac{3}{2}$
(b) $-\frac{3}{2}$
(c) $\frac{2}{3}$
(d) $-\frac{2}{3}$

Sol :
line 3x – 4y + 7 = 0 and 2x + ky + 5 = 0
are perpendicular to each other

$\therefore$ Product of their slopes $=\left(m_{1} \times m_{2}\right)=-1$

Slope of 3x-4y+7=0 
$\Rightarrow 4 y=3 x+7$

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

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

and slope of 2x+ky+5=0

ky=-2x-5
$y=\frac{-2}{k} x-\frac{5}{k}$

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

Since the given lines are perpendicular to each other

$\therefore \frac{3}{4} \times \frac{-2}{k}=-1$
$ \Rightarrow \frac{-6}{4 k}=-1$
$-k=\frac{-6}{4} $
$\Rightarrow k=\frac{3}{2}$

Ans (a)

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