A straight line L through the point (3,-2) is inclined at an angle of 60^ to the line √3x+…

Question

A straight line $L$ through the point $(3,-2)$ is inclined at an angle of $60^\circ$ to the line $\sqrt3x+y=1$. If $L$ also intersects the $x$-axis, then the equation of $L$ is:

Accepted Answer

Line $\sqrt3x+y=1$ has slope $-\sqrt3$ (inclination $120^\circ$). $L$ makes $60^\circ$ with this, giving possible inclinations $180^\circ$ (slope 0) or $60^\circ$ (slope $\sqrt3$).

For $L$ to intersect the $x$-axis (not be parallel/coincide trivially with special cases), take slope $\sqrt3$:
$$y-(-2) = \sqrt3(x-3) \implies y+2=\sqrt3x-3\sqrt3 \implies y-\sqrt3x+2+3\sqrt3=0$$

Solutions · 1