The distance of the point (2,3) from the line 2x-3y+28=0, measured parallel to the line √3…

Question

The distance of the point $(2,3)$ from the line $2x-3y+28=0$, measured parallel to the line $\sqrt3x-y+1=0$, is equal to:

Accepted Answer

The line $\sqrt3x-y+1=0$ has slope $\sqrt3$, i.e. inclination $60^\circ$. A line through $(2,3)$ with this direction: parametrize as $(2+r\cos60^\circ,\ 3+r\sin60^\circ)$.

This point must lie on $2x-3y+28=0$:
$$2\left(2+\frac r2\right)-3\left(3+\frac{r\sqrt3}2\right)+28=0$$
$$4+r-9-\frac{3\sqrt3r}2+28=0 \implies r\left(1-\frac{3\sqrt3}2\right) = -23$$

Solving this (as per the source computation) yields $r = 4+6\sqrt3$.

Solutions · 1