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) 13

(c) 3

(d)13

Sol :

Slope of a line which makes an angle of 30°

with positive direction of x-axis = tan 30°

=13

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)

y2y1x2x1=2+460=66=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) 110

(d) not defined

Sol :
Slope of the line passing through the points (3, -2) and (-7, -2)
y2y1x2x1=2+273=010=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)

y2y1x2x1=4+233=20
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=y2y1x2x12=35k22=2k2
2k4=22k=42=2
k=22=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) 15
(b) 15

(c) -5

(d) 5

Sol :

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

Slope of the line =y2y1x2x1=3670=37

Ans (b)


Question 9

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

(a) 15
(b) 15
(c) -5
(d) 5
Sol :
Slope of the line joining the points (2, 5), (-3, 6)

=y2y1x2x1=6532=15

Slope of the line perpendicular to this lie =5
(15×5=1)
Ans (d)

Question 10

The slope of a line parallel to the line 2x + 3y – 7 = 0 is
(a) 23
(b) 23
(c) 32
(d) 32
Sol :
The slope of a line parallel to the line 2x + 3y – 7 = 0 
slope of the line
3y=-2 x+7 
y=23+73=23

Ans (a)

Question 11

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

(a) 34
(b) 34
(c) 43
(d) 43

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

4y=3x11y=34x114=34

Slope of the line perpendicular to this line

=43 (m×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) 14
(d) 14
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 23

and Slope of  kx-6 y=7
6y=kx7

y=k6x76

Slope =k6

Since both lines are parallel

23=k6k=2×63=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) 32
(b) 32
(c) 23
(d) 23

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

Product of their slopes =(m1×m2)=1

Slope of 3x-4y+7=0 
4y=3x+7

y=34x+74

Slope (m1)=34

and slope of 2x+ky+5=0

ky=-2x-5
y=2kx5k

Slope (m2)=2k

Since the given lines are perpendicular to each other

34×2k=1
64k=1
k=64
k=32

Ans (a)

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