Question

The distance of the point $(3,5)$ from the line $2x+3y-14=0$ measured parallel to the line $x-2y=1$ is

• A. $\frac{7}{\sqrt{5}}$
• B. $\frac{7}{\sqrt{13}}$
• C. $\sqrt{5}$
• D. $\sqrt{13}$

Answer 1

Answer: (C)

$\sqrt{5}$

Slope of the line $x-2y=1$ is $\frac{1}{2}$

$\Rightarrow \tan \theta =\frac{1}{2}$

$\therefore \cos \theta =\frac{2}{\sqrt{5}},\sin \theta =\frac{1}{\sqrt{5}}$
Let $r$ be the required distance. Then the equation of line through $(3,5)$ and parallel to $x-2y=1$ is

$\frac{x-3}{\cos \theta }=\frac{y-5}{\sin \theta }=r$

$\Rightarrow x=3+r\cos \theta ,y=5+r\sin \theta$

Since this point lies on the line $2x+3y-14=0$ it will satisfy equation of a line

$\therefore 2(3+r\frac{2}{\sqrt{5}})+3(5+r\frac{1}{\sqrt{5}})-14=0$

$\Rightarrow r=-\sqrt{5}$

Required distance $=\sqrt{5}$