Answer
$\text{Area}=28\text{ square units}$
Work Step by Step
The area of a triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\text{ and }\left( {{x}_{3}},{{y}_{3}} \right)$ is:
$\text{Area}=\pm \frac{1}{2}\left| \begin{matrix}
{{x}_{1}} & {{y}_{1}} & 1 \\
{{x}_{2}} & {{y}_{2}} & 1 \\
{{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \right|$
The area of a triangle whose vertices are $\left( 3,-5 \right),\left( 2,6 \right)\text{ and }\left( -3,5 \right)$ is:
$\text{Area}=\pm \frac{1}{2}\left| \begin{matrix}
3 & -5 & 1 \\
2 & 6 & 1 \\
-3 & 5 & 1 \\
\end{matrix} \right|$
Calculate the above determinant by expanding along the first row as follows:
$\begin{align}
& \text{Area}=\frac{1}{2}\left[ 3\left| \begin{matrix}
6 & 1 \\
5 & 1 \\
\end{matrix} \right|+5\left| \begin{matrix}
2 & 1 \\
-3 & 1 \\
\end{matrix} \right|+1\left| \begin{matrix}
2 & 6 \\
-3 & 5 \\
\end{matrix} \right| \right] \\
& =\frac{1}{2}\left[ 3\left( 6-5 \right)+5\left( 2+3 \right)+\left( 10+18 \right) \right] \\
& =\frac{1}{2}\left[ 3\left( 1 \right)+5\left( 5 \right)+\left( 28 \right) \right] \\
& =\frac{1}{2}\left[ 3+25+28 \right]
\end{align}$
Next, we simplify it,
$\begin{align}
& \text{Area}=\frac{1}{2}\left[ 3+25+28 \right] \\
& =\frac{1}{2}\times 56 \\
& =28
\end{align}$
Therefore, $\text{Area}=28\text{ square units}$.
