Find the equation of the line joining the points $\left(1, \dfrac{\pi}{4}\right)$ and $\left(2, \dfrac{\pi}{2}\right)$.
Let $\;$ $A \left(r_1, \theta_1\right) = \left(1, \dfrac{\pi}{4}\right)$, $\;$ $B \left(r_2, \theta_2\right) = \left(2, \dfrac{\pi}{2}\right)$
Let $P \left(r, \theta\right)$ be any point on the line $AB$.
Then the equation of line $AB$ is
$r_1 r \sin \left(\theta - \theta_1\right) + r r_2 \sin \left(\theta_2 - \theta\right) + r_2 r_1 \sin \left(\theta_1 - \theta_2\right) = 0$
i.e. $\;$ $1 \times r \sin \left(\theta - \dfrac{\pi}{4}\right) + r \times 2 \sin \left(\dfrac{\pi}{2} - \theta\right) + 1 \times 2 \sin \left(\dfrac{\pi}{4} - \dfrac{\pi}{2}\right) = 0$
i.e. $\;$ $r \left(\sin \theta \cos \dfrac{\pi}{4} - \cos \theta \sin \dfrac{\pi}{4}\right) + 2r \cos \theta + 2 \sin \left(\dfrac{- \pi}{4}\right) = 0$
i.e. $\;$ $\dfrac{r}{\sqrt{2}} \left(\sin \theta - \cos \theta\right) + 2 r \cos \theta - \dfrac{2}{\sqrt{2}} = 0$
i.e. $\;$ $r \sin \theta + \dfrac{r}{\sqrt{2}} \left(2 \sqrt{2} - 1\right) \cos \theta - \sqrt{2} = 0$