Find graphically the coordinates of the vertices of the triangle formed by the lines y - 2 = 0, 2y + x = 0 and y + 1 = 3 (x - 2). Hence, find the area of the triangle formed by these lines.

Given,

⇒ y - 2 = 0

⇒ y = 2 ……..(1)

Given,

⇒ 2y + x = 0

⇒ 2y = -x

⇒ y = -$\dfrac{x}{2}$ …………(2)

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

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

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

Table of equation (2) :

Steps of construction :

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

2. Join the points.

Given,

⇒ y + 1 = 3(x - 2)

⇒ y + 1 = 3x - 6

⇒ y = 3x - 6 - 1

⇒ y = 3x - 7 ………..(3)

When x = 1, y = 3 × 1 - 7 = 3 - 7 = -4,

x = 2, y = 3 × 2 - 7 = 6 - 7 = -1,

x = 3, y = 3 × 3 - 7 = 9 - 7 = 2.

Table of equation (3) :

Steps of construction :

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

2. Join the points.

From graph,

A(-4, 2), B(3, 2), C(2, -1) are the vertices of the triangle.

From C, draw CD perpendicular to AB.

As, 1 block = 1 unit.

AB = 7 units and CD = 3 units

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

= $\dfrac{1}{2}$ × AB × CD

= $\dfrac{1}{2}$ × 7 × 3

= $\dfrac{21}{2}$ = 10.5 sq. units

Hence, coordinates of the vertices of the triangle are (-4, 2), (3, 2), (2, -1) and area = 10.5 sq. units.