in the given figure, $\mathrm{OABC}$ is a rectangle. slope of OB is 

Clearly, $\mathrm{OB}=5,$

So, $\frac{\mathrm{OA}}{\mathrm{OB}}=\frac{3}{5}=\frac{1.5}{2.5}=\frac{\mathrm{AD}}{\mathrm{DB}}$

Thus, OD bisects $\angle \mathrm{BOC}=\mathrm{\theta }.$ Then $\mathrm{tan\theta }=\frac{3}{4}$.

Let $\therefore \angle \mathrm{BOC}=\frac{1}{2}\angle \mathrm{AOB}=\frac{1}{2}\left({90}^0-\mathrm{\theta }\right)$

$\therefore$ Slope of $\mathrm{OB}=\tan \frac{1}{2}\left({90}^0-\mathrm{\theta }\right)=\frac{\tan {45}^0-\tan \frac{\mathrm{\theta }}{2}}{1+\tan {45}^0.\tan \frac{\mathrm{\theta }}{2}}$

Now, $\mathrm{tan\theta }=\frac{3}{4}$$\Rightarrow \frac{2\tan \frac{\mathrm{\theta }}{2}}{1-{\tan }^2\frac{\mathrm{\theta }}{2}}=\frac{3}{4}$

$3-3{\tan }^2\frac{\mathrm{\theta }}{2}=8\tan \frac{\mathrm{\theta }}{2}$

$\Rightarrow 3{\tan }^2\frac{\mathrm{\theta }}{2}+8\tan \frac{\mathrm{\theta }}{2}-3=0$

$\Rightarrow \left(3\tan \frac{\mathrm{\theta }}{2}-1\right)\left(\tan \frac{\mathrm{\theta }}{2}+3\right)=0$

so, $\tan \frac{\mathrm{\theta }}{2}=\frac{1}{3}$

$\therefore \mathrm{Slope}\mathrm{of}\mathrm{OB}=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}=\frac{1}{2}$