\(AB=\sqrt{\left(-1-2\right)^2+\left(4-3\right)^2}=\sqrt{10}\)
\(BC=\sqrt{\left(3+1\right)^2+\left(0-4\right)^2}=4\sqrt{2}\)
\(CA=\sqrt{\left(2-3\right)^2+\left(3-0\right)^2}=\sqrt{10}\)
\(\Rightarrow p=\frac{AB+BC+CA}{2}=\sqrt{10}+2\sqrt{2}\)
\(p-AB=2\sqrt{2};p-BC=\sqrt{10}-2\sqrt{2};p-CA=2\sqrt{2}\)
\(S_{ABC}=\sqrt{p\left(p-AB\right)\left(p-BC\right)\left(p-CA\right)}\)
\(=\sqrt{\left(\sqrt{10}+2\sqrt{2}\right).2\sqrt{2}.\left(10-2\sqrt{2}\right).2\sqrt{2}}=4\)
Vậy diện tích tam giác ABC là 4
