Co-ordinate Geometry — Exercise 23(C)

The inclination of a line is 60°. The slope of the line is :

1. $\dfrac{1}{\sqrt3}$

2. $-\dfrac{1}{\sqrt3}$

3. ${\sqrt3}$

4. $-{\sqrt3}^{0}$

Answer

The inclination of a line is 60°, then θ = 60°.

The slope of the line = m = tan 60° = $\sqrt3$

Hence, option 3 is the correct option.

For the equation 2x - 5y = 8; slope is :

1. 5

2. $-\dfrac{2}{5}$

3. 8

4. $\dfrac{2}{5}$

Answer

⇒ 2x - 5y = 8

⇒ - 5y = 8 - 2x

⇒ 5y = 2x - 8

⇒ y = $\dfrac{2}{5}\text{x} - \dfrac{8}{5}$

∴ Slope = coefficient of x = $\dfrac{2}{5}$

Hence, option 4 is the correct option.

For the equation 5x - 6y = 9, the y-intercept is :

1. $\dfrac{3}{2}$

2. $\dfrac{5}{6}$

3. $\dfrac{6}{5}$

4. $-\dfrac{3}{2}$

Answer

⇒ 5x - 6y = 9

⇒ -6y = 9 - 5x

⇒ 6y = 5x - 9

⇒ y = $\dfrac{5}{6}\text{x} - \dfrac{9}{6}$

∴ y-intercept = constant term = -$\dfrac{9}{6} = - \dfrac{3}{2}$

Hence, option 4 is the correct option.

If the slope of a line is -2 and its y-intercept is -7, the equation of the line is:

1. 2x + y + 7 = 0

2. 2x - y + 7 = 0

3. 2x - y - 7 = 0

4. 2x + y - 7 = 0

Answer

slope = -2 ⇒ m = -2

y-intercept = -7 ⇒ c = -7

∴ Equation is : y = mx + c

⇒ y = (-2)x + (-7)

⇒ y = -2x - 7

⇒ 2x + y + 7 = 0

Hence, option 1 is the correct option.

For the equation x - y + 1 = 0; the values of slope (m) and y-intercept (c) are :

1. m = 1, c = 1

2. m = -1, c = 1

3. m = 1, c = -1

4. m = -1, c = -1

Answer

The given equation : x - y + 1 = 0

⇒ y = x + 1

General form of equation: y = mx + c

m = 1, c = 1

Hence, option 1 is the correct option.

In the following, find the inclination of line AB :

Answer

MO = NO

∠ NMO = ∠ MNO

∠ MON = 90°

Let ∠ MNO = ∠ NMO = x.

As we know that sum of all angles of triangle = 180°

⇒ ∠ NMO + ∠ MNO + ∠ MON = 180°

⇒ x + x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ x = $\dfrac{90°}{2}$

⇒ x = 45°

Hence, the inclination = 45°.

In the following, find the inclination of line AB :

Answer

MO = NO

∠ NMO = ∠ MNO

∠ MON = 90°

Let ∠ MNO = ∠ NMO = x.

As we know that sum of all angles of triangle = 180°

⇒ ∠ NMO + ∠ MNO + ∠ MON = 180°

⇒ x + x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ x = $\dfrac{90°}{2}$

⇒ x = 45°

Inclination = 180° - 45° = 135°

Hence, the inclination = 135°.

In the following, find the inclination of line AB :

Answer

∠ NMO = 2x

∠ MNO = x

∠ MON = 90°

As we know that sum of all angles of triangle = 180°

⇒ ∠ NMO + ∠ MNO + ∠ MON = 180°

⇒ 2x + x + 90° = 180°

⇒ 3x = 180° - 90°

⇒ x = $\dfrac{90°}{3}$

⇒ x = 30°

Hence, the inclination = 30°.

Write the inclination of a line which is :

(i) parallel to x-axis.

(ii) perpendicular to x-axis.

(iii) parallel to y-axis.

(iv) perpendicular to y-axis.

Answer

(i) A line parallel to the x-axis has an inclination of 0°, as it does not make an angle with the x-axis.

Hence, the inclination of the line = 0°.

(ii) A line perpendicular to the x-axis is parallel to the y-axis and makes a 90° angle with the x-axis.

Hence, the inclination of the line = 90°.

