JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Straight Lines

Set up coordinates with the two parallel lines horizontal: one at $y=0$ and the other at $y=5$, so their distance is 5 units. Place P at distance 1 from the line $y=0$, so $P=(0,1)$; then P is automatically at distance 4 from the line $y=5$.

Let $Q=(q,0)$ lie on the near line. Since $PQR$ is equilateral, R is obtained by rotating $Q-P$ about P through $60^{\circ }$:

gives


Since R lies on $y=5$:


Then


Since the triangle is equilateral, $(QR{)}^2=P{Q}^2=28.$