Question

Find the area of the quadrilateral whose vertices, taken in order, are $(-4,-2),(-3,-5),(3,-2)$ and $(2,3)$.

Answer 1

$A(-4,-2)$ $B=(-3,-5)$ $C=(3,-2)$ $D=(2,3)$

Diagonal AC is horizontal of length

$\sqrt{(3-(-4){)}^2+(-2-(-2){)}^2}$

$=7$

Vertical distance between B & D is

$3+5=8$

Area $=\frac{1}{2}(7\times 8)=28$ sq. units.