Straight Lines Question 8
Question 8 - 2024 (29 Jan Shift 2)
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
(1) $4 \sqrt{2}$
(2) $6 \sqrt{3}$
(3) $3+4 \sqrt{2}$
(4) $4+6 \sqrt{3}$
Show Answer
Answer (4)
Solution
Writing $P$ in terms of parametric co-ordinates $2+r$
$\cos \theta, 3+r \sin \theta$ as $\tan \theta=\sqrt{3}$
$P\left(2+\frac{r}{2}, 3+\frac{\sqrt{3} r}{2}\right)$
$P$ must satisfy $2 x-3 y+28=0$
So, $2\left(2+\frac{r}{2}\right)-3\left(3+\frac{\sqrt{3} r}{2}\right)+28=0$
We find $r=4+6 \sqrt{3}$
