## Calculus: Early Transcendentals 8th Edition

The triangle with vertices $A(0,2)$, $B(-3,-1)$ and $C(-4,3)$ is isosceles.
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.