ML Aggarwal Equation of Straight Line MCQs Class 10 ICSE Maths Solutions Ch-12

ML Aggarwal Equation of Straight Line MCQs Class 10 ICSE Maths Solutions Ch-12. We Provide Step by Step Answer of MCQs Questions for Ch-12 Equation of Straight Line as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

ML Aggarwal Equation of Straight Line MCQs Class 10 ICSE Maths Solutions Ch-12

Board ICSE
Publications Avichal Publishig Company (APC)
Subject Maths
Class 10th
Chapter-12 Equation of Straight Line
Writer ML Aggarwal
Book Name Understanding
Topics Solution of MCQs Questions
Edition 2022-2023

Equation of Straight Line MCQs

Class 10 ICSE Maths ML Aggarwal Solutions Ch-12

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

Page-246

Question- 1

The slope of a line parallel to y-axis is
(a) 0
(b) 1
(c) – 1
(d) not defined

Answer -1

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

Page-247

Question -2

The slope of a line which makes an angle of 30° with the positive direction of x-axis is
(a) 1
(b) 1/√3
(c) √3
(d) -1/√3

Answer -2

Slope of a line which makes an angle of 30°
with positive direction of x-axis = tan 30°
= 1/√3 (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

Answer -3

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

Question- 4

The slope of the line passing through the points (3, – 2) and ( – 7, – 2) is
(a) 0
(b) 1
(c) -1/10
(d) not defined

Answer- 4

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

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

Answer- 5

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

=(-4+2)/(3-3)

=-2/0

(d)


Equation of Straight Line MCQs

Class 10 ICSE Maths ML Aggarwal Solutions Ch-12

Question -6

The inclination of the line y = √3x – 5 is
(a) 30°
(b) 60°
(c) 45°
(d) 0°

Answer -6

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

Answer- 7

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

m = (y2 – y1)/(x2 – x1)

⇒ 2 = (3 – 5)/(k – 2)

⇒ 2 = (-2)/(k – 2)

⇒ 2k – 4 = -2

⇒ 2k = 4 – 2

= 2

⇒ k = 2/2 = 1

Question- 8

The slope of a line parallel to the line passing through the points (0, 6) and (7, 3) is
(a) -1/5
(b) 1/5
(c) -5
(d) 5

Answer -8

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

=(3-6)/(7-0)

=-3/7

(b)

Question -9

The slope of a line perpendicular to the line passing through the points (2, 5) and ( – 3, 6) is
(a) -1/5
(b) 1/5
(c) -5
(d) 5

Answer -9

Slope of the line joining the points (2, 5), (-3, 6)

= (y– y1)/(x– x1)

= (6 – 5)/(-3 – 2)

= 1/-5 = -1/5

∴ Slope of the line perpendicular to this line = 5

Question- 10

The slope of a line parallel to the line 2x + 3y – 7 = 0 is

(a) -2/3
(b) 2/3
(c) -3/2
(d) 3/2

Answer -10

The slope of a line parallel to the line 2x + 3y – 7 = 0
slope of the line

3y = – 2x + 7

⇒ y = -2/3 + 7/3

= -2/3


Class 10 ICSE Maths ML Aggarwal Solutions Ch-12

Equation of Straight Line MCQs

Question -11

The slope of a line perpendicular to the line 3x = 4y + 11 is
(a) 3/4
(b) -3/4
(c) 4/3
(d) -4/3

Answer -11

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

⇒ 4y = 3x – 11

⇒ y = 3/4 x – 11/4

Slope = 3/4

∴ Slope of the line perpendicular to this line = -4/3 (∵ m × n = -1)

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) 1/4
(d) -1/4

Answer -12

lines 2x + 3y = 5 and kx – 6y = 7 are parallel
Slope of 2x + 3y = 5 = Slope of kx – 6y = 7
⇒ 3y – 2x + 5

⇒ y = -2/3 ×x + 5/3

Slope of 2x + 3y = 5 is -2/3

And slope of kx – 6y = 7

6y = kx – 7

⇒ y = (k/6)×x – 7/6

∴ Slope = k/6

Since both lines are parallel

∴ -2/3 = k/6

⇒ k = (-2 × 6)/3

= -4

so option (b) is correct

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) 3/2
(b) -3/2
(c) 2/3
(d) -2/3

Answer-13

(a) 3/2

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 = 3/4 x + 7/4

Slope (m1) = 3/4

And slope of 2x + ky + 5 = 0

ky = -2x – 5

y = (-2/k) ×x – 5/k

∴ Slope (m2) = -2/k

Since the given lines are perpendicular to each other

∴ 3/4 × -2/k = -1

⇒ -6/4k = -1

⇒ -k = -6/4

⇒ k = 3/2

— : End ML Aggarwal Equation of Straight Line MCQs Class 10 ICSE Maths Ch-12 : –

Return to :- ML Aggarwal Solutions for ICSE Class-10

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