Answer
$d(A,B)=10\sqrt{2}$
$d(B,C)=10\sqrt{2}$
$d(A,C)=20$
Since $( (10\sqrt{2})^{2}+(10\sqrt{2})^{2}=(20)^{2}$,
the triangle is a right triangle
Area =$100 $ 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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$d(A,B)=\sqrt{(12-(-2))^{2}+(3-5)^{2}}=\sqrt{14^{2}+(-2)^{2}}$
$=\sqrt{196+4}=\sqrt{200}=10\sqrt{2}$
$d(B,C)=\sqrt{(10-12)^{2}+(-14)^{2}}=\sqrt{4+196}=\sqrt{200}=10\sqrt{2}$
$d(A,C)=\sqrt{(10-(-2))^{2}+(-11-5)^{2}}=\sqrt{12^{2}+(-16)^{2}}$
$=\sqrt{144+256}=\sqrt{400}=20$
Since $(10\sqrt{2})^{2}+(10\sqrt{2})^{2}=200+200=400=20^{2}$
the triangle is a right triangle
Area = $\displaystyle \frac{1}{2}ab=\frac{1}{2}(10\sqrt{2})(10\sqrt{2})=\frac{1}{2}(100\cdot 2)=100$
