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

## Exe-12.2 Equation of Straight Line ML Aggarwal Solutions ICSE Class-10 Maths

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#### Question- 1

State which one of the following is true : The straight lines y = 3x – 5 and 2y = 4x + 7 are
(i) parallel
(ii) perpendicular
(iii) neither parallel nor perpendicular.

Slope of line y = 3x – 5 = 3
and slope of line 2y = 4x + 7
⇒ y = 2x + 7/2 = 2.
∴ Slope of both the lines are neither equal nor their product is – 1.
∴ These line are neither parallel nor perpendicular.

#### Question- 2  Exe-12.2 Equation of Straight Line ML

If 6x + 5y – 7 = 0 and 2px + 5y + 1 = 0 are parallel lines, find the value of p.

In equation
6x + 5 y – 7 = 0
⇒ 5y = -6x + 7

y = (-6/5) x + 7/5

So, the slope of the line (m1) = -6/5

Again, in equation 2px + 5y + 1 = 0

5y = -2px – 1

y = (-2p/5) x – 1/5

So, the slope of the line (m2) = -2p/5

For these two lines to be parallel

m1 = m2

-6/5 = -2p/5

p = (-6/5) x (-5/2)

Thus, p = 3

#### Question -3

Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel. Find the relation connecting a and b.

In equation 2x – by + 5 = 0

and ax + 3y = 2

If two lines to be parallel then their slopes must be equal.

In equation 2x – by + 5 = 0,

by = 2x + 5

y = (2/b) x + 5/b

So, the slope of the line (m1) = 2/b

And in equation ax + 3y = 2,

3y = -ax + 2

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

So, the slope of the line (m2) = (-a/3)

As the lines are parallel

m1 = m2

2/b = -a/3

6 = -ab

Hence, the relation connecting a and b is ab + 6 = 0

#### Question -4

If the straight lines 3x – 5y = 7 and 4x + ay + 9 = O are perpendicular to one another, find the value of a (2018)

Given lines are
3x – 5y = 1 ……….(i) and 4x + ay + 9 = 0  …………(ii)
Slope of line (i) (m1) =  (3/5)=3/5
Slope of line (ii) (m2) = (4/𝑎)

Also, given that two lines are perpendicular to one and another
∴ (m1) (m2) = – 1

Hence, the value of a = 12/5 .

#### Question- 5  Exe-12.2 Equation of Straight Line ML

If the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other, find the relation connecting a and b.

Given
In the equation 3x + by + 5 = 0

and ax – 5y + 7 = 0 are perpendicular to each other

Then the product of their slopes must be -1.

Slope of line 3x + by + 5 = 0 is,

by = -3x – 5

y = (-3/b) – 5/b

So, slope (m1) = -3/b

And,

The slope of line ax – 5y + 7 = 0 is

5y = ax + 7

y = (a/5) x + 7/5

So, slope (m2) = a/5

As the lines are perpendicular, we have

m1 x m2 = -1

-3/b x a/5 = -1

-3a/5b = -1

-3a = – 5b

3a = 5b

Hence, the relation connecting a and b is 3a = 5b.

#### Question-6

Is the line through ( – 2, 3) and (4, 1) perpendicular to the line 3x = y + 1 ?
Does the line 3x = y + 1 bisect the join of ( – 2, 3) and (4, 1).

Slope of the line passing through the points
(-2, 3) and (4, 1)

m= y2 – y1/ x2 – x1

= (1 – 3)/ (4 + 2)

= -2/6

= -1/3

And, the slope of the line: 3x = y + 1

y = 3x -1

Slope (m2) = 3

Now,

m1 x m= -1/3 x 3 = -1

Thus, the lines are perpendicular to each other as the product of their slopes is -1.

Now,

Co-ordinates of the mid-point of the line joining the points (-2, 3) and (4, 1) is

([-2 + 4]/2, [3 + 1]/2) = (1, 2)

Now, if the line 3x = y + 1 passes through the mid-point then it will satisfy the equation

3(1) = (2) + 1

3 = 3

Hence, the line 3x = y + 1 bisects the line joining the points (– 2, 3) and (4, 1).

#### Question -7

The line through A ( – 2, 3) and B (4, b) is perpendicular to the line 2x – 4y = 5. Find the value of b. (2012)

Gradient (m1) of the line passing through the
points A (-2, 3) and B (4, b)

m1 = (b – 3)/ (4 + 2) = (b – 3)/ 6

Now, the gradient of the given line 2x – 4y = 5 is

4y = 2x + 5

y = (2/4) x + 5/4

y = ½ x + 5/4

So, m2 = ½

As the line are perpendicular to each other, we have

m1 x m2 = -1

(b – 3)/ 6 × ½ = -1

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