Chapter-Test 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  (Chapter Test) Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-12.1, Exe-12.2, MCQ and Chapter Test Questions Academic Session 2021-2022

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

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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.

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#### Question- 1

Find the equation of a line whose inclination is 60° and y-intercept is – 4.

Angle of inclination = 60°
Slope = tan θ = tan 60° = √3
Equation of the line will be,
y = mx + c = √3x + ( – 4)
⇒ y – √3x – 4

#### Question -2

Write down the gradient and the intercept on the y-axis of the line 3y + 2x = 12.

Slope of the line 3y + 2x = 12
⇒ 3y = 12 – 2x
⇒ 3y = -2x + 12

y = (-2/3) x + 12/3

y = (-2/3) x + 4

Hence, gradient = -2/3 and the intercept on the y-axis is 4.

#### Question -3

If the equation of a line is y – √3x + 1, find its inclination.

In the line
y = √3 x + 1
Slope = √3
⇒ tan θ = √3
⇒ θ = 60° (∵ tan 60° = √3)

#### Question -4

If the line y = mx + c passes through the points (2, – 4) and ( – 3, 1), determine the values of m and c.

The equation of line y = mx + c
∵ it passes through (2, – 4) and ( – 3, 1)
Now substituting the value of these points -4 = 2m + c …(i)
and 1 = -3m + c …(ii)
Subtracting we get,

-5 = 5m

⇒ m = -5/5 = -1

Substituting the value of m in (i)

-4 = 2(-1) + c

⇒ -4 = -2 + c

c = -4 + 2 = -2

∴ m = -1, c = -2

#### Question -5  Chapter-Test Equation of Straight Line ML

If the point (1, 4), (3, – 2) and (p, – 5) lie on a st. line, find the value of p.

Let the points to be A (1, 4), B (3, -2) and C (p, -5) are collinear and let B (3, -2)
divides AC in the ratio of m1 : m2

∴ x = (m1x2 + m2x1)/(m+ m2)

⇒ 3 = (m1p + m2×1)/(m+ m2)

3m+ 3m2 = m1p + m2

⇒ 3m1 – m1p = m2 – 3m2

⇒ m1(3 – p) = -2 m2

⇒ m1/m2 = – 2/(3 – p) …(i)

and –2 = {m1(-5) + m2×4}/(m1 + m2)

⇒ -2m– 2m= – 5m1 + 4m2

⇒ -2m1 + 5m1 = 4 m2 + 2m2

⇒ 3m1 = 6 m2

⇒ m1/m2 = 6/3 = 2 …(ii)

From (i) and (ii)

-2/(3–p) = 2

⇒ -2 = 6 – 2p

⇒ 2p = 6 + 2 = 8

⇒ p = 8/2 = 4

#### Question- 6

Find the inclination of the line joining the points P (4, 0) and Q (7, 3).

Slope of the line joining the points P (4, 0) and Q (7, 3)

= (y2 – y1)/(x2 – x1) = (3–0)/(7–4) = 3/3 = 1

∴ tan θ = 1

⇒ θ = 45° (∵ tan 45° = 1)

Hence inclination of line = 45°

#### Question -7

Find the equation of the line passing through the point of intersection of the lines 2x + y = 5 and x – 2y = 5 and having y-intercept equal to $- \frac { 3 }{ 7 }$

Equation of lines are
2x + y = 5 …(i)
x – 2y = 5 …(ii)
Multiply (i) by 2 and (ii) by 1, we get
4x + 2y = 10
x – 2y = 5

5x = 15

⇒ x = 15/5 = 3

Substituting the values of x in (i)

2×3 + y = 5

⇒ 6 + y = 5

⇒ y = 5 – 6

= -1

∴ Co-ordinates of point of intersection are (3, -1)

∵ the line passes through (3, -1)

∴ -1 = m×3 – 3/7 (y = mx + c)

3m = -1 + 3/7 = -4/7

m = -4/(7×3)

= -4/21

∴ Equation of line y = -4/21 ×x – 3/7

⇒ 21y = -4x – 9

⇒ 4x + 21y + 9 = 0

#### Question -8  Chapter-Test Equation of Straight Line ML

If the lines  x/3 + y/4 = 7 and 3x + ky = 11 are perpendicular to each other, find the value of k.

Given Equation of lines are

x/3 + y/4 = 7

⇒ 4x + 3y = 84

⇒ 3y = -4x + 84

⇒ y = -4/3 ×x + 28 …(i)

And 3x + ky = 11

⇒ ky = -3x + 11

⇒ y = (-3/k)×x + 11/k …(ii)

Let slope of line (i) be m1 and of (ii) be m2

∴ m1 = -4/3 and m2 = -(3/k)

∵ These lines are perpendicular to each other

∴ m1m2 = – 1

⇒ -4/3 × (-3/k) = -1

⇒ 4/k = -1

⇒ – k = 4

⇒ k = – 4

#### Question -9

Write down the equation of a line parallel to x – 2y + 8 = 0 and passing through the point (1, 2).

The equation of the line is x – 2y + 8 = 0
⇒ 2y = x + 8

⇒ y = (1/2)×x + 4

∴ Slope of the line = 1/2

∴ Slope of the line parallel to the given line passing through (1, 2) = 1/2

∴ Equation of the lines will be,

y – y1 = m(x – x1)

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