JEE MAIN - Mathematics (2023 - 29th January Morning Shift - No. 2)
Let $$B$$ and $$C$$ be the two points on the line $$y+x=0$$ such that $$B$$ and $$C$$ are symmetric with respect to the origin. Suppose $$A$$ is a point on $$y-2 x=2$$ such that $$\triangle A B C$$ is an equilateral triangle. Then, the area of the $$\triangle A B C$$ is :
$$\frac{10}{\sqrt{3}}$$
$$2 \sqrt{3}$$
$$3 \sqrt{3}$$
$$\frac{8}{\sqrt{3}}$$
Explanation
Origin $(O)$ is mid-point of $B C(x+y=0)$.
$A$ lies on perpendicular bisector of $B C$, which is $x-y=0$
A is point of intersection of $x-y=0$ and $y-2 x=2$
$\therefore A \equiv(-2,-2)$
Let $h=A O=\frac{-2-2}{\sqrt{1^{2}+1^{2}}}=2 \sqrt{2}$
$$ \text { Area }=\frac{h^{2}}{\sqrt{3}}=\frac{8}{\sqrt{3}} $$
$A$ lies on perpendicular bisector of $B C$, which is $x-y=0$
A is point of intersection of $x-y=0$ and $y-2 x=2$
$\therefore A \equiv(-2,-2)$
Let $h=A O=\frac{-2-2}{\sqrt{1^{2}+1^{2}}}=2 \sqrt{2}$
$$ \text { Area }=\frac{h^{2}}{\sqrt{3}}=\frac{8}{\sqrt{3}} $$
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