Question

Find the area of the quadrilateral (in sq. units) whose vertices taken in the order are $A(-3,2),B(5,4),C(7,-6)$ and $D(-5,-4)$.

Answer 1

Let
the points be $A(-3,2),B(5,4),C(7,-6)$ and $D(-5,-4)$
The quadrilateral ABCD can be divided into triangles $ABC$ and $ACD$ and hence the area of the quadrilateral is the sum of the areas of the two triangles.

Area of a triangle with vertices $(x_1,y_1)$ ; $(x_2,y_2)$ and $(x_3,y_3)$ is $\left|\frac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2}\right|$
Hence, Area of triangle ABC $=\left|\frac{(-3)(4+6)+\left(5\right)(-6-2)+\left(7\right)(2-4)}{2}\right|=\left|\frac{-30-40-14}{2}\right|=\frac{84}{2}=42sq.units$
And, Area of triangle ACD $=\left|\frac{(-3)(-6+4)+\left(7\right)(-4-2)+(-5)(2+6)}{2}\right|=\left|\frac{6-42-40}{2}\right|=\frac{76}{2}=38sq.units$
Hence,
Area of quadrilateral =Area of triangle ACD + Area of triangle ABC

$=42+38=80sq.units$