SELINA Solution Class 9 Chapter 27 Graphical Solution (Solution of Simultaneous Linear Equations, Graphically) Exercise 27 A

Question 1

Draw the graph for the equation, given below :
x = 5

Sol:

The graph x= 5 in the following figure is a straight line AB which is parallel to y-axis at a distance of 5 units from it.

Question 1.2

Draw the graph for the equation, given below :
x + 5 = 0

Sol:

x + 5 = 0
x = - 5
The graph x = - 5 in the following figure is a straight line AB which is parallel to y-axis at a distance of 5 units from it in the negative x-direction.

Question 1.3

Draw the graph for the equation, given below :
y = 7

Sol:

The graph y = 7 in the following figure is a straight line AB which is parallel to the x-axis at a distance of 7 units from it.

Question 1.4

Draw the graph for the equation, given below :
y + 7 = 0

Sol:

y + 7 = 0
y = -7
The graph y = -7 in the following figure is a straight line AB which is parallel to the x-axis at a distance of 7 units from it in the negative y-direction.

Question 1.5

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

Sol:

2x + 3y = 0
⇒ 3y = - 2x
∴ y = ${\displaystyle \frac{-2x}{3}}$

When x = - 3;  y = ${\displaystyle \frac{-2(-3)}{3}=\frac{6}{3}}$ = 2

When x = 3;  y = ${\displaystyle \frac{-2\left(3\right)}{3}=\frac{-6}{3}}$ = - 2

When x = 6;  y = ${\displaystyle \frac{-2\left(6\right)}{3}=\frac{-12}{3}}$ = - 4

X	- 3	3	6
Y	2	- 2	- 4

Plotting these points we get the required graph as shown below:

Question 1.6

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

Sol:

3x + 2y = 6
⇒ 2y = 6 - 3x
∴ y = ${\displaystyle \frac{6-3x}{2}}$

When x = 0;  
y = ${\displaystyle \frac{6-3\times 0}{2}}$
= ${\displaystyle \frac{6-0}{2}}$
= 3

When x = 2; 
y = ${\displaystyle \frac{6-3\times 2}{2}}$
= ${\displaystyle \frac{6-6}{2}}$
= 0

When x = 4; 
y = ${\displaystyle \frac{6-3\times 4}{2}}$
= ${\displaystyle \frac{6-12}{2}}$
= -3

X	0	2	4
Y	3	0	- 3

Plotting these points we get the required graph as shown below :

Question 1.7

Draw the graph for the equation, given below :
x - 5y + 4 = 0

x - 5y + 4 = 0
⇒ 5y = 4 + x
∴ y = ${\displaystyle \frac{x+4}{5}}$

When x = 1; 
y = ${\displaystyle \frac{1+4}{5}}$
= ${\displaystyle \frac{5}{5}}$
= 1

When x = 6; 
y = ${\displaystyle \frac{6+4}{5}}$
= ${\displaystyle \frac{10}{5}}$
= 2

When x = - 4; 
y = ${\displaystyle \frac{-4+4}{5}}$
= ${\displaystyle \frac{0}{5}}$
= 0

X	1	6	- 4
Y	1	2	0

Plotting these points we get the required graph as shown below :

Sol:

x - 5y + 4 = 0
⇒ 5y = 4 + x
∴ y = ${\displaystyle \frac{x+4}{5}}$

When x = 1; 
y = ${\displaystyle \frac{1+4}{5}}$
= ${\displaystyle \frac{5}{5}}$
= 1

When x = 6; 
y = ${\displaystyle \frac{6+4}{5}}$
= ${\displaystyle \frac{10}{5}}$
= 2

When x = - 4; 
y = ${\displaystyle \frac{-4+4}{5}}$
= ${\displaystyle \frac{0}{5}}$
= 0

X	1	6	- 4
Y	1	2	0

Plotting these points we get the required graph as shown below :

Question 1.8

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

Sol:

5x + y + 5 = 0
⇒ y = - 5x - 5

When x = 0; 
y = - 5x x - 5
= - 0 - 5
= - 5

When x = - 1; 
y = - 5 x (- 1) - 5
= 5 - 5
= 0

When x = - 2; 
y = - 5 x (- 2) - 5
= 10 - 5
= 5

X	0	- 1	- 2
Y	- 5	0	5

Plotting these points we get the required graph as shown below:

Question 2.1

Draw the graph for the equation given below; hence find the co-ordinates of the points where the graph is drawn meets the co-ordinates axes:
${\displaystyle \frac{1}{3}x+\frac{1}{5}y=1}$.

Sol:

${\displaystyle \frac{1}{3}x+\frac{1}{5}y=1}$

⇒ ${\displaystyle \frac{5x+3y}{15}=1}$

⇒ 5x + 3y = 15

⇒ 3y = 15 - 5x

⇒ y = ${\displaystyle \frac{15-5x}{3}}$

When x = 0;  y  = ${\displaystyle \frac{15-5\times 0}{3}=\frac{15-0}{3}}$ = 5

When x = 3;  y = ${\displaystyle \frac{15-5\times 3}{3}=\frac{15-15}{3}}$  = 0

When x = -3;  y = ${\displaystyle \frac{15-5\times (-3)}{3}=\frac{15+15}{3}}$ = 10

X	0	3	- 3
Y	5	0	10

Plotting these points we get the required graph as shown below:

From the figure it is clear that, the graph meets the coordinate axes at (3,0) and (0,5).

