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

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

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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?

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

That is its definition, holds well, makes sense is it not?

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$$ y = mx + b $$ 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.

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