JEE Mains · Maths · STD 11 - 9. straight line
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
- A 48
- B 32
- C 16
- D 24
Answer & Solution
Correct Answer
(A) 48
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Slope 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…
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