\(AB=\sqrt{\left(2-1\right)^2+\left(1-2\right)^2}=\sqrt{2}\)
\(AC=\sqrt{\left(4-1\right)^2+\left(5-2\right)^2}=3\sqrt{2}\)
\(BC=\sqrt{\left(4-2\right)^2+\left(5-1\right)^2}=\sqrt{4^2+2^2}=2\sqrt{5}\)
Vì AB^2+AC^2=BC^2
nên ΔABC vuông tại A
\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC=\dfrac{1}{2}\cdot3\cdot2=3\)