Exams Joint Entrance Examination Mathematics Straight Lines and Pair of Straight Lines Question 92 Question 92 of 303 Medium MCQ
Let a a a be the length of a side of a square O A B C OABC O A B C with O O O being the origin. Its side O A OA O A makes an acute angle with the positive x x x -axis and the equations of its diagonals are ( 3 + 1 ) x + ( 3 − 1 ) y = 0 (\sqrt{3}+1)x+(\sqrt{3}-1)y=0 ( 3 + 1 ) x + ( 3 − 1 ) y = 0 and ( 3 − 1 ) x − ( 3 + 1 ) y + 8 3 = 0 (\sqrt{3}-1)x-(\sqrt{3}+1)y+8\sqrt{3}=0 ( 3 − 1 ) x − ( 3 + 1 ) y + 8 3 = 0 . Then a 2 a^2 a 2 is equal to:
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\[ \begin{aligned}
&\begin{aligned}
& \text { Slope of diagonal } \mathrm{OB}=\frac{\sqrt{3}+1}{1-\sqrt{3}} \\
& =\tan 105^{\circ} \\
& \therefore \alpha=60^{\circ} \\
& \therefore \mathrm{A}\left(\operatorname{acos} 60^{\circ}, \mathrm{asin} 60^{\circ}\right) \\
& \therefore \mathrm{A}\left(\frac{\mathrm{a}}{2}, \frac{\sqrt{3} \mathrm{a}}{2}\right)
\end{aligned} \\
\&\text { A Lies on other diagonal }\\
\& ∴ ( 3 − 1 2 ) a − ( 3 + 1 2 ) ⋅ 3 a + 8 3 = 0 a [ 3 − 1 − 3 − 3 2 ] = − 8 3 \begin{aligned}
& \therefore\left(\frac{\sqrt{3}-1}{2}\right) a-\left(\frac{\sqrt{3}+1}{2}\right) \cdot \sqrt{3} a+8 \sqrt{3}=0 \\
& a\left[\frac{\sqrt{3}-1-3-\sqrt{3}}{2}\right]=-8 \sqrt{3}
\end{aligned} ∴ ( 2 3 − 1 ) a − ( 2 3 + 1 ) ⋅ 3 a + 8 3 = 0 a [ 2 3 − 1 − 3 − 3 ] = − 8 3
\end{aligned}\]
\[ a = 4 3 ∴ a 2 = 48 \begin{aligned}
& \mathrm{a}=4 \sqrt{3} \\
& \therefore \mathrm{a}^2=48
\end{aligned} a = 4 3 ∴ a 2 = 48 \] Answer the question to unlock