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

The slope of the line $x - y + 1 = 0$ is 1. So, it makes an angle of $45°$ with $x-$axis.

The equation of a line passing through $(2, 3)$ and making an angle of $45°$ is given by

   $\frac{x-2}{\cos {45}^{\circ }}=\frac{y-3}{\sin {45}^{\circ }}=r$

Co-ordinates 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 $2x-3y+9=0$, then

    $4+r\sqrt{2}-9-\frac{3r}{\sqrt{2}}+9=0\Rightarrow r=4\sqrt{2}$

Hence, required distance $=4\sqrt{2}$ units .