Let a be the length of a side of a square O A B C with O being the origin. Its side O A makes an acute angle α with the positive x -axis and the equations of its diagonals are (√(3)+1) x+(√(3)-1) y=0 and (√(3)-1) x-(√(3)+1) y+8 √(3)=0 . Then a 2 is equal to

Mathematics · Straight lines

JEE Main 2025 — 8 April, Evening Shift — Question 37

Let $a$ be the length of a side of a square $O A B C$ with $O$ being the origin. Its side $O A$ makes an acute angle $\alpha$ with the positive $x$ -axis and the equations of its diagonals are $(\sqrt{3}+1) x+(\sqrt{3}-1) y=0$ and $(\sqrt{3}-1) x-(\sqrt{3}+1) y+8 \sqrt{3}=0$ . Then $a^{2}$ is equal to

Option A:
24
Option B:
32
Option C:
16
Option D:
48
Correct

Answer: D

Step-by-step solution

$O B:(\sqrt{3}+1) x+(\sqrt{3}-1) y=0$

$A C:(\sqrt{3}-1) x-(\sqrt{3}+1) y+8 \sqrt{3}=0$

$\Rightarrow(x, y) \equiv P(3-\sqrt{3}, 3+\sqrt{3})$

Let $A B=a=O A$

$\Rightarrow O A^{2}+A B^{2}=O B^{2}$

$2 a^{2}=4\left[(3-\sqrt{3})^{2}+(3-\sqrt{3})^{2}\right]$

$a^{2}=2 \times 24=48$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 8 April, Evening Shift
Subject: Mathematics
Chapter: Straight lines
Topic: Various forms of lines

A line passing through the point P(a, 0) makes an acute angle alpha with the positive x -axis. Let this line be rotated about the point P…

Answer: D

Step-by-step solution

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