## College Algebra 7th Edition

The triangle is a right triangle. Area =$\displaystyle \frac{41}{2}$
$d(A, B) = \sqrt{(11-6)^{2}+(-3-(-7))^{2}} = \sqrt{5^{2}+4^{2}}$ $= \sqrt{25+16} = \sqrt{41}.$ $d(A, C) = \sqrt{(2-6)^{2}+(-2-(-7))^{2}} = \sqrt{(-4)^{2}+5^{2}}$ $= \sqrt{16+25} = \sqrt{41}$; $d(B, C)=\sqrt{(2-11)^{2}+(-2-(-3))^{2}}=\sqrt{(-9)^{2}+1^{2}}$ $=\sqrt{81+1}=\sqrt{82}$. $[d(A, B)]^{2}+[d(A, C)]^{2}=41+41=82,$ $[d(B, C)]^{2}=82$, 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). $A = \displaystyle \frac{1}{2}(\sqrt{41}\cdot\sqrt{41})=\frac{41}{2}$.