Question

Find the area of a triangle whose vertices is (1, –1), (–4, 6) and (–3, –5).

Answer 1

Given: Vertices A(1, –1), B(–4, 6), C(–3, –5).

Use the coordinate area formula: 

ar(ΔABC) = `1/2 | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|`.

Let x₁ = 1, y₁ = –1

x₂ = –4, y₂ = 6

x₃ = –3, y₃ = –5

Compute each term:

x₁(y₂ – y₃) = 1(6 – (–5)) 

= 1(11)

= 11

x₂(y₃ – y₁) = –4((–5) – (–1)) 

= −4(–4)

= 16

x₃(y₁ – y₂) = –3((–1) – 6) 

= −3(–7)

= 21

Sum = 11 + 16 + 21

= 48

Area = `1/2 |48|`

= 24

The area of the triangle is 24 square units.