Question

Find the area of ΔABC whose vertices is A(–5, 7), B(–4, –5) and C(4, 5).

Find the area of a triangle whose vertices is (–5, 7), (–4, –5) and (4, 5).

Answer 1

A(–5, 7), B(–4, –5) and C(4, 5) are the vertices of ΔABC. Then

`(x_1=-5, y_1=7) . (x_2= -4, y_2=-5) and (x_3=4,y_3=5)`

Area of triangle ABC

`=1/2 { x_1(y_2-y_3) + x_2 (y_3-y_1) +x_3 (y_1-y_2)}`

`=1/2 {(-5)(-5-5) +(-4)(5-7)+4(7-(5))}`

`=1/2 {(-5) (-10) -4(-2)+4(12)}`

`=1/2 {50+8+48}`

`=1/2 (106)`

= 53 sq . units