已知$s_{\triangle ABC}$三点的坐标，如何求它的面积.

公式如下：

$$a=\sqrt{|XA-XB|^2+|YA-YB|^2}$$ $$b=\sqrt{|XA-XC|^2+|YA-YC|^2}$$ $$c=\sqrt{|XB-XC|^2+|YB-YC|^2}$$ $$S=\frac{a+b+c}{2}$$ $$s_{\triangle abc}=\sqrt{S(S-a)(S-b)(S-c)}$$

合起来就是：

$$s_{\triangle ABC}=\sqrt{\frac{\sqrt{|XA-XB|^2+|YA-YB|^2}+\sqrt{||XA-XC|^2+|YA-YC|^2}+\sqrt{|XB-XC|^2+|YB-YC|^2}}{2}(\frac{\sqrt{|XA-XB|^2+|YA-YB|^2}+\sqrt{|XA-XC|^2+|YA-YC|^2}+\sqrt{|XB-XC|^2+|YB-YC|^2}}{2}-\sqrt{|XA-XB|^2+|YA-YB|^2})(\frac{\sqrt{|XA-XB|^2+|YA-YB|^2}+\sqrt{|XA-XC|^2+|YA-YC|^2}+\sqrt{|XB-XC|^2+|YB-YC|^2}}{2}-\sqrt{|XA-XC|^2+|YA-YC|^2})(\frac{\sqrt{|XA-XB|^2+|YA-YB|^2}+\sqrt{|XA-XC|^2+|YA-YC|^2}+\sqrt{|XB-XC|^2+|YB-YC|^2}}{2}-\sqrt{|XB-XC|^2+|YB-YC|^2})} $$

s_{\triangle abc}=\sqrt{\frac{\sqrt{|ax-bx|^2+|ay-by|^2}+\sqrt{|ax-cx|^2+|ay-cy|^2}+\sqrt{|bx-cx|^2+|by-cy|^2}}{2}(\frac{\sqrt{|ax-bx|^2+|ay-by|^2}+\sqrt{|ax-cx|^2+|ay-cy|^2}+\sqrt{|bx-cx|^2+|by-cy|^2}}{2}-\sqrt{|ax-bx|^2+|ay-by|^2})(\frac{\sqrt{|ax-bx|^2+|ay-by|^2}+\sqrt{|ax-cx|^2+|ay-cy|^2}+\sqrt{|bx-cx|^2+|by-cy|^2}}{2}-\sqrt{|ax-cx|^2+|ay-cy|^2})(\frac{\sqrt{|ax-bx|^2+|ay-by|^2}+\sqrt{|ax-cx|^2+|ay-cy|^2}+\sqrt{|bx-cx|^2+|by-cy|^2}}{2}-\sqrt{|bx-cx|^2+|by-cy|^2})}

a=\sqrt{|ax-bx|^2+|ay-by|^2}
b=\sqrt{|ax-cx|^2+|ay-cy|^2}
c=\sqrt{|bx-cx|^2+|by-cy|^2}
s=\frac{a+b+c}{2}
s_{\triangle abc}=\sqrt{s(s-a)(s-b)(s-c)}