Answer
$d(A,B)=\sqrt{145}$
$d(B,C)=2\sqrt{29}$
$d(A,C)=\sqrt{29}$
Since $(\sqrt{116})^{2}+(\sqrt{29})^{2}=(\sqrt{145})^{2}$,
the triangle is a right triangle, hypotenuse is $\overline{AB}.$
Area =$29$ sq. units
Work Step by Step
$d(P_{1},P_{2})=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
In a right triangle with legs a and b, hypotenuse c,
$ a^{2}+b^{2}=c^{2}\quad$ (Pythagorean Th.)
Area = $\displaystyle \frac{1}{2}ab$
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$A=(-6,3),B=(3,-5),C=(-1,5)$
$d(A,B)=\sqrt{(3-(-6))^{2}+(-5-3)^{2}}=\sqrt{9^{2}+(-8)^{2}}$
$=\sqrt{81+64}=\sqrt{145}$
$d(B,C)=\sqrt{(-1-3)^{2}+(5-(-5))^{2}}=\sqrt{(-4)^{2}+10^{2}}$
$=\sqrt{16+100}=\sqrt{116}=2\sqrt{29}$
$d(A,C)=\sqrt{(-1-(-6))^{2}+(5-3)^{2}}=\sqrt{5^{2}+2^{2}}$
$=\sqrt{25+4}=\sqrt{29}$
Since
$a^{2}+b^{2}=(\sqrt{116})^{2}+(\sqrt{29})^{2}=145$
$c^{2}=(\sqrt{145})^{2}=145$
the triangle is a right triangle, hypotenuse is $\overline{AB}.$
Area = $\displaystyle \frac{1}{2}ab=\frac{1}{2}(2\sqrt{29})(\sqrt{29})=29$
