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

Question

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

Accepted Answer

Writing $P$ in terms of parametric co-ordinates $2+r$

$$\begin{aligned}
& \cos \theta, 3+\mathrm{r} \sin \theta \text { as } \tan \theta=\sqrt{3} \
& \mathrm{P}\left(2+\frac{\mathrm{r}}{2}, 3+\frac{\sqrt{3} \mathrm{r}}{2}\right)
\end{aligned}$$

$$\mathrm{P}$$ must satisfy $$2 \mathrm{x}-3 \mathrm{y}+28=0$$

So, $$2\left(2+\frac{r}{2}\right)-3\left(3+\frac{\sqrt{3} \mathrm{r}}{2}\right)+28=0$$

We find $$r=4+6 \sqrt{3}$$

Solutions · 1