Question

Find the area of a quadrilateral $ABCD$, the coordinates of whose vertices are $A(-3,2),B(5,4),C(7,-6)$ and $D(-5,-4)$.

Answer 1

|Area of triangle with vertices $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ is given by $\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

Area of quadrilateral $ABCD=$ Area of triangle $ABC+$Area of triangle $ADC$

$=\frac{1}{2}|-3(4+6)+5(-6-2)+7(2-4)|+\frac{1}{2}|-3(-4+6)-5(-6-2)+7(2+4)|$
$=\frac{1}{2}[|-3(10)+5(-8)+7(-2)|+-3(2)-5(-8)+7(6\left)|\right]$
$=\frac{1}{2}[|-30-40-14|+|-6+40+42|]$
$=\frac{1}{2}[|-84|+|76|]$
$=\frac{1}{2}[84+76]$
$=\frac{160}{2}$
$=80sq.units$