Mathematics · 3D Geometry

JEE Main 2024 — 29 January, Shift 2 — Question 11

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

Answer: D

Writing $P$ in terms of parametric co-ordinates $2+\mathrm{r}$$\cos \theta, 3+\mathrm{r} \sin \theta$ as $\tan \theta=\sqrt{3}$

$\mathrm{P}\left(2+\frac{\mathrm{r}}{2}, 3+\frac{\sqrt{3} \mathrm{r}}{2}\right)$

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

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

Solve and we get $\boxed{r=4+6 \sqrt{3}}$

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.