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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Exe-12.1 ,

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#### How to Solve Equation of Straight Line Problems/Questions / Exercise of ICSE Class-10 Mathematics

Before viewing Answer of Chapter-12 Equation of Straight Line of ML Aggarwal Solution. Read the Chapter Carefully and then solve all example given in  your text book.

For more practice on Equation of Straight Line related problems /Questions / Exercise try to solve Equation of Straight Line  exercise of other famous publications also such as Goyal Brothers Prakshan (RS Aggarwal ICSE)  / Concise Selina Publications ICSE  Mathematics. Get the formula of Equation of Straight Line for ICSE Class 10 Maths  to understand the topic more clearly in effective way.

#### 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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