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

coordinates of any point on this line are $\left(2+r\cos {45}^{\circ },3+r\sin {45}^{\circ }\right)$ $\text{ or }\left(2+\frac{r}{\sqrt{2}},3+\frac{r}{\sqrt{2}}\right)$

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

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

$\Rightarrow r=4$