Bài 2:
\(AB=\sqrt{\left(-5-2\right)^2+\left(0-1\right)^2}=5\sqrt{2}\)
\(AC=\sqrt{\left(1-2\right)^2+\left(-3-1\right)^2}=\sqrt{17}\)
\(BC=\sqrt{\left(1+5\right)^2+\left(-3\right)^2}=\sqrt{45}\)
\(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{50+17-45}{2\cdot5\sqrt{2}\cdot\sqrt{17}}=\dfrac{11}{5\sqrt{34}}\)
=>\(sinA=\dfrac{27}{5\sqrt{34}}\)
\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinA=\dfrac{1}{2}\cdot\dfrac{27}{5\sqrt{34}}\cdot5\sqrt{2}\cdot\sqrt{17}\)
\(=13.5\)