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

Given:

Vertices of the quadrilateral $ABCD$ are $A(-3, 2), B(5, 4), C(7, -6)$ and $D(-5, -4)$.

To do:

We have to find the area of the quadrilateral.

Solution:

Join $A$ and $C$ to get two triangles $ABC$ and $ADC$.

This implies,

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

We know that,

Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by, 

Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

Therefore,

Area of triangle \( ABC=\frac{1}{2}[-3(4+6)+5(-6-2)+7(2-4)] \)

\( =\frac{1}{2}[-3(10)+5(-8)+7(-2)] \)

\( =\frac{1}{2}[-30-40-14] \)

\( =\frac{1}{2} \times (-84) \)

\( =42 \) sq. units.

Area of triangle \( ADC=\frac{1}{2}[-3(-6+4)+7(-4-2)+-5(2+6)] \)

\( =\frac{1}{2}[-3(-2)+7(-6)+(-5)(8)] \)

\( =\frac{1}{2}[6-42-40] \)

\( =\frac{1}{2} \times (-76) \)

\( =38 \) sq. units.

Therefore,

The area of the quadrilateral $ABCD=42+38=80$ sq. units.

The area of the quadrilateral $ABCD$ is $80$ sq. units.