Answer
The area of the triangle is $26$ 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 given by:
$\text{Area}=\pm \frac{1}{2}\left| \begin{matrix}
{{x}_{1}} & {{y}_{1}} & 1 \\
{{x}_{2}} & {{y}_{2}} & 1 \\
{{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \right|$
The area of a triangle whose vertices are $\left( 3,-5 \right),\left( 2,6 \right)\text{ and }\left( -3,5 \right)$ is given by:
$\text{Area}=\pm \frac{1}{2}\left| \begin{matrix}
1 & 1 & 1 \\
-2 & -3 & 1 \\
11 & -3 & 1 \\
\end{matrix} \right|$
Calculate the above determinant by expanding along the first row as follows:
$\begin{align}
& \text{Area}=\frac{1}{2}\left[ 1\left| \begin{matrix}
-3 & 1 \\
-3 & 1 \\
\end{matrix} \right|-1\left| \begin{matrix}
-2 & 1 \\
11 & 1 \\
\end{matrix} \right|+1\left| \begin{matrix}
-2 & -3 \\
11 & -3 \\
\end{matrix} \right| \right] \\
& =\frac{1}{2}\left[ 1\left( -3+3 \right)-1\left( -2-11 \right)+\left( 6+33 \right) \right] \\
& =\frac{1}{2}\left[ 1\left( 0 \right)-1\left( -13 \right)+39 \right] \\
& =\frac{1}{2}\left[ 13+39 \right]
\end{align}$
Next, simplify it,
$\begin{align}
& \text{Area}=\frac{1}{2}\left[ 13+39 \right] \\
& =\frac{1}{2}\times 52 \\
& =26
\end{align}$
Hence, $\text{Area}=26\text{ square units}$.
