Answer
$26$ square units.
Work Step by Step
The given coordinates are
$\Rightarrow (x_1,y_1)=(1,1)$
$\Rightarrow (x_2,y_2)=(-2,-3)$
$\Rightarrow (x_3,y_3)=(11,-3)$
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.
$\Rightarrow A=\pm \frac{1}{2}\begin{vmatrix}
1& 1& 1\\
-2& -3& 1\\
11& -3& 1
\end{vmatrix}$
First solve the determinant.
$\Rightarrow \begin{vmatrix}
1& 1& 1\\
-2& -3& 1\\
11& -3& 1
\end{vmatrix}$
Perform $R_2\rightarrow R_2+2R_1$ and $R_3\rightarrow R_3-11R_1$.
$\Rightarrow \begin{vmatrix}
1& 1& 1\\
-2+2(1)& -3+2(1)& 1+2(1)\\
11-11(1)& -3-11(1)& 1-11(1)
\end{vmatrix}$
Simplify.
$\Rightarrow \begin{vmatrix}
1& 1& 1\\
0& -1& 3\\
0& -14& -10
\end{vmatrix}$
Perform $R_3\rightarrow R_3-14R_2$.
$\Rightarrow \begin{vmatrix}
1& 1& 1\\
0& -1& 3\\
0-14(0)& -14-14(-1)& -10-14(3)
\end{vmatrix}$
Simplify.
$\Rightarrow \begin{vmatrix}
1& 1& 1\\
0& -1& 3\\
0& 0& -52
\end{vmatrix}$
All elements below the diagonal are zero.
Hence, the determinant is the multiplication of the main diagonal elements.
$\Rightarrow 1\cdot (-1)\cdot (-52) $
Simplify.
$\Rightarrow 52 $
Substitute back the above value into the formula.
$\Rightarrow \pm \frac{1}{2}(52)$
Simplify.
$\Rightarrow \pm 26$
We take the positive value.
$\Rightarrow 26$
Hence, the area of the triangle is $26$ square units.
