Find the area of the quadrilateral whose angular points are $\left(-1, 6\right)$, $\left(-3, -9\right)$, $\left(5, -8\right)$ and $\left(3, 9\right)$.
Let $\;$ $A \left(x_1, y_1\right) = \left(-1, 6\right)$, $\;$ $B \left(x_2, y_2\right) = \left(-3, -9\right)$, $\;$ $C \left(x_3, y_3\right) = \left(5, -8\right)$, $\;$ $D \left(x_4, y_4\right) = \left(3, 9\right)$
Area of quadrilateral $ABCD$ is
$\begin{aligned}
\Delta & = \dfrac{1}{2} \left(x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_1 - y_1 x_2 - y_2 x_3 - y_3 x_4 - y_4 x_1\right) \\\\
& = \dfrac{1}{2} \left[-1 \times \left(-9\right) - 3 \times \left(-8\right) + 5 \times 9 + 3 \times 6 -6 \times \left(-3\right) + 9 \times 5 + 8 \times 3 - 9 \times \left(-1\right) \right] \\\\
& = \dfrac{1}{2} \left[9 + 24 + 45 + 18 + 18 + 45 + 24 + 9\right] \\\\
& = \dfrac{1}{2} \times 192 = 96
\end{aligned}$