JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Straight Lines

Since $B$ and $C$ are symmetric about the origin on $x+y=0$, the origin is the midpoint of $BC$, and for an equilateral triangle $A$ lies on the perpendicular bisector of $BC$, the line $y=x$. Intersecting $y=x$ with $y-2x=2$ gives $A(-2,-2)$. The altitude is the distance from $A$ to $x+y=0$, namely $h=\frac{|-2-2|}{\sqrt{2}}=2\sqrt{2}$, so the side is $s=\frac{2h}{\sqrt{3}}=\frac{4\sqrt{2}}{\sqrt{3}}$. The area is $\frac{\sqrt{3}}{4}{s}^2=\frac{\sqrt{3}}{4}\cdot \frac{32}{3}=\frac{8}{\sqrt{3}}$.