Answer
$A_{ABC}=12$
Work Step by Step
$A(4,2)$
$B(4, 8)$
$C(8, 5)$
$A_{ABC}=\pm\frac{1}{2}\begin{vmatrix}
4 & 2 & 1\\
4 & 8 & 1 \\
8 & 5 & 1
\end{vmatrix}$
$=\pm\frac{1}{2}[4(8-5)-2(4-8)+1(20-64)]$
$=\pm\frac{1}{2}(12+8-44)$
$=\pm\frac{1}{2}.(-24)$
$=\pm 12$
Hence $A_{ABC}=12$
