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

Question 1.1

In the following, find the inclination of line AB:

Sol:

The angle which a straight line makes with the positive direction of the x-axis (measured in an anticlockwise direction) is called inclination o the line.
The inclination of a line is usually denoted by θ
The inclination is θ = 45°.

Question 1.2

In the following, find the inclination of line AB:

Sol:

The angle which a straight line makes with the positive direction of x-axis (measured in anticlockwise direction) is called inclination o the line.
The inclination of a line is usually denoted by θ
The inclination is θ = 135°

Question 1.3

In the following, find the inclination of line AB:

Sol:

The angle which a straight line makes with the positive direction of x-axis (measured in anticlockwise direction) is called inclination o the line.
The inclination of a line is usually denoted by θ
The inclination is θ = 30°

Question 2.1

Write the inclination of a line which is: Parallel to the x-axis.

Sol:

The inclination of a line parallel to x-axis is θ = 0°

Question 2.2

Write the inclination of a line which is: Perpendicular to the x-axis.

Sol:

The inclination of a line perpendicular to x-axis is θ = 90°

Question 2.3

Write the inclination of a line which is: Parallel to the y-axis.

Sol:

The inclination of a line parallel to y-axis is θ = 90°

Question 2.4

Write the inclination of a line which is: Perpendicular to the y-axis.

Sol:

The inclination of a line perpendicular to the y-axis is θ = 0°.

Question 3.1

Write the slope of the line whose inclination is: 0°. 

 Sol:

If θ is the inclination of a line; the slope of the line is tan θ and is usually denoted by letter m.

Here the inclination of a line is 0°, then θ = 0°

Therefore the slope of the line is m = tan 0° = 0

Question 3.2

Write the slope of the line whose inclination is: 30°

Sol:

If θ is the inclination of a line; the slope of the line is tan θ and is usually denoted by letter m.

Here the inclination of a line is 30°, then θ = 30°

Therefore the slope of the line is m = tan θ = 30° = ${\displaystyle \frac{1}{\sqrt{3}}}$.

Question 3.3

Write the slope of the line whose inclination is: 45°.

Sol:

If θ is the inclination of a line; the slope of the line is tan θ and is usually denoted by letter m.

Here the inclination of a line is 45°, then θ = 45°

Therefore the slope of the line is m = tan 45° = 1

Question 3.4

Write the slope of the line whose inclination is: 60°

Sol:

If θ is the inclination of a line; the slope of the line is tan θ and is usually denoted by letter m.

Here the inclination of a line is 60°, then θ = 60°

Therefore the slope of the line is m = tan 60° = ${\displaystyle \sqrt{3}}$

Question 4.1

Find the inclination of the line whose slope is: 0

Sol:

If tan θ is the slope of a line; then inclination of the line is θ
Here the slope of line is 0; then tan θ = 0
Now
tan θ = 0
tan θ = tan 0°
θ = 0°
Therefore the inclination of the given line is θ = 0°

Question 4.2

Find the inclination of the line whose slope is: 1

Sol :

If tan θ is the slope of a line; then the inclination of the line is tan θ
Here the slope of the line is 1; then tan θ = 1
Now
tan θ = 1
tan θ = tan 45°
θ = 45°

Therefore the inclination of the given line is θ = 45°.

Question 4.3

Find the inclination of the line whose slope is: ${\displaystyle \sqrt{3}}$

Sol:

If tan θ is the slope of a line; then the inclination of the line is tan θ

Here the slope of line is ${\displaystyle \sqrt{3}}$; then tan θ = ${\displaystyle \sqrt{3}}$
Now
tan θ  = ${\displaystyle \sqrt{3}}$
tan θ  = tan 60°
θ = 60°
Therefore the inclination of the given line is θ = 60°

Question 4.4

Find the inclination of the line whose slope is: ${\displaystyle \frac{1}{\sqrt{3}}}$

Sol:

If tan θ is the slope of a line; then inclination of the line is tan θ
Here the slope of line is ${\displaystyle \frac{1}{\sqrt{3}}}$; then tan θ = ${\displaystyle \frac{1}{\sqrt{3}}}$
Now
tan θ = ${\displaystyle \frac{1}{\sqrt{3}}}$
tan θ = tan 30°
θ = 30°
Therefore the inclination of the given line is θ = 30°

Question 5.1

Write the slope of the line which is: Parallel to the x-axis.

