Solve graphically the following equations :

x + 2y = 4, 3x - 2y = 4.

Take 2 cm = 1 unit on each axis. Write down the area of the triangle formed by the lines and the x-axis.

Given,

⇒ x + 2y = 4

⇒ 2y = 4 - x

⇒ y = $\dfrac{4 - x}{2}$ ………..(1)

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

x = 2, y = $\dfrac{4 - 2}{2} = \dfrac{2}{2}$ = 1,

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

Table of values for equation (1)

Steps of construction :

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

2. Join the points.

Given,

⇒ 3x - 2y = 4

⇒ 2y = 3x - 4

⇒ y = $\dfrac{3x - 4}{2}$ ……….(2)

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

x = 2, y = $\dfrac{3 \times 2 - 4}{2} = \dfrac{6 - 4}{2} = \dfrac{2}{2}$ = 1,

x = 4, y = $\dfrac{3 \times 4 - 4}{2} = \dfrac{12 - 4}{2} = \dfrac{8}{2}$ = 4.

Table of values for equation (2)

Steps of construction :

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

2. Join the points.

From graph,

A(2, 1) is the point of intersection of lines.

ABC is the triangle between lines and x-axis.

From A, draw AD perpendicular to BC.

AD = 1 unit

BC = 2.7 unit

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

= $\dfrac{1}{2} \times AD \times BC$

= $\dfrac{1}{2} \times 1 \times 2.7$

= 1.35 sq units.

Hence, point of intersection is x = 2, y = 1 and area = 1.35 sq units.