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