Answer
The triangle with vertices $A(0,2)$ $,$ $B(-3,-1)$ and $C(-4,3)$ is isosceles.
Work Step by Step
The vertices of the triangle are $A(0,2)$ $,$ $B(-3,-1)$ and $C(-4,3)$
Use the distance between two points formula, which is $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$, to find the distance between each point.
Find the distance between points $A$ and $B$:
$d_{AB}=\sqrt{(-3-0)^{2}+(-1-2)^{2}}=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$
Find the distance between points $B$ and $C$:
$d_{BC}=\sqrt{(-4+3)^{2}+(3+1)^{2}}=\sqrt{(-1)^{2}+4^{2}}=\sqrt{1+16}=\sqrt{17}$
Find the distance between points $C$ and $A$:
$d_{CA}=\sqrt{(-4-0)^{2}+(3-2)^{2}}=\sqrt{(-4)^{2}+1^{2}}=\sqrt{16+1}=\sqrt{17}$
Since the sides $BC$ and $CA$ have the same length and the length of side $AB$ is different from these other two lenghts, the triangle is isosceles.
