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

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

$\frac{x-2}{\cos {45}^{\circ }}=\frac{y-3}{\sin {45}^{\circ }}=r\left[\text{ Using }\frac{x-x_1}{\cos \theta }=\frac{y-y_1}{\sin \theta }=r\right]$

co-ordinates of any point on this line are $\left(2+r\cos {45}^{\circ },3+r\sin {45}^{\circ }\right)$

$\Rightarrow \left(2+\frac{r}{\sqrt{2}},3+\frac{r}{\sqrt{2}}\right)$

If this point lies on the line $2x-3y+9=0$,

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

$$\begin{gathered} \Rightarrow r=4\sqrt{2} \\ \Rightarrow k=4 \end{gathered}$$