JEE Main 2026 (Online) 21st January Morning ShiftQuestion 1
Let a point $A$ lie between the parallel lines $L_1$ and $L_2$ such that its distances from $L_1$ and $L_2$ are $6$ and $3$ units, respectively. Then the area (in sq. units) of the equilateral triangle $ABC$, where the points $B$ and $C$ lie on the lines $L_1$ and $L_2$, respectively, is :
Points and Straight Lines
Solution
Set-up. Take $A$ as the origin and the parallel lines horizontal: $L_1:\ y=6$ and $L_2:\ y=-3$. Let the side of the triangle be $s$, with $AB$ making angle $\theta$ with the $x$-axis; then $AC$ makes angle $\theta-60^\circ$.Step 1 - $B$ on $L_1$. $$s\sin\theta=6 .$$ Step 2 - $C$ on $L_2$. $$s\sin(\theta-60^\circ)=-3\ \Longrightarrow\ \tfrac12\,s\sin\theta-\tfrac{\sqrt3}{2}\,s\cos\theta=-3 .$$ Using $s\sin\theta=6$: $\ 3-\tfrac{\sqrt3}{2}s\cos\theta=-3\Rightarrow s\cos\theta=\tfrac{12}{\sqrt3}=4\sqrt3 .$ Step 3 - Side. $$s^2=(s\cos\theta)^2+(s\sin\theta)^2=48+36=84 .$$ Step 4 - Area. $$\text{Area}=\frac{\sqrt3}{4}s^2=\frac{\sqrt3}{4}\cdot84=21\sqrt3 .$$ Answer: (B)