Question

Find the distance of the point $(1,3)$ from the line $2x-3y+9=0$ measured along a line $x-y+1=0$.

Answer 1

Given point $(1,3)$ from the line $2x-3y+9=0$

Measured along a line $x-y+1=0\Rightarrow x-y+1=c$ put $(1,3)$ in the line equation as it is a point on the line.

$x-y+1=0$

$1-3+1-c=0$

$c=-1$

Equation of the line $x-y+1=-1\Rightarrow x-y=-2$ now find the point of intersecting of $(x-y=-2)$ & $(2x-3y+9=0)$

$x-y+2=0---\left(1\right)$

$2x-3y+9=0---\left(2\right)$

$2x-2y+4=0---\left(3\right)$ (mul with $2$ in $\left(1\right)$

But subtracting $\left(3\right)$ & $\left(2\right)$

$y+5=0$

$y=5$ put $5=y$ in $\left(1\right)$

$x-5+2=0$

$x=3$

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

Distance $=\sqrt{{(1-5)}^2+{(3-3)}^2}=\sqrt{16}=4$