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

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:

Accepted Answer

Since the perpendicular makes $60^\circ$ with $x+y=0$ (inclination $135^\circ$), the perpendicular's inclination is $135^\circ-60^\circ=75^\circ$ or $135^\circ+60^\circ=195^\circ$; taking the angle giving positive intercepts, $75^\circ$ (or equivalently $15^\circ$).

$$L: x\cos75^\circ+y\sin75^\circ=4$$

Using $\cos75^\circ=\dfrac{\sqrt6-\sqrt2}4,\ \sin75^\circ=\dfrac{\sqrt6+\sqrt2}4$, multiplying through by 4 and rationalizing gives the form:
$$(\sqrt3-1)x+(\sqrt3+1)y=8\sqrt2 \quad\text{or}\quad(\sqrt3+1)x+(\sqrt3-1)y=8\sqrt2$$

Matching option B.

Solutions · 1