question_answer

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: [JEE Main 12-4-2019 Afternoon]

A) \[\left( \sqrt{3}+1 \right)x+\left( \sqrt{3}-1 \right)y=8\sqrt{2}\]

B) \[\left( \sqrt{3}-1 \right)x+\left( \sqrt{3}+1 \right)y=8\sqrt{2}\]

C) \[\sqrt{3}x+y=8\]

D) \[x+\sqrt{3}y=8\]

Correct Answer: A

Solution :

\[{{m}_{1}}=\tan {{75}^{o}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\] or\[m=\tan {{15}^{o}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\] \[{{m}_{2}}=\frac{-1}{{{m}_{1}}}=\frac{-\left( \sqrt{3}-1 \right)}{\sqrt{3}+1}\] or\[{{m}_{2}}=\frac{-1}{{{m}_{1}}}=\frac{-\left( \sqrt{3}+1 \right)}{\sqrt{3}-1}\] \[\Rightarrow y={{m}_{2}}x+C\Rightarrow y=\frac{-\left( \sqrt{3}-1 \right)x}{\sqrt{3}+1}+C\Rightarrow L\] or\[y=\frac{-\left( \sqrt{3}+1 \right)x}{\sqrt{3}-1}+C\Rightarrow L\] Distance from origin = 4 \[\therefore \]\[\left| \frac{C}{\sqrt{1+\frac{{{\left( \sqrt{3}-1 \right)}^{2}}}{{{\left( \sqrt{3}+1 \right)}^{2}}}}} \right|=4\]or\[\left| \frac{C}{\sqrt{1+\frac{{{\left( \sqrt{3}+1 \right)}^{2}}}{{{\left( \sqrt{3}-1 \right)}^{2}}}}} \right|=4\] \[\Rightarrow C=\frac{8\sqrt{2}}{\left( \sqrt{3}+1 \right)}\]or\[C=\frac{8\sqrt{2}}{\left( \sqrt{3}-1 \right)}\] \[\Rightarrow \left( \sqrt{3}-1 \right)y+\left( \sqrt{3}+1 \right)x=8\sqrt{2}\] or\[\left( \sqrt{3}-1 \right)x+\left( \sqrt{3}+1 \right)y=8\sqrt{2}\]