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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2026 · 21 January · Morning Shift · Mathematics · Straight Lines

Let a point A lie between the parallel lines L_ 1 and L_ 2 such...

Let a point AA lie between the parallel lines L1L_{1} and L2L_{2} such that its distances from L1L_{1} and L2L_{2} are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABCA B C, where the points BB and C lie on the lines L1\mathrm{L}_{1} and L2\mathrm{L}_{2}, respectively, is :

Options

  1. A15615 \sqrt{6}
  2. B21321 \sqrt{3}
  3. C12212 \sqrt{2}
  4. D27

Correct answer

B. 21321 \sqrt{3}

Step-by-step solution

Let L2:y=0L_2: y = 0, L1:y=9L_1: y = 9.
Point A at (0,3)(0, 3), B at (xB,9)(x_B, 9), C at (xC,0)(x_C, 0).
AB2=xB2+36AB^2 = x_B^2 + 36, AC2=xC2+9AC^2 = x_C^2 + 9, BC2=(xB−xC)2+81BC^2 = (x_B - x_C)^2 + 81.
From AB=ACAB = AC: xB2−xC2=−27x_B^2 - x_C^2 = -27.
From AB=BCAB = BC: 2xBxC=xC2+452x_Bx_C = x_C^2 + 45.
Solving: 3xC4−198xC2−2025=03x_C^4 - 198x_C^2 - 2025 = 0.
Let u=xC2u = x_C^2: u2−66u−675=0⇒u=75u^2 - 66u - 675 = 0 \Rightarrow u = 75.
a2=xC2+9=84a^2 = x_C^2 + 9 = 84.
Area =34×84=213= \frac{\sqrt{3}}{4} \times 84 = 21\sqrt{3}.
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