Question

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

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

Answer 1

The correct option is C $\sqrt{5}$

Let the equation of the line parallel to $x-2y=1$ is $x-2y+\lambda =0$
Since, it passes through $(3,5)$
$\Rightarrow 3-10+\lambda =0$

$\Rightarrow \lambda =7$
Therefore, the line is $x-2y+7=0$.
The point of intersection of $x-2y+7=0$ and $2x+3y-14=0$ is $(1,4)$.
The distance between $(3,5)$ and $(1,4)$
$=\sqrt{(3-1{)}^2+(5-4{)}^2}=\sqrt{4+1}=\sqrt{5}$.