JEE Main 2025 — Straight Lines Question with Solution

$\begin{aligned}
& \mathrm{m}_{\mathrm{PR}}=2-\sqrt{3}=\tan 15^{\circ} \\ & \therefore \frac{\alpha}{2}=15^{\circ} \quad \Rightarrow \alpha=30^{\circ}
\end{aligned}$<br/>equationofPR:<br/>$\begin{aligned}
& y=\tan 15^{\circ}(x-a) \\ & y=(2-\sqrt{3})(x-a)
\end{aligned}$<br/>$\perp$distancefromorigin$=\frac{1}{\sqrt{2}}$<br/>$\begin{aligned} & \left|\frac{\sqrt{3} a-2 a}{\sqrt{4+3-4 \sqrt{3}+1}}\right|=\frac{1}{\sqrt{2}} \\ & \frac{|a|(2-\sqrt{3})}{2 \sqrt{(2-\sqrt{3})}}=\frac{1}{\sqrt{2}} \\ & |a|=\frac{\sqrt{2}}{\sqrt{2-\sqrt{3}}}=\sqrt{2}(\sqrt{2+\sqrt{3}}) \\ & a^2=2(2+\sqrt{3}) \\ & 3 a^2 \tan ^2 \alpha-2 \sqrt{3} \\ & 3 \times(4+2 \sqrt{3}) \cdot \frac{1}{3}-2 \sqrt{3}=4\end{aligned}$