Answer
A right isosceles triangle
(both).
Work Step by Step
$d(P_{1},P_{2})=\sqrt{(0-(-2))^{2}+(7-(-1))^{2}}$
$=\sqrt{2^{2}+8^{2}}=\sqrt{4+64}=\sqrt{68}=2\sqrt{17}$
$d(P_{1},P_{3})=\sqrt{(3-(-2))^{2}+(2-(-1))^{2}}$
$=\sqrt{5^{2}+3^{2}}=\sqrt{25+9}=\sqrt{34}$
$d(P_{2},P_{3})=\sqrt{(3-0)^{2}+(2-7)^{2}}$
$=\sqrt{3^{2}+(-5)^{2}}=\sqrt{9+25}=\sqrt{34}$
Two sides have the same length - isosceles triangle.
To check whether it is a right triangle,
the longest side is $c$=$2\sqrt{17}=\sqrt{68}$ . Check whether $a^{2}+b^{2}=c^{2}$
$a^{2}+b^{2}=(\sqrt{34})^{2}+(\sqrt{34})^{2}=34+34=68=c^{2}$.
A right isosceles triangle
(both).
