JEE Mathematics 2023 Question with Solution

Given: $B$ and $C$ lie on $x+y=0$ and are symmetric about the origin. Point $A$ lies on $y-2x=2$. Triangle $ABC$ is equilateral.

Find: Area of triangle $ABC$.

From the solution working, take the midpoint of $BC$ at the origin. Since $BC$ lies on $x+y=0$, the altitude from $A$ to $BC$ is perpendicular to $x+y=0$, hence it lies along $y=x$.

So $A$ is the intersection of

$$y=x$$

and

$$y-2x=2$$

Substituting,

$$x-2x=2$$ $$-x=2$$ $$x=-2, y=-2$$

Thus, $A=(-2,-2)$.

Now the height of the equilateral triangle is the perpendicular distance from $A(-2,-2)$ to the line $x+y=0$:

$$d=\frac{|1(-2)+1(-2)+0|}{\sqrt{1^2+1^2}}$$ $$d=\frac{4}{\sqrt{2}}=2\sqrt{2}$$

For an equilateral triangle,

$$h=\frac{\sqrt{3}}{2}s$$

So,

$$s=\frac{2h}{\sqrt{3}}=\frac{4\sqrt{2}}{\sqrt{3}}$$

Area is

$$\frac{\sqrt{3}}{4}s^2$$

Substituting $s$,

$$\text{Area}=\frac{\sqrt{3}}{4}\left(\frac{4\sqrt{2}}{\sqrt{3}}\right)^2$$ $$=\frac{\sqrt{3}}{4}\cdot\frac{32}{3}=\frac{8\sqrt{3}}{3}$$ $$=\frac{8}{\sqrt{3}}$$

Therefore, the area of triangle $ABC$ is $\frac{8}{\sqrt{3}}$. This matches option D.

The source options show option (3) in the question block, but the extracted solution concludes option D. The geometric working supports D.