Mathematics · Straight lines

JEE Main 2025 — 22 January, Evening Shift — Question 23

Let the distance between two parallel lines be 5 units and a point $P$ lie between the lines at a unit distance from one of them. An equilateral triangle PQR is formed such that Q lies on one of the parallel lines, while R lies on the other. Then $(\mathrm{QR})^{2}$ is equal to $\_\_\_\_\_$ .

Answer: 28

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{PR}=\operatorname{cosec} \theta, \mathrm{PQ}=4 \sec (30+\theta)$

For equilateral

$\mathrm{d}=\mathrm{PR}=\mathrm{PQ}$

$\Rightarrow \cos \left(\theta+30^{\circ}\right)=4 \sin \theta$

$\Rightarrow \frac{\sqrt{3}}{2} \cos \theta-\frac{1}{2} \sin \theta=4 \sin \theta$

$\Rightarrow \tan \theta=\frac{1}{3 \sqrt{3}}$

$\mathrm{QR}^{2}=\mathrm{d}^{2}=\operatorname{cosec}^{2} \theta=28$

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Exam: JEE Main 2025
Paper: 22 January, Evening Shift
Subject: Mathematics
Chapter: Straight lines
Topic: Angle between lines, perpendicular distance & distance between parallel lines, foot, image

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Answer: 28

Step-by-step solution

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