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

The correct option is B $(\sqrt{3}-1)x+(\sqrt{3}+1)y=8\sqrt{2}$



Given $OA=p=4$

$x+y=0$ makes an angle of ${135}^{\circ }$ with the positive $x$-axis because

$\tan \theta =$slope of line $x+y=0\Rightarrow y=-x$ comparing with $y=mx$,

we get slope $m=-1$

$\Rightarrow \tan {\theta }_1=-1$

$\Rightarrow \tan {\theta }_1=\tan {135}^{\circ }$

$\therefore {\theta }_1={135}^{\circ }$

Thus, line $OA$ makes an angle of $135-60={75}^{\circ }$ with the x-axis.

Equation of the line $L$ in perpendicular form is $x\cos \theta +y\sin \theta =p$

$\Rightarrow x\cos {75}^{\circ }+y\sin {75}^{\circ }=p$

$\Rightarrow x\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)+y\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)=4$

$\therefore x(\sqrt{3}-1)+y\left(\sqrt{3}+1\right)=8\sqrt{2}$ is the required equation.