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

Answer 1

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


$B$ is a point on the line $2x+3y-14=0$
such that line $AB$ is parallel to $x-2y=1$
slope of line $AB$ is $m=\frac{1}{2}$
Now equation of line $AB:$
$y-5=\frac{1}{2}(x-3)$
$\Rightarrow 2y-10=x-3$
$\Rightarrow x-2y+7=0$
Intersection of line $AB$ and $2x+3y-14=0$
$B\equiv (1,4)$
Now, distance $AB=\sqrt{(1-3{)}^2+(4-5{)}^2}=\sqrt{5}$ units