(iii) A line parallel to the y-axis is perpendicular to the x-axis, so its inclination is 90°.

Hence, the inclination of the line = 90°.

(iv) A line perpendicular to the y-axis is parallel to the x-axis, so its inclination is 0°.

Hence, the inclination of the line = 0°.

Write the slope of the line whose inclination is:

(i) 0°

(ii) 30°

(iii) 45°

(iv) 60°

Answer

(i) 0°

The inclination of a line is 0°, then θ = 0°.

The slope of the line = m = tan 0° = 0

Hence, the slope of the line whose inclination is 0° is 0.

(ii) 30°

The inclination of a line is 30°, then θ = 30°.

The slope of the line = m = tan 30° = $\dfrac{1}{\sqrt3}$

Hence, the slope of the line whose inclination is 30° is $\dfrac{1}{\sqrt3}$.

(iii) 45°

The inclination of a line is 45°, then θ = 45°.

The slope of the line = m = tan 45° = 1

Hence, the slope of the line whose inclination is 45° is 1.

(iv) 60°

The inclination of a line is 60°, then θ = 60°.

The slope of the line = m = tan 60° = $\sqrt3$

Hence, the slope of the line whose inclination is 60° is $\sqrt3$.

Find the inclination of the line whose slope is:

(i) 0

(ii) 1

(iii) ${\sqrt3}$

(iv) $\dfrac{1}{\sqrt3}$

Answer

(i) The slope of the line = m = 0 ⇒ tan θ = 0

⇒ tan θ = tan 0°

⇒ θ = 0°

Hence, the inclination is 0°.

(ii) The slope of the line = m = 1 ⇒ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°

Hence, the inclination is 45°.

(iii) The slope of the line = m = ${\sqrt3}$ ⇒ tan θ = ${\sqrt3}$.

⇒ tan θ = tan 60°

⇒ θ = 60°

Hence, the inclination is 60°.

(iv) The slope of the line = m = $\dfrac{1}{\sqrt3}$ ⇒ tan θ = $\dfrac{1}{\sqrt3}$

⇒ tan θ = tan 30°

⇒ θ = 30°

Hence, the inclination is 30°.

Write the slope of the line which is :

(i) parallel to x-axis.

(ii) perpendicular to x-axis.

(iii) parallel to y-axis.

(iv) perpendicular to y-axis.

Answer

(i) A line parallel to the x-axis has an inclination of 0°, as it does not make an angle with the x-axis.

Therefore, the inclination of the line is 0°.

The slope of the line = m = tan θ = tan 0° = 0

Hence, the slope of the line is 0.

(ii) A line perpendicular to the x-axis is parallel to the y-axis and makes a 90° angle with the x-axis.

Therefore, the inclination of the line is 90°.

The slope of the line = m = tan θ = tan 90° = not defined

Hence, the slope of the line is not defined.

(iii) A line parallel to the y-axis is perpendicular to the x-axis, so its inclination is 90°.

Therefore, the inclination of the line is 90°.

The slope of the line = m = tan θ = tan 90° = not defined

Hence, the slope of the line is not defined.

(iv) A line perpendicular to the y-axis is parallel to the x-axis, so its inclination is 0°.

Therefore, the inclination of the line is 0°.

The slope of the line = m = tan θ = tan 0° = 0

Hence, the slope of the line is 0.

For each of the equations given below, find the slope and the y-intercept :

(i) x + 3y + 5 = 0

(ii) 3x - y - 8 = 0

(iii) 5x = 4y + 7

(iv) x = 5y - 4

(v) y = 7x - 2

(vi) 3y = 7

(vii) 4y + 9 = 0

Answer

(i) x + 3y + 5 = 0

⇒ 3y = -x - 5

⇒ y = -$\dfrac{1}{3}x - \dfrac{5}{3}$

∴ Slope = coefficient of x = $\dfrac{-1}{3}$

And, y-intercept = constant term = $\dfrac{-5}{3}$

Hence, the slope = $\dfrac{-1}{3}$ and y-intercept = $\dfrac{-5}{3}$.

(ii) 3x - y - 8 = 0

⇒ y = 3x - 8

∴ Slope = coefficient of x = 3

And, y-intercept = constant term = -8

Hence, the slope = 3 and y-intercept = -8.

