Question

Find the distance of the point $(3,5)$ from the line $2x+3y-14=0$ along the line $x-2y=1$

Answer 1

The line parallel to $x-2y=1$ passing through $(3,5)$

It will be $x-2y=c$

Put $(3,5)$ in the line equation as it is a point on the line.

$\begin{array}{c}3-2\times 5=c\\ \Rightarrow c=-7\\ Equationofline:x-2y=-7\end{array}$

Now find point of intersection of $(x-2y=7)$ and $(2x+3y-14=0)$.

$\begin{array}{c}x-2y+7=\mathrm{0..........}\left(1\right)\\ 2x+3y-14=\mathrm{0..........}\left(2\right)\\ 2x-4y+14=\mathrm{0.........}\left(3\right)\left[bymultiply2withequation\left(1\right)\right]\\ Bysubtracting\left(3\right)from\left(2\right)\\ 7y-28=0\\ \Rightarrow y=4\\ puty=4inequation\left(1\right)\\ x=1\end{array}$

So the point of inersection is $(1,4)$. Find the distance between $(3,5)$

and $(1,4)$. That is the required answer.

Distance $=\sqrt{{\left(3-1\right)}^2+{\left(5-4\right)}^2}=\sqrt{2^2+1^2}=\sqrt{5}$