Sol:

For any line which is parallel to x-axis, the inclination is θ = 0°

Therefore, Slope(m) = tan θ = tan 0° = 0.

Question 5.2

Write the slope of the line which is: Perpendicular to the x-axis.

Sol :

For any line which is perpendicular to x-axis, the inclination is θ = 90°

Therefore, Slope(m) = tan θ = tan 90° = ∞ (not defined).

Question 5.3

Write the slope of the line which is: Parallel to the y-axis.

Sol :

For any line which is parallel to y-axis, the inclination is θ = 90°

Therefore, Slope(m) = tan θ = tan 90° = ∞ (not defined).

Question 5.4

Write the slope of the line which is: Perpendicular to the y-axis.

Sol:

For any line which is perpendicular to y-axis, the inclination is θ = 0°

Therefore, Slope(m) = tan θ = tan 0° = 0.

Question 6.1

For the equation given below, find the slope and the y-intercept:
x + 3y + 5 = 0

Sol:

Equation of any straight line in the form y = mx + c, where slope = m(co-efficient of x) and y-intercept = c(constant term)
x + 3y + 5 = 0

x + 3y + 5 = 0

3y = - x - 5

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

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

Therefore,
slope = co-efficient of x = ${\displaystyle -\frac{1}{3}}$ 

y-intercept = constant term = ${\displaystyle -\frac{5}{3}}$.

Question 6.2

Equation of any straight line in the form y = mx + c, where slope = m(co-efficient of x) and y-intercept = c(constant term)
x + 3y + 5 = 0

x + 3y + 5 = 0

3y = - x - 5

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

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

Therefore,
slope = co-efficient of x = ${\displaystyle -\frac{1}{3}}$ 

y-intercept = constant term = ${\displaystyle -\frac{5}{3}}$.

Sol:

Equation of any straight line in the form y = mx + c, where slope = m(co-efficient of x) and y-intercept = c(constant term)
x + 3y + 5 = 0

x + 3y + 5 = 0

3y = - x - 5

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

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

Therefore,
slope = co-efficient of x = ${\displaystyle -\frac{1}{3}}$ 

y-intercept = constant term = ${\displaystyle -\frac{5}{3}}$.

Question 6.3

For the equation given below, find the slope and the y-intercept:
5x = 4y + 7

Sol:

Equation of any straight line in the form y = mx + c, where slope = m(co-efficient of x) and y-intercept = c(constant term)
5x = 4y + 7

5x = 4y + 7
4y = 5x - 7
y = ${\displaystyle \frac{5x-7}{4}}$
y = ${\displaystyle \frac{5}{4}x+(-\frac{7}{4})}$
Therefore,
slope - co-efficient of x = ${\displaystyle \frac{5}{4}}$
y-intercept = constant term = ${\displaystyle -\frac{7}{4}}$

Question 6.4

For the equation given below, find the slope and the y-intercept:
x= 5y - 4

Sol:

Equation of any straight line in the form y = mx + c, where slope = m(co-efficient of x) and y-intercept = c(constant term)
x= 5y - 4

x= 5y - 4
5y = x + 4
y = ${\displaystyle \frac{x+4}{5}}$
y = ${\displaystyle \frac{1}{5}x+\frac{4}{5}}$
Therefore,
slope = co-efficient of x = ${\displaystyle \frac{1}{5}}$
y-intercept = constant term = ${\displaystyle \frac{4}{5}}$

Question 6.5

For the equation given below, find the slope and the y-intercept:
y = 7x - 2

Sol:

Equation of any straight line in the form y = mx + c, where slope = m(co-efficient of x) and y-intercept = c(constant term)
y = 7x - 2

y = 7x - 2
y = 7x + (- 2)
Therefore,
slope = co-efficient of x = 7
y-intercept = constant term = - 2

Question 6.6

For the equation given below, find the slope and the y-intercept:
3y = 7

Sol:

Equation of any straight line in the form y = mx + c, where slope = m(co-efficient of x) and y-intercept = c(constant term)
3y = 7

3y = 7

3y = 0 · x + 7

y = ${\displaystyle \frac{0}{7}x+\frac{7}{3}}$

y = 0 · x + ${\displaystyle \frac{7}{3}}$

Therefore,

slope = co-efficient of x = 0

y-intercept = constant term = ${\displaystyle \frac{7}{3}}$.

