Answer
$d(A,B)=\sqrt{130}$
$d(B,C)=\sqrt{26}$
$d(A,C)=2\sqrt{26}$
Since $( \sqrt{26})^{2}+(2\sqrt{26})^{2}=(\sqrt{130})^{2}$,
the triangle is a right triangle, and the hypotenuse is $\overline{AB}.$
Area =$26$ 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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$A=(-5,3),B=(6,0),C=(5,5)$
$d(A,B)=\sqrt{(6-(-5))^{2}+(0-3)^{2}}=\sqrt{11^{2}+(-3)^{2}}=\sqrt{121+9}=\sqrt{130}$
$d(B,C)=\sqrt{(5-6)^{2}+(5-0)^{2}} =\sqrt{(-1)^{2}+5^{2}}=\sqrt{1+25} =\sqrt{26}$
$d(A,C)=\sqrt{(5-(-5))^{2}+(5-3)^{2}}=\sqrt{10^{2}+2^{2}}=\sqrt{100+4}=\sqrt{104}=2\sqrt{26}$
Since
$a^{2}+b^{2}=(\sqrt{104})^{2}+(\sqrt{26})^{2}=104+26=130$
$c^{2}=(\sqrt{130})^{2}=130$
the triangle is a right triangle, hypotenuse is $\overline{AB}.$
Area = $\displaystyle \frac{1}{2}ab=\frac{1}{2}(\sqrt{26})(2\sqrt{26})=26$
