The triangle formed by $x^2-3y^2=0andthelinex=ais$

Given Equations is

$\begin{array}{l}{x}^2-{\left(\sqrt{3}y\right)}^2=0\\ \Rightarrow \left(x+\sqrt{3}y\right)\left(x-\sqrt{3}y\right)=0\\ \Rightarrow x+\sqrt{3}y=0--\left(1\right),x-\sqrt{3}y=0--\left(2\right)andx-a=0--\left(3\right)\\ Angleb/w\left(1\right)\&\left(3\right)\cos A=\frac{\left|1\left(1\right)+\sqrt{3}\left(0\right)\right|}{\sqrt{1+3}\sqrt{1+0}}=\frac{1}{2}\\ \therefore \angle A={60}^0\\ Angleb/w\left(2\right)\&\left(3\right)\cos B=\frac{\left|1\left(1\right)+\left(-3\right)\left(0\right)\right|}{\sqrt{1+3}\sqrt{1+0}}=\frac{1}{2}=\angle B={60}^0\\ \therefore \angle C={180}^0-\left(\angle A+\angle B\right)\\ \angle c={60}^0\\ \therefore {\Delta }^{le}isEquilateral\end{array}$