SELINA Solution Class 9 Chapter 26 Co-ordinate Geometry Exercise 26B

Question 1.01

Draw the graph for the linear equation given below:
x = 3

Sol:

Since x = 3, therefore the value of y can be taken as any real no.
First prepare a table as follows:

x	3	3	3
y	-1	0	1

Thus the graph can be drawn as follows:

Question 1.02

Draw the graph for the linear equation given below:
x + 3 = 0

Sol:

First prepare a table as follows:

x	- 3	- 3	- 3
y	- 1	0	1

Thus the graph can be drawn as follows:

Question 1.03

Draw the graph for the linear equation given below:
x - 5 = 0

Sol:

First, prepare a table as follows:

x	5	5	5
y	-1	0	1

Thus the graph can be drawn as follows:

Question 1.04

Draw the graph for the linear equation given below:
2x - 7 = 0

Sol:

The equation can be written as:
x = ${\displaystyle \frac{7}{2}}$
First prepare a table as follows:

x	${\displaystyle \frac{7}{2}}$	${\displaystyle \frac{7}{2}}$	${\displaystyle \frac{7}{2}}$
y	-1	0	1

Thus the graph can be drawn as follows:

Question 1.05

Draw the graph for the linear equation given below:
y = 4

Sol:

First, prepare a table as follows:

x	- 1	0	1
y	4	4	4

Thus the graph can be drawn as follows:

Question 1.06

Draw the graph for the linear equation given below:
y + 6 = 0

Sol:

First, prepare a table as follows:

x	- 1	0	1
y	- 6	- 6	- 6

Thus the graph can be drawn as follows:

Question 1.07

Draw the graph for the linear equation given below:
y - 2 = 0

Sol:

First, prepare a table as follows:

x	- 1	0	1
y	2	2	2

Thus the graph can be drawn as follows:

Question 1.08

Draw the graph for the linear equation given below:
3y + 5 = 0

Sol:

First prepare a table as follows:

x	- 1	0	1
y	- 6	- 6	- 6

Thus the graph can be drawn as follows:

Question 1.09

Draw the graph for the linear equation given below:
2y - 5 = 0

Sol:

First prepare a table as follows:

x	-1	0	1
y	${\displaystyle \frac{5}{2}}$	${\displaystyle \frac{5}{2}}$	${\displaystyle \frac{5}{2}}$

Thus the graph can be drawn as follows:

Question 1.1

Draw the graph for the linear equation given below:
y = 0

Sol:

First prepare a table as follows:

x	-1	0	1
y	0	0	0

Thus the graph can be drawn as follows:

Question 1.11

Draw the graph for the linear equation given below:
x = 0

Sol:

First prepare a table as follows:

x	0	0	0
y	-1	0	1

Thus the graph can be drawn as follows:

Question 2.1

Draw the graph for the linear equation given below:
y = 3x

Sol:

First, prepare a table as follows:

x	- 1	0	1
y	- 3	0	3

Thus the graph can be drawn as follows:

Question 2.2

Draw the graph for the linear equation given below:
y = - x

Sol:

First prepare a table as follows:

x	- 1	0	1
y	1	0	- 1

Thus the graph can be drawn as follows:

Question 2.3

Draw the graph for the linear equation given below:
y = - 2x

Sol:

First prepare a table as follows:

x	- 1	0	1
y	2	0	- 2

Thus the graph can be drawn as follows:

Question 2.4

Draw the graph for the linear equation given below:
y = x

Sol:

First, prepare a table as follows:

x	- 1	0	1
y	- 1	0	1

Thus the graph can be drawn as follows:

Question 2.5

Draw the graph for the linear equation given below:
5x+ y = 0.

