Let B and C be the two points on the line y + x =0 such that B and...

On plotting the diagram of the given data we get,

Now given, $x = y$ and $y - 2x = 2$

On solving we get, $A ( - 2, - 2 )$

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

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

$⇒ h = 2\sqrt{2}$

Given $∆ ABC$ is an equilateral triangle,

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

$⇒ \sin \left({60}^{°}\right) = \dfrac{h}{AB} ( ∵ ∠ A = ∠ B = ∠ C = {60}^{° )}$

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

$⇒$Area of $\Delta = \dfrac{\sqrt{3}}{4}\dfrac{{h}^{2}}{\sin^{2}{60}^{∘}} = \dfrac{\sqrt{3}}{4} \times \dfrac{32}{3}$

So, Area of $∆ ABC = \dfrac{8}{\sqrt{3}}$