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

The slope of the line $x-y+1=0$ is 1 . So, it makes an angle of $45^{\circ}$ with $x$-axis. The equation of a line passing through $(2,3)$ and making an angle of $45^{\circ}$ $\frac{x-2}{\cos 45^{\circ}}=\frac{y-3}{\sin 45^{\circ}}=r$ Coordinates of any point on this line are $\left(2+r \cos 45^{\circ}, 3+r \sin 45^{\circ}\right)$ or, $\left(2+\frac{r}{\sqrt{2}}, 3+\frac{r}{\sqrt{2}}\right)$ If this point lies on the line $2 x-3 y+9=0$, then $4+\sqrt{2} r-9-\frac{3 r}{\sqrt{2}}+9=0$ $\Rightarrow r=4 \sqrt{2}$ Alternate Method
Since the point $(2,3)$ lies on the line $x-y+1=0$. Therefore the distance from $(2,3)$ to the line $2 x-3 y+9=0$ along the line $x-y+1=0$ is equal to the distance between the points $(2,3)$ and intersection point of $2 x-3 y+9=0$ and $x-y+1=0$ ie, $(6,7)$. Hence required distance, $\begin{aligned} & d=\sqrt{(6-2)^2+(7-3)^2}=\sqrt{32} \\ & d=4 \sqrt{2} \end{aligned}$