Answer
The triangle with vertices $A(0,2)$, $B(-3,-1)$ and $C(-4,3)$ is isosceles.
Work Step by Step
Given: The triangle with vertices $A(0,2)$, $B(-3,-1)$ and $C(-4,3)$ is isosceles.
An isosceles is a triangle that having at least two sides of equal length.
Consider AB, BC and AC are three sides of a triangle.
Use the distance formula $d=\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ to calculate each side.
Therefore,
For side AB: $AB=\sqrt {(-3-0)^{2}+(-1-2)^{2}}=\sqrt {9+9}=3\sqrt 2$
For side BC: $BC=\sqrt {(-4-(-3))^{2}+(3-(-1))^{2}}=\sqrt {1+16}=\sqrt {17}$
For side AC: $AC=\sqrt {(-4-0)^{2}+(3-2)^{2}}=\sqrt {16+1}=\sqrt {17}$
From the above calculations, we conclude that $BC=AC$
Hence, the triangle with vertices $A(0,2)$, $B(-3,-1)$ and $C(-4,3)$ is isosceles.