Question 2.2

Draw the graph for the equation given below; hence find the co-ordinates of the points where the graph is drawn meets the co-ordinates axes:
${\displaystyle \frac{2x+15}{3}=y-1}$

Sol :

 

Sol:

Draw the graph for the equation given below; hence find the co-ordinates of the points where the graph is drawn meets the co-ordinates axes:
${\displaystyle \frac{2x+15}{3}=y-1}$

Question 3

Draw the graph of the straight line given by the equation 4x - 3y + 36 = 0
Calculate the area of the triangle formed by the line drawn and the co-ordinate axes.

Sol:

4x - 3y + 36 = 0
⇒ 4x - 3y = -36
⇒ -3y = -36 - 4x
⇒ 3y = 36 + 4x
⇒ y = ${\displaystyle \frac{36+4x}{3}}$

When x = - 6,
y = ${\displaystyle \frac{36+4\times (-6)}{3}}$
= ${\displaystyle \frac{36-24}{3}}$
= 4

When x = - 3,
y = ${\displaystyle \frac{36+4\times (-3)}{3}}$
= ${\displaystyle \frac{36-12}{3}}$
= 8

When x = - 9,
y = ${\displaystyle \frac{36+4\times (-9)}{3}}$
= ${\displaystyle \frac{36-36}{3}}$
= 0

X	- 9	- 3	- 6
Y	0	8	4

Plotting these points we get the required graph as shown below:

The straight line cuts the co-ordinates axis at A(0,12) and B(-9,0).
∴ The triangle ΔAOB is formed.
Area of the triangle AOB
= ${\displaystyle \frac{1}{2}}$ x AO x OB

= ${\displaystyle \frac{1}{2}}$ x 12 x 9
 = 54 sq . units
∴ Area of the triangle is 54 sq . units

Question 4

Draw the graph of the equation 2x - 3y - 5 = 0
From the graph, find:
(i) x₁, the value of x, when y = 7
(ii) x₂, the value of x, when y = - 5.

Sol:

2x - 3y - 5 = 0
⇒ 2x = 3y + 5
⇒ x = ${\displaystyle \frac{3y+5}{2}}$

When y = 1,
x = ${\displaystyle \frac{3\left(1\right)+5}{2}}$
= ${\displaystyle \frac{8}{2}}$
= 4

When y = 3,
x = ${\displaystyle \frac{3\left(3\right)+5}{2}}$
= ${\displaystyle \frac{9+5}{2}}$
= 7

When y = - 1,
x = ${\displaystyle \frac{3(-1)+5}{2}}$
= ${\displaystyle \frac{5-3}{2}}$
= 1

X	4	7	1
Y	1	3	- 1

Plotting these points we get the required graph as shown below:

(i) The value of x, when y = 7:

We have the equation of the line as 

x = ${\displaystyle \frac{3y+5}{2}}$

Now substitute y = 7 and x = x₁:
x₁ = ${\displaystyle \frac{3\left(7\right)+5}{2}}$

= ${\displaystyle \frac{21+5}{2}}$

= ${\displaystyle \frac{26}{2}}$
= 13

(ii) The value of x, when y = - 5:

Now substitute y = - 5 and x = x₂

x₂ = ${\displaystyle \frac{3(-5)+5}{2}}$

= ${\displaystyle \frac{-15+5}{2}}$

= ${\displaystyle \frac{-10}{2}}$

 = -5.

Question 5

Draw the graph of the equation
4x + 3y + 6 = 0
From the graph, find :
(i) y₁, the value of y, when x = 12.
(ii) y₂, the value of y, when x = - 6.

Sol:

4x + 3y + 6 = 0
⇒ 3y = -4x - 6
⇒ y = ${\displaystyle \frac{-4x-6}{3}}$

When x = 0,
y = ${\displaystyle \frac{-4\left(0\right)-6}{3}}$
= ${\displaystyle \frac{-6}{3}}$
= -2
When x = 3,
y = ${\displaystyle \frac{-4\left(3\right)-6}{3}}$
= ${\displaystyle \frac{-12-6}{3}}$
= -6
When x = -3,
y = ${\displaystyle \frac{-4\left(3\right)-6}{3}}$
= ${\displaystyle \frac{12-6}{3}}$
= 2

X	0	3	-3
Y	-2	-6	2

Plotting these points we get the required graph as shown below:

The value of y, when x = 12:
We have the equation of the line as
y = ${\displaystyle \frac{-4x-6}{3}}$

Now substitute x = 12 and y = y₁:
y₁ = ${\displaystyle \frac{-4\left(12\right)-6}{3}}$

= ${\displaystyle \frac{-48-6}{3}}$

= ${\displaystyle \frac{-54}{3}}$
= - 18

The value of y, when x = - 6:
Now substitute x = - 6 and y = y₂:
y₂ = ${\displaystyle \frac{-4(-6)-6}{3}}$

= ${\displaystyle \frac{24-6}{3}}$

= ${\displaystyle \frac{18}{3}}$
= 6

Question 6

Use the table given below to draw the graph.

