On graph paper, take 2 cm to represent one unit on both axes, draw the lines :

x + 3 = 0, y - 2 = 0, 2x + 3y = 12.

Write down the coordinates of the vertices of the triangle formed by these lines.

Given,

1st equation :

⇒ x + 3 = 0

⇒ x = -3 …………(1)

2nd equation :

⇒ y - 2 = 0

⇒ y = 2 ………….(2)

3rd equation :

⇒ 2x + 3y = 12

⇒ 3y = 12 - 2x

⇒ y = $\dfrac{12 - 2x}{3}$ ………(3)

When, x = -3, y = $\dfrac{12 - 2 \times (-3)}{3} = \dfrac{12 + 6}{3} = \dfrac{18}{3}$ = 6,

x = 0, y = $\dfrac{12 - 2 \times 0}{3} = \dfrac{12}{3}$ = 4,

x = 3, y = $\dfrac{12 - 2 \times 3}{3} = \dfrac{12 - 6}{3} = \dfrac{6}{3}$ = 2.

Table of values of equation (3) :

Steps of construction :

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

2. Join the points.

From graph,

ABC is the required triangle.

Hence, coordinates of vertices of triangle formed by these lines = (-3, 6), (-3, 2) and (3, 2).