Let a be the length of a side of a square OABC with O being the origin. Its side OA 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

Correct answer: A. 48 aligned & Slope of diagonal OB = 3 +1 1- 3 & = 105^ & α=60^ & A(acos 60^ , asin 60^ ) & A( a 2 , 3 a 2 ) aligned A Lies on other diagonal aligned & ( 3 -1 2 ) a-( 3 +1 2 ) · 3 a+8 3 =0 & a[ 3 -1-3- 3 2 ]=-8 3 & a=4 3 & a^2=48 aligned

JEE Main 20258 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 8 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Straight Lines

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Straight Lines

Let a be the length of a side of a square OABC with O being the...

Let a be the length of a side of a square OABC with

O

being the origin. Its side OA 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

\mathrm{a}^2

is equal to

Options

Correct answer

A. 48

Step-by-step solution

\begin{aligned} & \text{ Slope\text{ of } diagonal OB }=\frac{\sqrt{3}+1}{1-\sqrt{3}} \\ & \therefore=\tan 105^{\circ} \\ & \therefore \alpha=60^{\circ} \\ & \therefore \mathrm{A}\left(\operatorname{acos} 60^{\circ}, \operatorname{asin} 60^{\circ}\right) \\ & \therefore \mathrm{A}\left(\frac{\mathrm{a}}{2}, \frac{\sqrt{3} \mathrm{a}}{2}\right)\end{aligned}

A Lies on other diagonal

\begin{aligned}& \therefore\left(\frac{\sqrt{3}-1}{2}\right) a-\left(\frac{\sqrt{3}+1}{2}\right) \cdot \sqrt{3} a+8 \sqrt{3}=0 \\ & a\left[\frac{\sqrt{3}-1-3-\sqrt{3}}{2}\right]=-8 \sqrt{3} \\ & a=4 \sqrt{3} \\ & \therefore a^2=48\end{aligned}

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