X	- 5	- 1	3	b	13
Y	- 2	a	2	5	7

From your graph, find the values of 'a' and 'b'.
State a linear relationship between the variables x and y.

Sol:

The table is:

X	- 5	- 1	3	b	13
Y	- 2	a	2	5	7

Plotting the points as shown in the above table, we get the following required graph:

When x = - 1, then y = 0
⇒ a = 0
When y = 5, then x = 9
⇒  b = 9

Let y = px + q                   ....(1)
be a linear relation between x and y

Substitute x = 9 and y = 5 in the equation (1), we have,
5 = 9p + q                          ....(2)

Substitute x = - 1 and y = 0 in the equation (1), we have,
0 = - p + q                          ....(3)

Subtracting (3) from (2), we have,
5 = 10p

⇒ p = ${\displaystyle \frac{5}{10}}$

⇒ p = ${\displaystyle \frac{1}{2}}$

From (3), we have,
p = q

∴ q = ${\displaystyle \frac{1}{2}}$

Thus, the linear relation is

y = px + q

⇒ y = ${\displaystyle \frac{1}{2}x+\frac{1}{2}}$

⇒ y = ${\displaystyle \frac{x+1}{2}}$

Question 7

Draw the graph obtained from the table below:

X	a	3	- 5	5	c	- 1
Y	- 1	2	b	3	4	0

Use the graph to find the values of a, b and c. State a linear relation between the variables x and y.

Sol:

The table is:

X	a	3	- 5	5	c	- 1
Y	- 1	2	b	3	4	0

Plotting the points as shown in the above table,
we get the following required graph:

When y = - 1, then x = - 3 
⇒ a = - 3
When x = - 5, then y = - 2
⇒ b = - 2
When y = 4, then x = 7
⇒ c = 7

Let y = px + q                     ....(1)
be a linear relation between x and y

Substitute x = - 3 and y = - 1 in the equation (1), we have,
-1 = - 3p + q                       ....(2)
Substitute x = - 5 and y = - 2 in the equation (1), we have,
-2 = - 5p + q                       ....(3)

Subtracting (3) from (2), we have,
1 = 2p
⇒ p = ${\displaystyle \frac{1}{2}}$

From (3), we have,
-2 = -5p + q
-2 = -5${\displaystyle \left(\frac{1}{2}\right)}$+q
⇒ -4 = -5 + 2q
⇒ 2q = 5 - 4
⇒ 2q = 1
∴ q = ${\displaystyle \frac{1}{2}}$
Thus, the linear relation is
y = px + q
⇒ y = ${\displaystyle \frac{1}{2}x+\frac{1}{2}}$

⇒ y = ${\displaystyle \frac{x+1}{2}}$

Question 8

A straight line passes through the points (2, 4) and (5, - 2). Taking 1 cm = 1 unit; mark these points on a graph paper and draw the straight line through these points. If points (m, - 4) and (3, n) lie on the line drawn; find the values of m and n.

Sol:

The table is:

X	2	3	5	m
Y	4	n	- 2	- 4

Plotting the points as shown in the above table,
we get the following required graph:

Plotting the points in the graph we get the above required graph.
Now draw a line x = 3, parallel to y-axis to meet the line
It meets the line at y = 2 and therefore, n = 2
Now draw a line y = -4, parallel to x-axis to meet the line
It meets the line at x = 6 and therefore, m = 6
Thus the values of m and n are 6 and 2 respectively.

Question 9

Draw the graph (straight line) given by equation x - 3y = 18. If the straight line is drawn passes through the points (m, - 5) and (6, n); find the values of m and n.

Sol:

Consider the equation
x - 3y = 18
⇒ - 3y = 18 - x
⇒ 3y = x - 18
⇒  y = ${\displaystyle \frac{x-18}{3}}$
The table for x - 3y = 18 is

X	9	0	6	3
Y	- 3	- 6	- 4	- 5

Plotting the above points, we get the following required graph:

From the above figure, we have m = 3 and n = - 4.

Question 10.1

Use the graphical method to find the value of k, if:
(k, -3) lies on the straight line 2x + 3y = 1

Sol:

2x + 3y = 1
⇒ 3y = 1 - 2x
⇒ y = ${\displaystyle \frac{1-2x}{3}}$
The table for 2x + 3y = 1 is

X	- 1	2	5
Y	1	- 1	- 3

Plotting the above points in a graph, we get the following graph:

From the above graph, it is clear that k = 5

Question 10.2

Use the graphical method to find the value of k, if:
(5, k - 2) lies on the straight line x - 2y + 1 = 0

Sol:

x - 2y + 1 = 0
⇒ 2y = x + 1
⇒ y = ${\displaystyle \frac{x+1}{2}}$
The table for x - 2y + 1 = 0 is

X	1	3	5
Y	1	2	3

Plotting the above points in a graph, we get the following graph:

From the above graph, it is clear that
k - 2 = 3
⇒ k = 5.