Answer
$|AB|=|AC|=|BC|=2$
so the triangle is equilateral.
Work Step by Step
Apply the distance formula
$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
$|AB|=\sqrt{(x_{B}-x_{A})^{2}+(y_{B}-y_{A})^{2}}=\sqrt{(1-0)^{2}+(\sqrt{3}-0)^{2}}=\sqrt{4} =2$
$|AC|=\sqrt{(x_{C}-x_{A})^{2}+(y_{C}-y_{A})^{2}}=\sqrt{(2-0)^{2}+(0-0)^{2}}=\sqrt{4} =2$
$|BC|=\sqrt{(x_{C}-x_{B})^{2}+(y_{C}-y_{B})^{2}}=\sqrt{(2-1)^{2}+(0-\sqrt{3})^{2}}=\sqrt{1+3} =2$
