Mathematics · Straight lines

JEE Main 2024 — 30 January, Shift 1 — Question 1

A line passing through the point $A(9,0)$ makes an angle of $30^{\circ}$ with the positive direction of $x$ -axis. If this line is rotated about A through an angle of $15^{\circ}$ in the clockwise direction, then its equation in the new position is

Option A:
$\frac{y}{\sqrt{3}-2}+x=9$
Correct
Option B:
$\frac{x}{\sqrt{3}-2}+y=9$
Option C:
$\frac{x}{\sqrt{3}+2}+y=9$
Option D:
$\frac{y}{\sqrt{3}+2}+x=9$

Answer: A

Step-by-step solution

$\mathrm{Eq}^{\mathrm{n}}: y-0=\tan 15^{\circ}(x-9)$

$\Rightarrow y=(2-\sqrt{3})(x-9)$

$\Rightarrow \frac{y}{\sqrt{3}-2}+x=9$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 30 January, Shift 1
Subject: Mathematics
Chapter: Straight lines
Topic: Various forms of lines