JEE MAIN - Mathematics (2024 - 29th January Evening Shift - No. 4)
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
$$3+4 \sqrt{2}$$
$$6 \sqrt{3}$$
$$4+6 \sqrt{3}$$
$$4 \sqrt{2}$$
Explanation
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 $$r=4+6 \sqrt{3}$$
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