Question

A straight line $L$ at a distance of $4$ units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of $60^{\circ }$ with the line $x+y=0$. Then an equation of the line $L$ is :

• A. $(\sqrt{3}+1)x+(\sqrt{3}-1)y=8\sqrt{2}$
• B. $(\sqrt{3}-1)x+(\sqrt{3}+1)y=8\sqrt{2}$
• C. $\sqrt{3}x+y=8$
• D. $x+\sqrt{3}y=8$

Answer 1

Answer: (B)

$m_1=tan75^{\circ }=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
$m_2=\frac{-1}{m_1}=\frac{-(\sqrt{3}-1)}{\sqrt{3}+1}$
$\Rightarrow y=m_{2}x+C$
$\Rightarrow y=\frac{-(\sqrt{3}-1)x}{\sqrt{3}+1}+C\Rightarrow L$
Distance from origin = 4
$\therefore \frac{C}{\sqrt{1+\frac{(\sqrt{3}-1{)}^2}{(\sqrt{3}+1{)}^2}}}$
$\Rightarrow \frac{(\sqrt{3}+1)C}{\sqrt{8}}=4$
$\Rightarrow C=\frac{8\sqrt{2}}{(\sqrt{3}+1)}$
Hence
$\Rightarrow (\sqrt{3}-1)y+(\sqrt{3}+1)x=8\sqrt{2}$