Selina Concise Class 9 Maths Chapter 26 Co-ordinate Geometry Exercise 26B Solutions

Selina Concise Class 9 Maths Chapter 26 Co-ordinate Geometry Exercise 26B Solutions

Selina Concise Class 9 Maths Chapter 26 Co-ordinate Geometry 26B Solutions

EXERCISE – 26B

(Q1) Draw the graph for each linear equation given below.

(i) x =  3

Solution:

X 3 3 3
y 4 0 -3

x = 3 is parallel to y-axis.

(ii) x + 3 = 0

Solution:

x + 3 = 0

x = -3

X -3 -3 -3
y 0 4 -4

(iii) x – 5 = 0

Solution:

x – 5 = 0

x = 5

X 5 5 5
y

(iv) 2x – 7 = 0

Solution:

2x – 7 = 0

2x = 7

x = 7/2

x = 3.5

X 3.5 3.5 3.5
y 0 4 -4

(v) y = 4

Solution:

X 0 -4 5
y 4 4 4

 y = 4 is parallel to x-axis.

(vi) y + 6 = 0 

Solution:

y + 6 = 0

y = -6

X 0 -5 4
y -6 -6 -6

y = -6 is parallel to x-axis.

(vii) y – 2 = 0

Solution:

y – 2 = 0

y = 2

X 0 -3 4
y 2 2 2

(viii) 3x + 5 = 0

Solution:

3x + 5 = 0

3x = -5

x = -5/3

x = – 1.6

X -1.6 -1.6 -1.6
y 0 3 -4

(ix) 2y – 5 = 0

Solution:

2y – 5 = 0

2y =

y = 5/2

y = 2.5

X 0 -4 4
y 2.5 2.5 2.5

(x) y = 0

Solution:

X 0 -2 3
y 0 0 0

(xi) x = 0

Solution:

X 0 0 0
y 0 3 -3

 

(Q2) Draw the graph for each linear equation given below:

(i) y = 3x

Solution:

y = 3x

Put x = 0

y = 3 × 0

y = 0

∴ (x, y) = (0, 0)

Put x = 1

y = 3 (1)

y = 3

Put x = -2

y = 3 (-2)

y = -6

X 0 1 -2
y 0 3 -6

(ii) y = -x

Solution:

y = -x

Put = -1

y = -1

Put x = 2

y = -2

Put x = -3

y = +3

X 1 2 -3
y -1 -2 3

(iii) y = – 2x 

Solution:

y = -2x

Put x = 1

y = (-2) × 1

y = -2

Put x = -1

y = (-2) × (-1)

y = 2

Put x = 3

y = (-2) × 3

y = -6

X 1 -1 3
y -2 2 -6

(iv) y = x

Solution:

X 1 -2 4
y 1 -2 4

(v) 5x + y = 0

Solution:

5x + y = 0

y = – 5x

Put x = 1

y = (-5) × 1

y = -5

Put x = -2

y = -5 (-2)

y = 10

Put x = 0

y = 0

X 1 -2 0
y -5 10 0

(vi) x + 2y = 0

Solution:

x + 2y = 0

x = -2y

Put y = 0

x = 0

Put y = 1

x = -2

Put y = 3

x = (-2) × 3

x = -6

X 0 -2 -6
y 0 1 3

(vii) 4x – y = 0

Solution:

4x – y = 0

4x = y

y = 4x

Put x = 0

Put x = 1

y = 4 ×1

y = 4

Put x = -2

y = 4 ×(-2)

y = -8

X 0 1 -2
y 4 4 -8

(viii) 3x + 2y = 0

Solution:

3x + 2y = 0

3x = -2y

x = -2y/3

Put y = 0

x = 0

Put y = 3

x = -2/3 × 3

x = -2

Put y = 6

x = -2/3 × 6

x = -2 ×2

x = -4

X 0 -2 -4
y 0 3 6

(ix) x = -2y

Solution:

x = -2y

Put y = 1

x = 0

Put y = -2

x = (-2) × (-2)

x = 4

X -2 0 4
y 1 0 -2

(Q3) Draw the graph for each linear equation given below.