Sol:

First, prepare a table as follows:

x	- 1	0	1
y	5	0	- 5

Thus the graph can be drawn as follows:

Question 2.6

Draw the graph for the linear equation given below:
x + 2y = 0

Sol:

First prepare a table as follows:

x	-1	0	1
y	${\displaystyle \frac{1}{2}}$	0	-${\displaystyle \frac{1}{2}}$

Thus the graph can be drawn as follows:

Question 2.7

Draw the graph for the linear equation given below:
4x - y = 0

Sol:

First, prepare a table as follows:

x	- 1	0	1
y	- 4	0	4

Thus the graph can be drawn as follows:

Question 2.8

Draw the graph for the linear equation given below:
3x + 2y = 0

Sol:

First prepare a table as follows:

x	-1	0	1
y	${\displaystyle \frac{3}{2}}$	0	-${\displaystyle \frac{3}{2}}$

Thus the graph can be drawn as follows:

Question 2.9

Draw the graph for the linear equation given below:
x = - 2y

Sol:

First prepare a table as follows:

x	-1	0	1
y	${\displaystyle \frac{1}{2}}$	0	-${\displaystyle \frac{1}{2}}$

Thus the graph can be drawn as follows:

Question 3.1

Draw the graph for the linear equation given below:
y = 2x + 3

Sol:

First, prepare a table as follows:

x	-1	0	1
y	${\displaystyle -\frac{5}{3}}$	3	5

Thus the graph can be drawn as follows:

Question 3.2

Draw the graph for the linear equation given below:
y = ${\displaystyle \frac{2x}{3}-1}$

Sol:

First prepare a table as follows:

x	-1	0	1
y	${\displaystyle -\frac{5}{3}}$	-1	${\displaystyle -\frac{1}{3}}$

Thus the graph can be drawn as follows:

Question 3.3

Draw the graph for the linear equation given below:
y = - x + 4

Sol:

First, prepare a table as follows:

x	-1	0	1
y	5	4	3

Thus the graph can be drawn as follows:

Question 3.4

Draw the graph for the linear equation given below:
y = ${\displaystyle 4x-\frac{5}{2}}$

Sol:

First prepare a table as follows:

x	-1	0	1
y	${\displaystyle -\frac{13}{2}}$	${\displaystyle -\frac{5}{2}}$	${\displaystyle \frac{3}{2}}$

Thus the graph can be drawn as follows:

Question 3.5

Draw the graph for the each linear equation given below:
y = ${\displaystyle \frac{3x}{2}+\frac{2}{3}}$

Sol:

First prepare a table as follows:

x	-1	0	1
y	${\displaystyle -\frac{5}{6}}$	${\displaystyle \frac{2}{3}}$	${\displaystyle \frac{13}{6}}$

Thus the graph can be drawn as follows:

Question 3.6

Draw the graph for the linear equation given below:
2x - 3y = 4

Sol:

First prepare a table as follows:

x	- 1	0	1
y	- 2	${\displaystyle -\frac{4}{3}}$	${\displaystyle -\frac{2}{3}}$

Thus the graph can be drawn as follows

:

Question 3.7

Draw the graph for the linear equation given below:
${\displaystyle \frac{x-1}{3}-\frac{y+2}{2}=0}$

Sol:

The equation will become:
2x - 3y = 8
First prepare a table as follows:

x	-1	0	1
y	${\displaystyle -\frac{10}{3}}$	${\displaystyle -\frac{8}{3}}$	-2

Thus the graph can be drawn as follows:

Question 3.8

Draw the graph for the linear equation given below:
x - 3 = ${\displaystyle \frac{2}{5}(y+1)}$

Sol:

The equation will become:
5x - 2y = 17
First prepare a table as follows:

x	- 1	0	1
y	- 11	${\displaystyle -\frac{17}{2}}$	- 6

Thus the graph can be drawn as follows:

Question 3.9

Draw the graph for the linear equation given below:
x + 5y + 2 = 0

Sol:

First prepare a table as follows:

x	- 1	0	1
y	${\displaystyle -\frac{1}{5}}$	${\displaystyle -\frac{2}{5}}$	${\displaystyle -\frac{3}{5}}$

Thus the graph can be drawn as follows:

Question 4.1

Draw the graph for the equation given below:
3x + 2y = 6

Sol:

To draw the graph of 3x + 2y = 6 follows the steps:
First prepare a table as below:

X	- 2	0	2
Y	6	3	0

Now sketch the graph as shown:

From the graph it can verify that the line intersects the x-axis at (2,0) and y at (0,3).

Question 4.2

Draw the graph for the equation given below:
2x - 5y = 10

Sol:

To draw the graph of 2x - 5y = 10 follows the steps:
First, prepare a table as below:

X	-1	0	1
Y	${\displaystyle -\frac{12}{5}}$	-2	${\displaystyle -\frac{8}{5}}$

Now sketch the graph as shown

:
From the graph it can verify that the line intersects the x-axis at (5,0) and y at (0,-2).

