Answer
A right triangle.
Work Step by Step
$d(P_{1},P_{2})=\sqrt{(-4-7)^{2}+(0-2)^{2}}$
$=\sqrt{(-11)^{2}+(-2)^{2}}=\sqrt{121+4}=\sqrt{125}=5\sqrt{5}$
$d(P_{1},P_{3})=\sqrt{(4-7)^{2}+(6-2)^{2}}$
$=\sqrt{(-3)^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5$
$d(P_{2},P_{3})=\sqrt{(4-(-4))^{2}+(6-0)^{2}}$
$=\sqrt{8^{2}+6^{2}}=\sqrt{64+36}=\sqrt{100}=10$
No two sides have the same length - not an isosceles triangle.
To check whether it is a right triangle,
the longest side is $c$=$5\sqrt{5}=\sqrt{125}$ . Check whether $a^{2}+b^{2}=c^{2}$
$a^{2}+b^{2}=(10)^{2}+(5)^{2}=100+25=125=c^{2}$.
A right triangle.
