Answer
The sides $AC$ and $CB$ have the same length, so the triangle is isosceles.
Work Step by Step
Find the lengths of all three sides.
$d(A, B)=\sqrt{(-3-0)^{2}+(-1-2)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$.
Find the lengths of all three sides.
$d(A, B)=\sqrt{(-3-0)^{2}+(-1-2)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}$
$=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$.
$d(C, B)=\sqrt{(-3-(-4))^{2}+(-1-3)^{2}}=\sqrt{1^{2}+(-4)^{2}}$
$=\sqrt{1+16}=\sqrt{17}$.
$d(A, C)=\sqrt{(0-(-4))^{2}+(2-3)^{2}}=\sqrt{4^{2}+(-1)^{2}}$
$=\sqrt{16+1}=\sqrt{17}$.
The sides $AC$ and $CB$ have the same length, so the triangle is isosceles.
