A(0;4); B(-4;2); C(1;2)
\(AB=\sqrt{\left(-4-0\right)^2+\left(2-4\right)^2}=\sqrt{4^2+2^2}=2\sqrt{5}\)
\(AC=\sqrt{\left(1-0\right)^2+\left(2-4\right)^2}=\sqrt{1+4}=\sqrt{5}\)
\(BC=\sqrt{\left(1+4\right)^2+\left(2-2\right)^2}=5\)
Vì \(AB^2+AC^2=BC^2\)
nên ΔABC vuông tại A