Q33 / 75 · JEE Main 2026 (21 January Shift 1) · Mathematics
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 $A B C$, where the points $B$ and C lie on the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$, respectively, is :
- A$15 \sqrt{6}$
- Correct answerB$21 \sqrt{3}$
- C$12 \sqrt{2}$
- D27
Solution
Let $L_2: y = 0$, $L_1: y = 9$.
Point A at $(0, 3)$, B at $(x_B, 9)$, C at $(x_C, 0)$.
$AB^2 = x_B^2 + 36$, $AC^2 = x_C^2 + 9$, $BC^2 = (x_B - x_C)^2 + 81$.
From $AB = AC$: $x_B^2 - x_C^2 = -27$.
From $AB = BC$: $2x_Bx_C = x_C^2 + 45$.
Solving: $3x_C^4 - 198x_C^2 - 2025 = 0$.
Let $u = x_C^2$: $u^2 - 66u - 675 = 0 \Rightarrow u = 75$.
$a^2 = x_C^2 + 9 = 84$.
Area $= \frac{\sqrt{3}}{4} \times 84 = 21\sqrt{3}$.
Point A at $(0, 3)$, B at $(x_B, 9)$, C at $(x_C, 0)$.
$AB^2 = x_B^2 + 36$, $AC^2 = x_C^2 + 9$, $BC^2 = (x_B - x_C)^2 + 81$.
From $AB = AC$: $x_B^2 - x_C^2 = -27$.
From $AB = BC$: $2x_Bx_C = x_C^2 + 45$.
Solving: $3x_C^4 - 198x_C^2 - 2025 = 0$.
Let $u = x_C^2$: $u^2 - 66u - 675 = 0 \Rightarrow u = 75$.
$a^2 = x_C^2 + 9 = 84$.
Area $= \frac{\sqrt{3}}{4} \times 84 = 21\sqrt{3}$.
Hence, the correct answer is option: 'B'.