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JEE Main 202226 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
26 June
Shift
Morning Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2022 · 26 June · Morning Shift · Mathematics · Straight Lines

Let R be the point (3, 7) and let P and Q be two points on the line...

Let RR be the point (3,7)\left(3, 7\right) and let PandQP and Q be two points on the line x+y=5x + y = 5 such that PQRPQR is an equilateral triangle. Then the area of ∆PQR∆ PQR isLet R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle.

Options

  1. A2543\dfrac{25}{4\sqrt{3}}
  2. B2532\dfrac{25\sqrt{3}}{2}
  3. C253\dfrac{25}{\sqrt{3}}
  4. D2523\dfrac{25}{2\sqrt{3}}

Correct answer

D. 2523\dfrac{25}{2\sqrt{3}}

Step-by-step solution

Given PQRPQR is an equilateral triangle.

Let the side of the triangle be aa

Perpendicular distance from R(3,7)R\left(3, 7\right) to the line x+y=5x + y = 5 is

∣3+7−5∣1+1=52\dfrac{\left|3 + 7 - 5\right|}{\sqrt{1 + 1}} = \dfrac{5}{\sqrt{2}}

Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle.

So sin⁡60°=52a\sin 60 ° = \dfrac{\dfrac{5}{\sqrt{2}}}{a}

a=523a = \dfrac{5\sqrt{2}}{\sqrt{3}}

Area of ∆PQR=34a2=2523∆ PQR = \dfrac{\sqrt{3}}{4}{a}^{2} = \dfrac{25}{2\sqrt{3}}

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