(i) y = 2x + 3

Solution:

y = 2x + 3

Put x = 1

y = 2 ×1 + 3

y = 5

Put x = 0

y = 2 (0) + 3)

y = 3

Put x = 2

y = 4 + 3

y = 7

X 1 0 2
y 5 3 7

(ii) y = 2x/3 – 1

Solution:

y = 2x/3 – 1

Put x = 0

y = 2/3 (0) – 1

y = -1

Put x = 3

y = 2/3 × 3 – 1

y = 2-1

y = 1

Put x = 6

y = 2/3 × 6 – 1

y = 2 ×2 – 1

y = 4-1

y = 3

X 0 3 6
y -1 1 3

 

(iii) y = -x + 4

Solution:

y = -x + 4

Put x = 0

y = 4

Put x = 1

y = -1 + 4

y = 3

Put x = -2

y = (-2) + 4 = 2+4

y = 6

X 0 1 -2
y 4 3 6

(iv) y = 4x – 5/2

Solution:

y = 4x – 5/2

Put x = 0

y = -5/2

y = -2.5

Put x = 1

y = 4 – 5/2

y = 8-5/2

= 3/2

y = 1.5

Put x = 2

y = 4 × 2 – 5/2

y = 8-5/2 = 16-5/2

y = 11/2

y = 5.5

X 0 1 2
y -2.5 1.5 5.5

(v) y = 3x/2 + 2/3

Solution:

y = 3x/2 + 2/3

Put x = 0

y = 2/3

y = 0.6

Put x = 1

y = 3/2 + 2/3

y = 9+4/6

y = 13/6

y = 2.1

Put x = 2

y = 3/2 × 2 + 2/3

y = 3 + 2/3

y = 9+2/3

y = 11/3

y = 3.6

X 0 1 2
y 0.6 2.1 3.6

(vi) 2x – 3y = 4

Solution:

2x – 3y = 4

2x = 4 + 3y

x = 4+3y/2

x = 4/2 + 3/2y

x = 2 + 3/2 y

Put y = 1

x = 2 + 3/2

x = 4 + 3/2

x = 7/2

x = 3.5

Put y = 0

x = 2

Put y = -1

x = 2 + 3/2 (-1)

= 2 – 3/2

x = 4-3/2

x = 1/2

x = 0.5

X 3.5 2 0.5
y 1 0 -1

(viii) x – 3 = 2/3 (y + 1)

Solution:

x – 3 = 2/3 (y + 1)

3 (x – 3) = 2 (y + 1)

3x – 9 = 2y + 2

3x – 2y = 2+9

3x – 2y = 11

3x = 11+2y

x = 11+2y/3

x = 11/3 + 2/3 y

Put y = 0

x = 11/3

x = 3.6

Put y = 1

x = 11/3 + 2/3

= 11+2/3

x = 13/3

x = 4.3

Put y = -1

x = 11/3 – 2/3

= 11-2/3

= 9/3

x = 3

X 3.6 4.3 3
y 0 1 -1

(ix) x + 5y + 2 = 0

x + 5y + 2 = 0

x = -5y – 2

Put y = (-1)

x = -5 (-1) – 2

= 5 – 2

x = 3

Put y = 1

x = -5-2

x = -7

X -2 3 -7
y 0 -1 1

(Q4) Draw the graph for each equation given below:

(i) 3x + 2y = 6

(ii) 2x – 5y = 10

(iii) 1/2 x + 2/3 y = 5

(iv) 2x-1/3 – y-2/5 = 0

In each case, find the co-ordinates of the points where the graph (line) drawn meets the co-ordinates axes.

