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

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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Exe-12.1 , 

  Exe-12.2 ,  

 MCQS ,

Chapter Test


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.


Exercise 12.2 , Equation of Straight Line ML Aggarwal ICSE Class-10 

page-245

Question- 1

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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.

Answer- 1

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.

Answer -2

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

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-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. 

Answer- 3

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)

Answer-4

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

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ML Aggarwal Class-10 Solutions Chapter-12 equation of straight line exe 12.2 Ans-4
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.

Answer -5

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). 

Answer -6

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)

Answer -7

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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