Có\(\overrightarrow{AB}\left(1;-3\right),\overrightarrow{AC}\left(6;2\right),\overrightarrow{BC}\left(5;5\right)\)
\(\left|\overrightarrow{AB}\right|=\sqrt{1^2+\left(-3\right)^2}=\sqrt{10}\)
tương tự \(\left|\overrightarrow{AC}\right|=2\sqrt{10},\left|\overrightarrow{BC}\right|=5\sqrt{2}\)
Có \(AB^2+AC^2=\left(\sqrt{10}\right)^2+\left(2\sqrt{10}\right)^2=50=BC^2\)
\(\Rightarrow\Delta ABC\) là tam giác vuông
\(P_{\Delta ABC}=2\sqrt{10}+\sqrt{10}+5\sqrt{2}=3\sqrt{10}+5\sqrt{2}\)
\(S_{\Delta ABC}=\frac{1}{2}.2\sqrt{10}.\sqrt{10}=10\)