Question 4.3

Draw the graph for the equation given below:
${\displaystyle \frac{1}{2}x+\frac{2}{3}y=5}$.

Sol:

To draw the graph of ${\displaystyle \frac{x}{2}+\frac{2y}{3}=5}$ follows the steps:

First, prepare a table as below:

X	-1	0	1
Y	5.25	4.5	3.75

Now sketch the graph as shown:

From the graph it can verify that the line intersect the x-axis at (10,0) and y at (0,7.5).

Question 4.4

Draw the graph for the equation given below:
${\displaystyle \frac{2x-1}{3}-\frac{y-2}{5}=0}$

Sol:

To draw the graph of  ${\displaystyle \frac{2x-1}{3}-\frac{y-2}{5}=0}$  follows the steps:
First prepare a table as below:

X	-1	0	1
Y	-3	${\displaystyle \frac{1}{3}}$	${\displaystyle \frac{11}{3}}$

Now sketch the graph as shown:

From the graph it can verify that the line intersects the x-axis at ${\displaystyle (-\frac{1}{10},0)}$ and y at (0,4.5).

Question 5.1

For the linear equation, given above, draw the graph and then use the graph drawn (in the following case) to find the area of a triangle enclosed by the graph and the co-ordinates axes:
3x - (5 - y) = 7

Sol:

First draw the graph as follows:

This is an right triangle.

Thus the area of the triangle will be:

= ${\displaystyle \frac{1}{2}\times \text{base}\times \text{altitude}}$

=  ${\displaystyle \frac{1}{2}\times 4\times 12}$

=  24 sq.units

Question 5.2

For the linear equation, given above, draw the graph and then use the graph drawn (in the following case) to find the area of a triangle enclosed by the graph and the co-ordinates axes:
7 - 3 (1 - y) = - 5 + 2x.

Sol:

First draw the graph as follows:

This is a right triangle.
Thus the area of the triangle will be:

A = ${\displaystyle \frac{1}{2}\times \text{base}\times \text{altitude}}$

 = ${\displaystyle \frac{1}{2}\times \frac{9}{2}\times 3}$

= ${\displaystyle \frac{27}{4}}$

= 6.75 sq.units

Question 6.1

For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other.
y = 3x - 1
y = 3x + 2

Sol:

To draw the graph of y = 3x - 1 and y = 3x + 2 follows the steps:
First, prepare a table as below:

X	- 1	0	1
Y = 3x -1	- 4	- 1	2
Y = 3x + 2	- 1	2	5

Now sketch the graph as shown

:

From the graph it can verify that the lines are parallel.

Question 6.2

For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other.
y = x - 3
y = - x + 5

Sol:

To draw the graph of y = x - 3 and y = - x + 5 follows the steps:

First, prepare a table as below:

X	- 1	0	1
Y = x - 3	- 4	-3	- 2
Y = - x + 5	6	5	4

Now sketch the graph as shown:

From the graph it can verify that the lines are perpendicular.

Question 6.3

For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other.
2x - 3y = 6
${\displaystyle \frac{x}{2}+\frac{y}{3}=1}$

Sol:

To draw the graph of 2x - 3y = 6 and ${\displaystyle \frac{x}{2}+\frac{y}{3}=1}$ follows the steps:
First prepare a table as below:

X	-1	0	1
Y = ${\displaystyle \frac{2}{3}\times 2}$	${\displaystyle -\frac{8}{3}}$	-2	${\displaystyle -\frac{4}{3}}$
Y = ${\displaystyle -\frac{3}{2}\times +3}$	${\displaystyle \frac{9}{2}}$	3	${\displaystyle \frac{3}{2}}$

Now sketch the graph as shown:

From the graph it can verify that the lines are perpendicular.