(i) 3x + 2y = 6

Solution:

3x = 6 – 2y

Put y = 0

x = 6-2y/3

x = 2 – 2/3 y

x = 2

Put y = 1

x = 2 – 2/3

x = 6-2/3

x = 4/3

x = 1.3

Put y = -1

x = 2 + 2/3

x = 8/3

x = 2.6

The co-ordinates of the points which meets the co-ordinates axes is –

(0, 3) and (2, 0)

X 2 1.3 2.6 0
y 0 1 -1 3

(Q4) (ii) 2x – 5y = 10

Solution:

2x = 10 + 5y

x = 10/2 + 5/2 y

x = 5 + 5/2 y

x – 5 = 5/2 y

2 (x-5)/5 = y

2x – 10/5 = y

2x/5 – 2 = y

Put x = -1

-2/5 – 2 = y

-2-10/5 = y

-12/5 = y

-2.4 = y

Put x = 0

y = -2

Put x = 1

y = 2/5 (1) – 2

y = 2/5 – 2

y = 2 – 10/5

y = -8/5

y = 1.6

X -1 0 1
y -2.4 -2 -1.6

The co-ordinates of the points which meets the co-ordinates axes is (0, -2), and (5, 0)

 

(iii) 1/2 x + 2/3 y = 5

Solution:

1/2 x + 2/3 y = 5

2/3 y = 5 – 1/2 x

Multiply by 6 on both sides,

6×2/3 y = 6 × 5 – 6 × 1/2 x

2 × 2y = 30-3x

4y = 30-3x

y = 30-3x/4

y = 30/4 – 3/4 x

y = 15/2 – 3/4 x

Put x = 4

y = 15/2 – 3

y = 15-6/2

= 9/2

y = 4.5

Put x = 0

y = 15/2

y = 7.5

Put x = (-1)

Put = 15/2 + 3/4

y = 7.5 + 0.75

y = 8.25

X 4 0 1==-1
y 4.5 7.5 8.25

The co-ordinates of the points which are meet the co-ordinates at x-axis is (10, 0) and on-axis is (0, 7.5)

(iv) 2x-1/3 – y-2/5 = 

Solution:

2x-1/3 – y-2/5 = 0

(2x-1)/3 = (y-2)/5

5 (2x – 1) = 3 (y – 2)

10x – 5 = 3y – 6

10x = 3y – 6 + 5

10x – 3y = -1

– 3y = -1 – 10x

y = -1-10x/-3

y = (-1) (1 + 10x)/-3

y = 1 + 10x/3

y = 1/3 + 10/3 x

Put x = 0

y = 1/3

y = 0.3

Put x = 3

y = 1/3 + 10

y = 1+30/3

y = 31/3 = y = 10.3

Put x = -1

y = 1/3 + 10/3 (-1)

y = 1/3 – 10/3

y = 1/3 – 10/3

y = 1-10/3

y = -9/3

y = -3

X 0 3 -1
y 0.3 10.3 -3

The co-ordinates meets at x-axis is (-0.1,0) and on y-axis is (0, 0.3)

(Q5) For each linear equation, given below, draw the graph and then use the graph drawn (in each case) to find the area of a triangle enclosed by the graph and the co-ordinates axes:

(i) 3x – (5 – y) = 7

(ii) 7 – 3 (1 – y) = -5 + 2x

solution:

(i) 3x – (5 – y) = 7

3x – 5 + y = 7

3x + y = 7 + 5

3x + y = 12

y = 12 – 3x

Put x = 0

y = 12

Put x = 4

y = 12 – 12

y = 0

∴ Area of △AOB = 1/2 × Base × height

= 1/2 × 4 × 12

= 4 × 6

= 24 sq. units

(ii) 7 – 3 (1 – y) = – 5 + 2x

Solution:

7 – 3 (1 – y) = -5 + 2x

7 – 3 + 3y = – 5 + 2x

4 + 3y = – 5 + 2x

3y = – 5 + 2x

3y = -9 + 2x

y = -9+2x/3

y = -9/3 + 2/3 x

y = -3 + 2/3 x

Put x = 0, y = -3

Put x = 3

y = -3 + 2

y = -1

Put x = +9/2

y = -3 + 2/3 (+9/2)

= -3 + 3

y = 0

Area of △AOB

1/2 × Base × Height

= 1/2 × AO × OB

= 1/2 × 4.5 × 3

= 3/2 × 4.5

= 1.5 × 4.5

= 675 sq.units

 

Here is your solution of Selina Concise Class 9 Maths Chapter 26 Co-ordinate Geometry Exercise 26B

Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.

 

For more solutions, See below ⇓

Leave a Reply

Your email address will not be published. Required fields are marked *