Answer
$28$ square units.
Work Step by Step
The given coordinates are
$\Rightarrow (x_1,y_1)=(3,-5)$
$\Rightarrow (x_2,y_2)=(2,6)$
$\Rightarrow (x_3,y_3)=(-3,5)$
The formula for the area of the triangle is
$\Rightarrow A=\pm \frac{1}{2}\begin{vmatrix}
x_1& y_1& 1\\
x_2& y_2& 1\\
x_3& y_3& 1
\end{vmatrix}$
Plug all the values.
$\Rightarrow A=\pm \frac{1}{2}\begin{vmatrix}
3& -5& 1\\
2& 6& 1\\
-3& 5& 1
\end{vmatrix}$
First solve the determinant.
$\Rightarrow \begin{vmatrix}
3& -5& 1\\
2& 6& 1\\
-3& 5& 1
\end{vmatrix}$
Perform $R_2\rightarrow R_2-R_1(2/3)$ and $R_3\rightarrow R_3+R_1$.
$\Rightarrow \begin{vmatrix}
3& -5& 1\\
2-3(2/3)& 6-(-5)(2/3)& 1-1(2/3)\\
-3+3& 5-5& 1+1
\end{vmatrix}$
Simplify.
$\Rightarrow \begin{vmatrix}
3& -5& 1\\
0& 28/3& 1/3\\
0& 0& 2
\end{vmatrix}$
All elements below the diagonal are zero.
Hence, the determinant is the multiplication of the main diagonal elements.
$\Rightarrow 3\cdot \frac{28}{3}\cdot 2 $
Simplify.
$\Rightarrow 56 $
Substitute back the above value into the formula.
$\Rightarrow \pm \frac{1}{2}(56)$
Simplify.
$\Rightarrow \pm 28$
We take the positive value.
$\Rightarrow 28$
Hence, the area of the triangle is $28$ square units.
