Question

A triangle is formed by the lines x + 2y – 3 = 0, 3x – 2y = –7 and y = –1. Find the area of this triangle.

Answer 1

Given: Lines: x + 2y – 3 = 0, 3x – 2y = –7, y = –1.

Step-wise calculation:

1. Find intersections with y = –1:

With x + 2y – 3 = 0:

x + 2(–1) – 3 = 0

⇒ x – 2 – 3 = 0

⇒ x = 5

⇒ A(5, –1)

With 3x – 2y = –7:

3x – 2(–1) = –7

⇒ 3x + 2 = –7

⇒ 3x = –9

⇒ x = –3

⇒ B(–3, –1)

2. Find intersection of the other two lines:

Solve x + 2y = 3 and 3x – 2y = –7.

Add: 4x = –4

⇒ x = –1

Then y: –1 + 2y = 3

⇒ 2y = 4

⇒ y = 2

⇒ C(–1, 2)

3. Base and height:

Base AB lies on y = –1:

Length AB = Distance between (5, –1)

And (–3, –1) = 5 – (–3) = 8

Height = Vertical distance from C(–1, 2) to line y = –1

= 2 – (–1)

= 3

4. Area = `1/2` × base × height 

= `1/2 xx 8 xx 3`

= 12

Area of the triangle = 12 square units.