Question

Find the area of the triangle whose vertices are $(-2,-3),(3,2)$ and $(-1,-8)$.

Answer 1

area of triangle $=|\frac{A_x(B_y-C_y)+B_x(C_y-A_y)+C_x(A_y-B_y)}{2}|$

$A(x,y)=(-2,-3),B(x,y)=(3,2),C(x,y)=(-1.-8)$

therefore substituting the values we get area $=|\frac{-2(2+8)+3(-8+2)-1(-3-2)}{2}|=|\frac{-20-18+5}{2}|=\frac{33}{2}$