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 ° with the line x + y = 0 . Then an equation of the line L is: Note: In actual JEE Main paper, two options were…

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 &#176; with the line x + y = 0 . Then an equation of the line L is: Note: In actual JEE Main paper, two options were correct for this question. Hence, we have changed one option.

Aicharya · Accepted Answer

Hence, equation of required line using normal form is

x c o s 15^{°} + y s i n 15^{°} = 4

or

x c o s 75^{°} + y s i n 75^{°} = 4

\Rightarrow (\sqrt{3} + 1) x + (\sqrt{3} - 1) y = 8 \sqrt{2}

or

(\sqrt{3} - 1) x + (\sqrt{3} + 1) y = 8 \sqrt{2}

A straight line L at a distance of 4 units from the origin makes positive intercepts on the coordinate axes…

Solution