Найди площадь треугольника ABC с вершинами : а)A(-2;4),B(4;9),C(5;-6)...

Решите задачу:

$a=|AB|=\sqrt{(4+2)^2+(9-4)^2}=\sqrt{36+25}=\sqrt{61}\\\\b=|AC|=\sqrt{(5+2)^2+(-6-4)^2}=\sqrt{49+100}=\sqrt{149}\\\\c=|BC|=\sqrt{(5-4)^2+(-6-9)^2}=\sqrt{1+225}=\sqrt{226}\\\\S=\sqrt{p(p-a)(p-b)(p-c)}\\\\p=\frac{1}{2}(\sqrt{61}+\sqrt{149}+\sqrt{226})\\\\S^2=\frac{1}{2}(\sqrt{61}+\sqrt{149}+\sqrt{226})\cdot \frac{1}{2}(-\sqrt{61}+\sqrt{149}+\sqrt{226})\cdot \\\\\cdot \frac{1}{2}(\sqrt{61}-\sqrt{149}+\sqrt{226})\cdot \frac{1}{2}(\sqrt{61}+\sqrt{149}-\sqrt{226})=$

$=\frac{1}{16}((\sqrt{149}+\sqrt{226})^2-61)(61-(\sqrt{149}-\sqrt{226})^2)=\\\\=\frac{1}{16}(149+226+2\sqrt{149\cdot 226}-61})(61-149-226+2\sqrt{149\cdot 226})=\\\\=\frac{1}{16}(314+2\sqrt{149\cdot 226})(-314+2\sqrt{149\cdot 226})=\\\\=\frac{1}{16}\cdot (4\cdot 149\cdot 226-314^2)=\frac{1}{16}(134696-98596)=\frac{36100}{16}=2256,25$