Answer
$d(A,B)=\sqrt{13}$
$d(B,C)=\sqrt{13}$
$d(A,C)=\sqrt{26}$
Since $(\sqrt{13})^{2}+(\sqrt{13})^{2}=(\sqrt{26})^{2}$,
the triangle is a right triangle
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 $(Pythagorean Th.)
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$A=(-2,5),B=(1,3),C=(-1,0)$
$d(A,B)=\sqrt{(1-(-2))^{2}+(3-5)^{2}}=\sqrt{3^{2}+(-2)^{2}}=\sqrt{9+4}=\sqrt{13}$
$d(B,C)=\sqrt{(-1-1)^{2}+(0-3)^{2}}=\sqrt{(-2)^{2}+(-3)^{2}}=\sqrt{4+9}=\sqrt{13}$
$d(A,C)=\sqrt{(-1-(-2))^{2}+(0-5)^{2}}=\sqrt{1^{2}+(-5)^{2}}=\sqrt{1+25}=\sqrt{26}$
Since $(\sqrt{13})^{2}+(\sqrt{13})^{2}=(\sqrt{26})^{2}$,
the triangle is a right triangle
