Question

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

Answer 1

A(-5,7) , B (-4,-5) and C (4,5)  are the vertices of .  Δ ABC 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