Question

Tính diện tích tam giác ABC biết A(3;-4),B(1;5),C(3;1).

Answer 1

\(AB=\sqrt{\left(1-3\right)^2+\left(5+4\right)^2}=\sqrt{85}\)

\(BC=\sqrt{\left(3-1\right)^2+\left(1-5\right)^2}=2\sqrt{5}\)

\(CA=\sqrt{\left(3-3\right)^2+\left(-4-1\right)^2}=5\)

\(p=\frac{\sqrt{85}+2\sqrt{5}+5}{2}\)

\(p-AB=\frac{2\sqrt{5}+5-\sqrt{85}}{2};p-BC=\frac{\sqrt{85}+5-2\sqrt{5}}{2};p-CA=\frac{\sqrt{85}+2\sqrt{5}-5}{2}\)

\(\Rightarrow S_{ABC}=\sqrt{p\left(p-AB\right)\left(p-BC\right)\left(p-CA\right)}=...\)