# Line equation in polar coordinates does not hold

I am having trouble understanding how linear equations in polar coordinates work.

Say I have two points on the same line: $(1,1)$ and $(3,3)$.

Then for the equation $$b=y-mx,$$ $b=0$ and $m=1$ holds for the two points.

But for the equation $$r=x\cos\theta+y\sin\theta,$$ with $\theta=45\pi/180$, $r$ is different; it's $1.4$ for $(1,1)$ and $4.2$ for $(3,3)$.

I don't understand; shouldn't $r$ be constant?

• Do you know what $r$ is supposed to represent in polar coordinates? – Jose27 Mar 31 '15 at 5:18

Given is $$y =x$$ Convert it to polar coordinates.
$$r \cos \theta = r \sin \theta$$ Now divide by $r$,
$$\tan \theta = m = 1$$ for all $r$, independent of $r$. Means in polar coordinates the straight line through origin does not involve radius $r$. It should not be a constant, it should not appear anywhere at all.
$$y = mx + b$$ Since $m=1$ and $b=0$, we have $$y = x$$ Converting to polar coordinates yields $$r\sin\phi = r\cos\phi$$ $$r\sin\phi - r\cos\phi = 0$$ $$r\left(\sin\phi - \cos\phi\right) = 0$$ At point $(1, 1)$, we have $$r = \sqrt{x^2+y^2} = \sqrt{1^2+1^2} = \sqrt{2}$$ $$\phi = \mathbb{atan2}\left(y, x\right) = \arctan\left(\frac{1}{1}\right)= \frac{\pi}{4}$$ So now $$\sqrt2\left(\sin \frac{\pi}{4} - \cos \frac{\pi}{4}\right) = \sqrt2\left(\frac{1}{\sqrt2}-\frac{1}{\sqrt2}\right)=0$$ Try it yourself with points $(3,3)$ and $(4,4)$. If you get a mismatch, then you likely evaluated $\mathbb{atan2}$ incorrectly. Have a look here.