How to convert a straight line into polar coordinates? The straight line $y=mx+b$ can be expressed in polar coordinates as:
$$\rho=x\cos(\theta) + y\sin(\theta)$$
Where $(\rho,\theta)$ defines a vector from the origin to the nearest point on the line. Thus the Hough transform of a straight line in $x-y$ space is a point in $(\rho,\theta)$ space.
Find $(\rho, \theta)$ for the following straight line $y=-x+5$.

I'm trying to go through a simple exercise for the Hough transform where I have a  simple straight line in the form of $\;y=-x+5\;$ and I want to obtain polar coordinates $\;(\rho,\theta)$. I know polar coordinates can be represented by $\;\rho = x⋅\cos(\theta) + y⋅\sin(\theta).$
What are the steps I'm supposed to take to solve this problem? I have searched around and couldn't really find any examples I can follow in this exact format.
 A: $$\frac{\left|c\right|}{\sqrt{a^2+b^2}}$$
Gives you the normal distance from the origin to the straight line which is $\rho$
So if you multiply $$ax+by=c$$
by $\rho/c$ you get
$$mx+ny=\rho$$ where
$m=\cos(\theta)$ and $n=\sin(\theta)$ and getting $\theta$ given this should be easy.
A: You can write any point $(x,y)$ on the line as $(r\cos \theta, r\sin \theta)$, where $r = \sqrt{x^2+y^2}$ and $\theta = \tan^{-1}(y/x)$.
For example, consider the point $(4,3)$, which is on the line. You have $r = \sqrt{4^2+3^2} = 5$ and $\theta = \tan^{-1}(3/4) = 0.6435$. This gives $\cos \theta = 0.8$ and $\sin \theta = 0.6$. You can see that $x = r \cos \theta$ and $y = r \sin \theta$.
A: Using polar coordinates, a line is represented as
$$ax+by+c=a\rho\cos\theta+b\rho\sin\theta+c=0$$
or
$$\rho=-\frac c{a\cos\theta+b\sin\theta}.$$

With $\theta_0:=\tan\dfrac ba$ and $p:=-\dfrac c{\sqrt{a^2+b^2}}$, it can be rewritten
$$\rho=\frac p{\cos(\theta-\theta_0)},$$
where $\theta_0$ is the direction of the normal to the line, and $p$ the distance from the line to the origin.
A: After the definition of the Hough transform is


*

*$\rho$ equal to the distance of the origin to the line. The nearest point on the line (to the origin) is $\;P=\left(\frac{5}{2},\frac{5}{2}\right)\;$ so  $\rho=\frac{5\sqrt 2}{2}.$

*$\theta=\frac \pi4\;$ is the angle between $x-$ axis and $OA.$