Question 6.4

For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other.
3x + 4y = 24
${\displaystyle \frac{x}{4}+\frac{y}{3}=1}$

Sol:

To draw the graph of 3x + 4y = 24 and ${\displaystyle \frac{x}{4}+\frac{y}{3}=1}$ follows the steps:
First prepare a table as below

X	-1	0	1
Y = ${\displaystyle -\frac{3}{4}\times +6}$	${\displaystyle \frac{27}{4}}$	6	${\displaystyle \frac{21}{4}}$
Y = ${\displaystyle -\frac{3}{4}\times +3}$	${\displaystyle \frac{15}{4}}$	3	${\displaystyle \frac{9}{4}}$

Now sketch the graph as shown:

From the graph it can verify that the lines are parallel.

Question 7

On the same graph paper, plot the graph of y = x - 2, y = 2x + 1 and y = 4 from x= - 4 to 3.

Sol:

First, prepare a table as follows:

X	-1	0	1
Y = x - 2	-3	-2	-1
Y = 2x + 1	-1	1	3
Y = 4	4	4	4

Now the graph can be drawn as follows:

Question 8

On the same graph paper, plot the graphs of y = 2x - 1, y = 2x and y = 2x + 1 from x = - 2 to x = 4. Are the graphs (lines) drawn parallel to each other?

Sol:

First, prepare a table as follows:

X	-1	0	1
Y = 2x - 1	-3	-1	1
Y = 2x	-2	0	2
Y = 2x + 1	-1	1	3

Now the graph can be drawn as follows:

The lines are parallel to each other.

Question 9

The graph of 3x + 2y = 6 meets the x=axis at point P and the y-axis at point Q. Use the graphical method to find the co-ordinates of points P and Q.

Sol:

To draw the graph of 3x + 2y = 6 follows the steps:
First, prepare a table as below:

X	- 2	0	2
Y	6	3	0

Now sketch the graph as shown:

From the graph it can verify that the line intersects the x-axis at (2,0) and y at (0,3), therefore the coordinates of P(x-axis) and Q(y-axis) are (2,0) and (0,3) respectively.

Question 10

Draw the graph of equation x + 2y - 3 = 0. From the graph, find:
(i) x₁, the value of x, when y = 3
(ii) x₂, the value of x, when y = - 2.

Sol:

First, prepare a table as follows:

X	-1	0	1
Y	2	${\displaystyle \frac{3}{2}}$	1

Thus the graph can be drawn as shown:

(i) For y = 3 we have x = - 3
(ii) For y = - 2 we have x = 7

Question 11

Draw the graph of the equation 3x - 4y = 12.
Use the graph drawn to find:
(i) y₁, the value of y, when x = 4.
(ii) y₂, the value of y, when x = 0.

Sol:

First, prepare a table as follows:

X	- 1	0	1
Y	${\displaystyle -\frac{15}{4}}$	- 3	${\displaystyle -\frac{9}{4}}$

The graph of the equation can be drawn as follows:

From the graph, it can verify that
If x = 4 the value of y = 0
If x = 0 the value of y = - 3.

Question 12

Draw the graph of equation ${\displaystyle \frac{x}{4}+\frac{y}{5}=1}$ Use the graph drawn to find:
(i) x₁, the value of x, when y = 10
(ii) y₁, the value of y, when x = 8.

Sol

First, prepare a table as follows:

X	-1	0	1
Y	${\displaystyle \frac{25}{4}}$	5	${\displaystyle \frac{15}{4}}$

The graph of the equation can be drawn as follows:

From the graph, it can be verified that:
for y = 10, the value of x = - 4.
for x = 8 the value of y = - 5.

Question 13

Use the graphical method to show that the straight lines given by the equations x + y = 2, x - 2y = 5 and ${\displaystyle \frac{x}{3}+y=0}$ pass through the same point.

Sol:

The equations can be written as follows:
y = 2 - x

y = ${\displaystyle \frac{1}{2}(x-5)}$

y = ${\displaystyle -\frac{x}{3}}$
First prepare a table as follows:

X	Y = 2 - x	Y = ${\displaystyle \frac{1}{2}(x-5)}$	Y = ${\displaystyle -\frac{x}{3}}$
- 1	3	- 3	${\displaystyle \frac{1}{3}}$
0	2	${\displaystyle -\frac{5}{2}}$	0
1	1	- 2	${\displaystyle -\frac{1}{3}}$

Thus the graph can be drawn as follows:

From the graph it is clear that the equation of lines are passes through the same point.