JEE Main 2026 — Jan 21, Shift 1 — Mathematics solution

JEE Main2026Jan 21, Shift 1Mathematics

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 :

A
$156$
B
27
C
$213$
D
$122$

Solution

Let side length be $a$ and let $θ$ be the angle that $A C$ makes with $L_{2}$ , so the perpendicular from $A$ to $L_{2}$ gives $sin θ = \frac{3}{a}$ .

Since $∠ B A C = 60°$ , line $A B$ makes angle $(60° + θ)$ with $L_{2}$ . The perpendicular distance from $B$ to $L_{2}$ equals the gap between the parallel lines plus 0 (since $B$ lies on $L_{1}$ , 6 above $A$ , and $A$ is 3 above $L_{2}$ ), giving distance $= 9$ and so $sin (60° + θ) = \frac{9}{a}$ .

Expanding: $\frac{3}{2} cos θ + \frac{1}{2} sin θ = \frac{9}{a}$ . Substitute $sin θ = 3/ a$ and $cos θ = 1 - 9/ a^{2}$ and solve:

$3 \cdot a^{2} - 9 = 15 \Rightarrow a^{2} = 84$ .

Area $= \frac{3}{4} a^{2} = 213$ .

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