# 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.

• Are you sure that’s what $p$ is supposed to be? – amd Nov 30 '18 at 20:15
• thats just the format, it says a equation that is y=mx+b can be expressed in polar coordinates like that equation. it says the hough transform of a line turns into a point and I need to find the point in (p,θ) space – dshawn Nov 30 '18 at 23:02
• "I know polar coordinates can be represented by p=x⋅cos(θ)+y⋅sin(θ)": er, no. – Yves Daoust Nov 30 '18 at 23:16
• Well the issue is this equation is given to me by the question, so I don't know am I misreading the question? I know the p is a greek letter and not a normal p. I will post the full question just in case – dshawn Dec 1 '18 at 1:57

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 $$OA.$$

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.

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 $$y = r \sin \theta$$.

• is it $\tan ^{-1}?$ – user376343 Dec 1 '18 at 20:58
• @user376343 Yes, fixed it. Thanks, – Aditya Dua Dec 2 '18 at 3:41