Coordinate Geometry — Exercise 18.2

Draw the graph of the following linear equation :

2x + y + 3 = 0

Answer

The given equation 2x + y + 3 = 0, can be written as :

⇒ y = -2x - 3

When x = 0, y = -2(0) - 3 = -3,

x = 1, y = -2(1) - 3 = -2 - 3 = -5,

x = 2, y = -2(2) - 3 = -4 - 3 = -7.

Table of values :

Steps of construction :

1. Plot the points (0, -3), (1, -5) and (2, -7) on the graph.

2. Connect any two points by a straight line.

Observe that the third point lies on the straight line.

Hence, the graph of the given equation is shown in the adjoining figure.

Draw the graph of the following linear equation :

x - 5y - 4 = 0

Answer

The given equation x - 5y - 4 = 0, can be written as :

⇒ 5y = x - 4

⇒ y = $\dfrac{1}{5}$(x - 4)

When x = -6, y = $\dfrac{1}{5}(-6 - 4) = -\dfrac{10}{5}$ = -2,

x = -1, y = $\dfrac{1}{5}(-1 - 4) = -\dfrac{5}{5}$ = -1,

x = 4, y = $\dfrac{1}{5}(4 - 4) = \dfrac{0}{5}$ = 0.

Table of values :

Steps of construction :

1. Plot the points (-6, -2), (-1, -1) and (4, 0) on the graph.

2. Connect any two points by a straight line.

Observe that the third point lies on the straight line.

Hence, the graph of the given equation is shown in the adjoining figure.

Draw the graph of 3y = 12 - 2x. Take 2 cm = 1 unit on both axes.

Answer

The above equation, 3y = 12 - 2x can be written as :

⇒ 3y = 12 - 2x

⇒ y = $\dfrac{12}{3} - \dfrac{2x}{3}$

⇒ y = 4 - $\dfrac{2x}{3}$.

When x = -3, y = 4 - $\dfrac{(2 \times -3)}{3} = 4 - \dfrac{-6}{3}$ = 4 - (-2) = 6,

x = 0, y = 4 - $\dfrac{2 \times 0}{3} = 4 - 0$ = 4,

x = 3, y = 4 - $\dfrac{2 \times 3}{3}$ = 4 - 2 = 2.

Table of values :

Steps of construction :

1. Plot the points (-3, 6), (0, 4) and (3, 2) on the graph.

2. Connect any two points by a straight line.

Observe that the third point lies on the straight line.

Hence, the graph of the given equation is shown in the adjoining figure.

Draw the graph of 5x + 6y - 30 = 0 and use it to find the area of the triangle formed by the line and coordinate axes.

Answer

The above equation, 5x + 6y - 30 = 0 can be written as :

⇒ 6y = -5x + 30

⇒ y = $\dfrac{1}{6}(-5x + 30)$

⇒ y = $-\dfrac{5x}{6} + 5$.

When x = 0, y = -$\dfrac{(5 \times 0)}{6} + 5$ = 0 + 5 = 5,

x = 6, y = -$\dfrac{5 \times 6}{6} + 5$ = -5 + 5 = 0,

x = 12, y = -$\dfrac{5 \times 12}{6}$ + 5 = -10 + 5 = -5.

Table of values :

Steps of construction :

1. Plot the points (0, 5), (6, 0) and (12, -5) on the graph.

2. Connect any two points by a straight line.

Observe that the third point lies on the straight line.

By formula,

Area of triangle = $\dfrac{1}{2} \times \text{ base} \times \text{ height}$

From graph,

Base = 6 units, Height = 5 units.

Area = $\dfrac{1}{2} \times 6 \times 5 = 15$ sq. units.

Hence, the graph of the given equation is shown in the adjoining figure and area of triangle = 15 sq. units.

Draw the graph of 4x - 3y + 12 = 0 and use it to find the area of the triangle formed by the line and co-ordinate axes. Take 2 cm = 1 unit on both axes.

Answer

The above equation, 4x - 3y + 12 = 0 can be written as :

⇒ 3y = 4x + 12

⇒ y = $\dfrac{4}{3}x + 4$

When x = -6, y = $\dfrac{4}{3} \times -6 + 4$ = -8 + 4 = -4,

x = -3, y = $\dfrac{4}{3} \times -3 + 4 =$ -4 + 4 = 0,

x = 0, y = $\dfrac{4}{3} \times 0 + 4 =$ 0 + 4 = 4.

Table of values :

Steps of construction :

1. Plot the points (-6, -4), (-3, 0) and (0, 4) on the graph.

2. Connect any two points by a straight line.

Observe that the third point lies on the straight line.

By formula,

Area of triangle = $\dfrac{1}{2} \times \text{ base} \times \text{ height}$

From graph,

Base = 3 units, Height = 4 units.

Area = $\dfrac{1}{2} \times 3 \times 4 = 6$ sq. units.

Hence, the graph of the given equation is shown in the adjoining figure and area of triangle = 6 sq. units.

Draw the graph of the equation y = 3x - 4. Find graphically

(i) the value of y when x = -1

(ii) the value of x when y = 5.

Answer

Equation :

⇒ y = 3x - 4

When x = 0, y = 3(0) - 4 = 0 - 4 = -4,

x = 2, y = 3(2) - 4 = 6 - 4 = 2,

Table of values :

Steps of construction :

1. Plot the points (0, -4) and (2, 2) on the graph.

2. Connect the two points by a straight line.

Hence, the graph of the given equation is shown in the adjoining figure.

(i) Steps of construction :

1. From point, P (x = -1) draw a line parallel to y-axis touching the graph. Mark the point as Q.

2. From Q draw a line parallel to x-axis, touching y-axis at point R (y = -7).

Hence, y = -7, when x = -1.

(ii) Steps of construction :

1. From point, S (y = 5) draw a line parallel to x-axis touching the graph. Mark the point as T.

2. From T draw a line parallel to y-axis, touching x-axis at point U (x = 3).

Hence, x = 3, when y = 5.

The graph of a linear equation in x and y passes through (4, 0) and (0, 3). Find the value of k if the graph passes through (k, 1.5).

Answer

Steps of construction :

1. Plot the points (4, 0) and (0, 3) on graph.

2. Connect the points through a straight line.

3. Take a point Q (y = 1.5) and draw a line parallel to x-axis touching the graph at point R.

4. From point R draw a line parallel to y-axis and touching x-axis at point S (x = 2).

From graph R = (2, 1.5)

Comparing point R with (k, 1.5), we get :

k = 2.

Hence, the value of k = 2.

Use the table given alongside to draw the graph of a straight line. Find, graphically, the values of a and b.

Answer

Steps of construction :

1. Plot the points (1, -2) and (3, 4) on graph paper.

2. Connect the points through a straight line.

3. Take a point C (x = 2) and draw a line parallel to y-axis touching the graph at point B.

4. From point B draw a line parallel to x-axis and touching y-axis at point A (y = 1).

From graph,

B = (2, 1)

Comparing it with (2, b) we get :

b = 1.

Also, the graph touches y-axis at (0, -5).

Comparing it with (a, -5) we get :

a = 0.

Hence, a = 0 and b = 1.