(iii) 5x = 4y + 7

⇒ 4y = 5x - 7

⇒ y = $\dfrac{5}{4}$ x - $\dfrac{7}{4}$

∴ Slope = coefficient of x = $\dfrac{5}{4}$

And, y-intercept = constant term = $\dfrac{-7}{4}$

Hence, the slope = $\dfrac{5}{4}$ and y-intercept = $\dfrac{-7}{4}$.

(iv) x = 5y - 4

⇒ 5y = x + 4

⇒ y = $\dfrac{1}{5}$ x + $\dfrac{4}{5}$

∴ Slope = coefficient of x = $\dfrac{1}{5}$

And, y-intercept = constant term = $\dfrac{4}{5}$

Hence, the slope = $\dfrac{1}{5}$ and y-intercept = $\dfrac{4}{5}$.

(v) y = 7x - 2

∴ Slope = coefficient of x = 7

And, y-intercept = constant term = -2

Hence, the slope = 7 and y-intercept = -2.

(vi) 3y = 7

⇒ 3y = 0 $\times$ x + 7

⇒ y = 0 $\times$ x + $\dfrac{7}{3}$

∴ Slope = coefficient of x = 0

And, y-intercept = constant term = $\dfrac{7}{3}$

Hence, the slope = 0 and y-intercept = $\dfrac{7}{3}$.

(vii) 4y + 9 = 0

⇒ 4y = 0 $\times$ x - 9

⇒ y = 0 $\times$ x - $\dfrac{9}{4}$

∴ Slope = coefficient of x = 0

And, y-intercept = constant term = $\dfrac{-9}{4}$

Hence, the slope = 0 and y-intercept = $\dfrac{-9}{4}$.

Find the equation of the line, whose :

(i) slope = 2 and y-intercept = 3

(ii) slope = 5 and y-intercept = - 8

(iii) slope = - 4 and y-intercept = 2

(iv) slope = - 3 and y-intercept = - 1

(v) slope = 0 and y-intercept = - 5

(vi) slope = 0 and y-intercept = 0

Answer

(i) slope = 2 ⇒ m = 2

y-intercept = 3 ⇒ c = 3

∴ Equation is : y = mx + c

⇒ y = 2x + 3

Hence, the equation of the line is y = 2x + 3.

(ii) slope = 5 ⇒ m = 5

y-intercept = -8 ⇒ c = -8

∴ Equation is : y = mx + c

⇒ y = 5x - 8

Hence, the equation of the line is y = 5x - 8.

(iii) slope = -4 ⇒ m = -4

y-intercept = 2 ⇒ c = 2

∴ Equation is : y = mx + c

⇒ y = -4x + 2

⇒ 4x + y = 2

Hence, the equation of the line is 4x + y = 2.

(iv) slope = -3 ⇒ m = -3

y-intercept = -1 ⇒ c = -1

∴ Equation is : y = mx + c

⇒ y = -3x - 1

⇒ 3x + y + 1 = 0

Hence, the equation of the line is 3x + y + 1 = 0.

(v) slope = 0 ⇒ m = 0

y-intercept = - 5 ⇒ c = - 5

∴ Equation is : y = mx + c

⇒ y = 0 $\times$ x - 5

⇒ y + 5 = 0

Hence, the equation of the line is y + 5 = 0.

(vi) slope = 0 ⇒ m = 0

y-intercept = 0 ⇒ c = 0

∴ Equation is : y = mx + c

⇒ y = 0 $\times$ x + 0

⇒ y = 0

Hence, the equation of the line is y = 0.

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

Answer

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 3 $\times$ 0 + 4y = 12 ⇒ y = 3

Let x = 1, then 3 $\times$ 1 + 4y = 12 ⇒ y = 2.2

Let x = 4, then 3 $\times$ 4 + 4y = 12 ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

From the graph, y-intercept of the line = OB = 3.

Hence, y-intercept of the line = 3.

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

Answer

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 2 $\times$ 0 - 3y - 18 = 0 ⇒ y = -6

Let x = 3, then 2 $\times$ 3 - 3y - 18 = 0 ⇒ y = -4

Let x = 6, then 2 $\times$ 6 - 3y - 18 = 0 ⇒ y = -2

Let x = 9, then 2 $\times$ 9 - 3y - 18 = 0 ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

From the graph, y-intercept of the line = OB = -6.

Hence, y-intercept of the line = -6.