Solve the following system of linear equations graphically :

2x - y - 4 = 0, x + y + 1 = 0.

Hence, find the area of the triangle formed by these lines and the y-axis.

Given,

⇒ 2x - y - 4 = 0

⇒ y = 2x - 4 …………(1)

When x = 1, y = 2(1) - 4 = 2 - 4 = -2,

x = 2, y = 2(2) - 4 = 4 - 4 = 0,

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

Table of values for equation (1)

Steps of construction :

1. Plot the points (1, -2), (2, 0) and (3, 2).

2. Join the points.

Given,

⇒ x + y + 1 = 0

⇒ y = -(x + 1) ………….(2)

When, x = -1, y = -(-1 + 1) = 0,

x = 0, y = -(0 + 1) = -1,

x = 1, y = -(1 + 1) = -2.

Table of values for equation (2)

Steps of construction :

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

2. Join the points.

From graph,

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

ABC are the vertices of triangle.

From A, draw AD perpendicular to BC.

AD = 1 unit and BC = 3 units.

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

= $\dfrac{1}{2}$ × BC × AD

= $\dfrac{1}{2}$ × 3 × 1

= $\dfrac{3}{2}$ sq. units

Hence, point of intersection = (1, -2) and area of triangle = $\dfrac{3}{2}$ sq. units