## Representation of a line in the polar coordinate system

Recently, I read a tutorial about Hough line transform at the OpenCV tutorials. It is a technique to find lines in an image using a parameter space. As explained in the tutorial, for this it is necessary to use the polar coordinate system. In the commonly used Cartesian coordinate system, a line would be represented by $$y=mx+b$$. In the polar coordinate system on the other hand, a line is represented by

$$y=-\frac{\cos{\theta}}{\sin{\theta}}x + \frac{\rho}{\sin{\theta}}. \label{eq:PolarCoordinateSystem_LineRepresentation}$$

In \eqref{eq:PolarCoordinateSystem_LineRepresentation} there are two new parameters: radius $$\rho$$ and angle $$\theta$$ as also depicted in the following figure. $$\rho$$ is the length of a vector which always starts at the pole $$(0,0)$$ (analogous term to the origin in the Cartesian coordinate system) and ends at the line (orange in the figure) so that $$\rho$$ will be orthogonal to the line. This is important because otherwise the following conclusions wouldn't work.

So, first start with the $$y$$-intercept $$b=\frac{\rho}{\sin{\theta}}$$. Note that the angle $$\theta$$ comes up twice: between the $$x$$-axis and the $$\rho$$ vector plus between the $$y$$-axis and the blue line (on the right side). We will use trigonometrical functions to calculate the $$y$$-intercept. This is simply done by using the $$\sin$$-function

\begin{equation*} \begin{split} \sin{\theta} &= \frac{\text{opposite}}{\text{hypotenuse}} \\ \sin{\theta} &= \frac{\rho}{b} \\ b &= \frac{\rho}{\sin{\theta}} \end{split} \end{equation*}

and that is exactly what the equation said for the $$y$$-intercept.

Now it is time for the slope $$m=-\frac{\cos{\theta}}{\sin{\theta}}$$. For this, the relation

\begin{equation*} m = \tan{\alpha} \end{equation*}

is needed, where $$\alpha$$ is the slope angle of the line. $$\alpha$$ can be calculated by using our known $$\theta$$ angle:

\begin{equation*} \alpha = 180^{\circ} - (180^{\circ} - 90^{\circ} - \theta) = 90^{\circ} + \theta. \end{equation*}

Now we have $$m=\tan{\left(90^{\circ} + \theta\right)}$$, which is equivalent to $$m=\frac{\sin{\left(90^{\circ} + \theta\right)}}{\cos{\left(90^{\circ} + \theta\right)}}$$. Because of $$\sin{x}=\cos{\left(90^{\circ}-x\right)}$$ and $$\cos{x}=\sin{\left(90^{\circ}-x\right)}$$ we can do a little bit of rewriting

\begin{equation*} m=\frac {\cos\left({90^{\circ} - (90^{\circ} + \theta)}\right)} {\sin\left({90^{\circ} - (90^{\circ} + \theta}\right))} = \frac {\cos\left({-\theta}\right)} {\sin\left({-\theta}\right)} = \frac {\cos\left({\theta}\right)} {-\sin\left({\theta}\right)} = -\frac {\cos\left({\theta}\right)} {\sin\left({\theta}\right)} \end{equation*}

and we have exactly the form we need. In the last step, I used the property that $$\sin(x) = -\sin(-x)$$ is an odd and $$\cos(x) = \cos(-x)$$ an even function.