JEE Main 2026 — Straight Lines Question with Solution

The given equations of the half-lines are $x-\sqrt{3}y=\alpha$ for $y\ge 0$ and $x+\sqrt{3}y=\alpha$ for $y\le 0$.

The point of intersection $P$ is obtained by setting $y=0$, which gives $x=\alpha$. Thus, $P\equiv (\alpha ,0)$.

The angle bisector of these half-lines is the x-axis ($y=0$).

The slope of the first half-line is $\frac{1}{\sqrt{3}}$, so it makes an angle of $30^{\circ }$ with the positive x-axis. The second half-line has a slope of $-\frac{1}{\sqrt{3}}$, making an angle of $-30^{\circ }$ with the positive x-axis.

Since $A$ and $B$ are at a distance $\alpha$ from $P$, their coordinates can be written using the parametric form of a straight line:
$A\equiv (\alpha +\alpha \cos 30^{\circ },0+\alpha \sin 30^{\circ })\equiv \left(\alpha +\frac{\alpha \sqrt{3}}{2},\frac{\alpha }{2}\right)$
$B\equiv (\alpha +\alpha \cos (-30^{\circ }),0+\alpha \sin (-30^{\circ }))\equiv \left(\alpha +\frac{\alpha \sqrt{3}}{2},-\frac{\alpha }{2}\right)$

The line segment $AB$ is vertical and intersects the angle bisector (x-axis) at $Q$. Thus, the coordinates of $Q$ are $\left(\alpha +\frac{\alpha \sqrt{3}}{2},0\right)$.

The distance $PQ$ is the difference in their x-coordinates:
$PQ=\left(\alpha +\frac{\alpha \sqrt{3}}{2}\right)-\alpha =\frac{\alpha \sqrt{3}}{2}$

Given $PQ=\frac{9}{2}$, we have:
$\frac{\alpha \sqrt{3}}{2}=\frac{9}{2}\Rightarrow \alpha \sqrt{3}=9\Rightarrow \alpha =3\sqrt{3}$

In $△PAB$, $PA=PB=\alpha$ and $\angle APB=30^{\circ }-(-30^{\circ })=60^{\circ }$. Therefore, $△PAB$ is an equilateral triangle with side length $\alpha$.

The circumradius $R$ of an equilateral triangle of side $\alpha$ is given by $R=\frac{\alpha }{\sqrt{3}}$.

We need to find the value of $\frac{{\alpha }^2}{R}$:
$\frac{{\alpha }^2}{R}=\frac{{\alpha }^2}{\frac{\alpha }{\sqrt{3}}}=\alpha \sqrt{3}$

Since we already found $\alpha \sqrt{3}=9$, we get:
$\frac{{\alpha }^2}{R}=9$

Answer: $9$