Question 6.7

For the equation given below, find the slope and the y-intercept:
4y + 9 = 0

Sol:

Equation of any straight line in the form y = mx + c, where slope = m(co-efficient of x) and y-intercept = c(constant term)
4y + 9 = 0

4y + 9 = 0

4y = 0 · x - 9

y = ${\displaystyle \frac{0}{4}x-\frac{9}{4}}$

y = 0 · x + ${\displaystyle (-\frac{9}{4})}$

Therefore,

slope = co-efficient of x = 0

y-intercept = constant term = ${\displaystyle -\frac{9}{4}}$

Question 7.1

Find the equation of the line whose:
Slope = 2 and y-intercept = 3

Sol:

Given
Slope is 2, therefore m = 2
Y-intercept is 3, therefore c = 3
Therefore,
y = mx + c
y = 2x + 3
Therefore the equation of the required line is y = 2x + 3

Question 7.2

Find the equation of the line whose:
Slope = 5 and y-intercept = - 8

Sol:

Given
Slope is 5, therefore m = 5
Y-intercept is - 8, therefore c = - 8
Therefore,
y = mx + c
y = 5x + - 8
Therefore the equation of the required line is y = 5x + (- 8)

Question 7.3

Find the equation of the line whose:
slope = - 4 and y-intercept = 2

Sol:

Given
Slope is - 4, therefore m = - 4
Y-intercept is 2, therefore c = 2
Therefore,
y = mx + c
y = - 4x + 2
Therefore the equation of the required line is y = - 4x + 2

Question 7.4

Find the equation of the line whose:
slope = - 3 and y-intercept = - 1

Sol:

Given
Slope is - 3, therefore m = - 3
Y-intercept is - 1, therefore c = - 1
Therefore,
y = mx + c
y = - 3x - 1
Therefore the equation of the required line is y = - 3x - 1

Question 7.5

Find the equation of the line whose:
slope = 0 and y-intercept = - 5

Sol:

Given
Slope is 0, therefore m = 0
Y-intercept is - 5, therefore c = - 5
Therefore,
y = mx + c
y = 0 · x + (- 5)
y = - 5
Therefore the equation of the required line is y = - 5

Question 7.6

Find the equation of the line whose:
slope = 0 and y-intercept = 0

Sol :

Given
Slope is 0, therefore m = 0
Y-intercept is 0, therefore c = 0
Therefore,
y = mx + c
y = 0 · x + 0
y = 0
Therefore the equation of the required line is y = 0

Question 8

Draw the line 3x + 4y = 12 on a graph paper. From the graph paper, read the y-intercept of the line.

Sol:

Given line is 3x + 4y = 12

The graph of the given line is shown below.

Clearly from the graph we can find the y-intercept.
The required y-intercept is 3.

Question 9

Draw the line 2x - 3y - 18 = 0 on a graph paper. From the graph paper, read the y-intercept of the line.

Sol:

Given line is
2x – 3y – 18 = 0
The graph of the given line is shown below.

Clearly from the graph we can find the y-intercept.
The required y-intercept is -6

Question 10

Draw the graph of the line x + y = 5. Use the graph paper drawn to find the inclination and the y-intercept of the line.

Sol:

Given line is
x + y = 5
The graph of the given line is shown below.

From the given line x + y = 5, we get
x + y = 5
y = - x + 5
y = (-1) . x + 5 .......(A)

Again we know that equation of any straight line in the form y = mx + c, where m is the gradient and c is the intercept. Again we have if slope of a line is tan θ then inclination of the line is θ 

Now from the equation (A), we have
m = - 1
tan θ = - 1
tan θ = tan 135°
θ  = 135°
And c = 5
Therefore the required inclination is θ = 135° and y-intercept is c = 5

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.