\(\left\{{}\begin{matrix}\overrightarrow{AB}=\left(\frac{3}{2};\frac{3}{2}\right)\\\overrightarrow{AC}=\left(\frac{1}{2};-\frac{1}{2}\right)\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{AB}.\overrightarrow{AC}=\left(\frac{3}{2}\right).\left(\frac{1}{2}\right)+\left(\frac{3}{2}\right).\left(-\frac{1}{2}\right)=0\)
\(\Rightarrow AB\perp AC\) hay tam giác vuông tại A
\(\left\{{}\begin{matrix}AB=\sqrt{\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^2}=\frac{3\sqrt{2}}{2}\\AC=\sqrt{\left(\frac{1}{2}\right)^2+\left(-\frac{1}{2}\right)^2}=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(S_{ABC}=\frac{1}{2}AB.AC=\frac{3}{4}\)
