Question

A straight line through the point $(2,2)$ intersects the lines $\sqrt{3}x+y=0$ and $\sqrt{3}x-y=0$ at the points $A$ and $B$. The equation to the line $AB$ so that the $△OAB$ is equilateral is

• A. $x-2=0$
• B. $y-2=0$
• C. $x+y-4=0$
• D. $x+y+4=0$

Answer 1

Answer: (B)

$y-2=0$

Given line is passing through point $(2,2)$

and intersect with $\sqrt{3}x+y=0$ and $\sqrt{3}x-y=0$

I am assuming that O is the origin AB so (or such) that the triangle ΔAOB is equilateral.

Given $\sqrt{3}x+y=0,slope=\sqrt{(}3),\Rightarrow \alpha ={\tan }^{-1}(\sqrt{3})=60$

similarly

$\sqrt{3}x-y=0,slope=-\sqrt{(}3),\Rightarrow \beta ={\tan }^{-1}(-\sqrt{3})=60$

$\Rightarrow \angle BOA=180-60-60=60$

For $\Delta OAB$ to be equilateral,Line AB has to parallel to x-axis

Hence, the eq of line AB that passes through (2,2) and parallel to x-axis is

$y=2$

$y-2=0$