Đáp án:
$\begin{array}{l}
\overrightarrow {AB} = \left( { - 4;4} \right);\overrightarrow {BC} = \left( {3; - 1} \right);\overrightarrow {AC} = \left( { - 1; - 3} \right)\\
\Rightarrow \overrightarrow {BC} .\overrightarrow {AC} = 3.\left( { - 1} \right) + \left( { - 1} \right).\left( { - 3} \right) = 0\\
\Rightarrow \overrightarrow {BC} \bot \overrightarrow {AC} \\
\Rightarrow BC \bot AC
\end{array}$
=> Tam giác ABC vuông tại C
$ \Rightarrow {S_{ABC}} = \frac{1}{2}.BC.AC = \frac{1}{2}.\sqrt {{3^2} + {{\left( { - 1} \right)}^2}} .\sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 3} \right)}^2}} = \frac{1}{2}.10 = 5$
