On plotting the diagram of the given data we get,

Now given, $\mathrm{x}=\mathrm{y}$ and $y-2x=2$

On solving we get, $A(-2,-2)$

Now height $h$ of $∆ABC$ $=$ Distance of point $A$ to origin,

$h=\sqrt{{\left(-2\right)}^2+(-2{)}^2}$

$\Rightarrow h=2\sqrt{2}$

Given $∆ABC$ is an equilateral triangle,

So, $\sin \left(B\right)=\frac{h}{AB}$

$\Rightarrow \sin \left({60}^{°}\right)=\frac{h}{AB} (\because \angle A=\angle B=\angle C={60}^{°)}$

Area of $\Delta =\frac{\sqrt{3}}{4}{\left(AB\right)}^2$

$\Rightarrow$Area of $\Delta =\frac{\sqrt{3}}{4}\frac{h^2}{{\sin }^2{60}^{\circ }}=\frac{\sqrt{3}}{4}\times \frac{32}{3}$

So, Area of $∆ABC=\frac{8}{\sqrt{3}}$