MCQ Equation of Straight Line ML Aggarwal Solutions ICSE Class-10 Maths Chapter-12. We Provide Step by Step Answer of Exercise-12.1 ,  Exercise-12.2 , Equation of Straight Line , with MCQs and Chapter-Test Questions  / Problems related  for ICSE Class-10 APC Understanding Mathematics  .  Visit official Website CISCE  for detail information about ICSE Board Class-10.

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 10th Chapter-12 Equation of Straight Line  (MCQ) Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-12.1, Exe-12.2, MCQ and Chapter Test Questions Academic Session 2021-2022

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

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MCQ of Equation of Straight Line of ML Aggarwal Solutions for ICSE Class-10

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

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

Slope of a line which makes an angle of 30°
with positive direction of x-axis = tan 30°
$\frac { 1 }{ \sqrt { 3 } }$ (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

Slope of the line passing through the points (0, -4) and (-6, 2)
$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 2+4 }{ -6-0 } =\frac { 6 }{ -6 } =-1$ (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

Slope of the line passing through the points (3, -2) and (-7, -2)
$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { -2+2 }{ -7-3 } =\frac { 0 }{ -10 } =0$ (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

The slope of the line passing through (3, -2) and (3, -4)
$\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { -4+2 }{ 3-3 } =\frac { -2 }{ 0 }$ (d)

Question -6

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

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

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) $- \frac { 1 }{ 5 }$
(b) $\\ \frac { 1 }{ 5 }$
(c) -5
(d) 5

Slope of the line parallel to the line passing through (0, 6) and (7, 3)
Slope of the line = $\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 3-6 }{ 7-0 } =\frac { -3 }{ 7 }$ (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

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

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

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) $- \frac { 4 }{ 3 }$

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) $\\ \frac { 1 }{ 4 }$
(d) $- \frac { 1 }{ 4 }$

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

(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 of Equation of Straight Line Solutions :–

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