A(2;4); B(0;-2); C(5;3)
\(AB=\sqrt{\left(0-2\right)^2+\left(-2-4\right)^2}=2\sqrt{10}\)
\(AC=\sqrt{\left(5-2\right)^2+\left(3-4\right)^2}=\sqrt{10}\)
\(BC=\sqrt{\left(5-0\right)^2+\left(3+2\right)^2}=5\sqrt{2}\)
Xét ΔABC có \(AB^2+AC^2=BC^2\)
nên ΔABC vuông tại A
=>\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC=\dfrac{1}{2}\cdot2\sqrt{10}\cdot\sqrt{10}=10\)