Answer
The triangle is a right triangle.
Area =$\displaystyle \frac{41}{2}$
Work Step by Step
$d(A, B) = \sqrt{(11-6)^{2}+(-3-(-7))^{2}} = \sqrt{5^{2}+4^{2}}$
$= \sqrt{25+16} = \sqrt{41}.$
$d(A, C) = \sqrt{(2-6)^{2}+(-2-(-7))^{2}} = \sqrt{(-4)^{2}+5^{2}}$
$= \sqrt{16+25} = \sqrt{41}$;
$d(B, C)=\sqrt{(2-11)^{2}+(-2-(-3))^{2}}=\sqrt{(-9)^{2}+1^{2}}$
$=\sqrt{81+1}=\sqrt{82}$.
$[d(A, B)]^{2}+[d(A, C)]^{2}=41+41=82,$
$[d(B, C)]^{2}=82$,
Since$\qquad [d(A, B)]^{2}+[d(A, C)]^{2}=[d(B, C)]^{2}$,
we conclude that the triangle is a right triangle.
The Area is$ =\displaystyle \frac{1}{2}$ (base)$\times$(height).
$A = \displaystyle \frac{1}{2}(\sqrt{41}\cdot\sqrt{41})=\frac{41}{2}$.
