Answer
$A_{ABC}=25$
Work Step by Step
$A(-6,1)$
$B(-2, -6)$
$C(0, 3)$
$A_{ABC}=\pm\frac{1}{2}\begin{vmatrix}
-6 &1 & 1\\
-2 & -6 & 1 \\
0 & 3 & 1
\end{vmatrix}$
$=\pm\frac{1}{2}[-6(-6-3)-1(-2-0)+1(-6-0)]$
$=\pm\frac{1}{2}(54+2-6)$
$=\pm\frac{1}{2}.(50)$
$=\pm25$
Hence $A_{ABC}=25$.
