# What are the standard equations for change of Cartesian coordinates in $\mathbb{R}^2$?

I'm working through the first section of Boothby's Introduction to Differentiable Manifolds, and one of the exercises reads as follows:

Using standard equations for change of Cartesian coordinates, verify that $$\frac{m_2 - m_1}{1 + m_1m_2}$$, where $$m_1, m_2$$ are the slope of two lines, is independent of the choice of coordinates.

It's mentioned that this can also be done by proving that its value is just the tangent of the angle between the two lines, but I believe the exercise isn't meant to be completed in this way.

I'm not familiar with the standard equations for change of Cartesian coordinates. It seems like a change of coordinates in $$\mathbb{R}^2$$ would be any affine transformation, though I'm not sure of this.

My question is: what are the standard equations for change of Cartesian coordinates?

• You are right with respect to affine transformation - shift and rotate. Aug 17 '20 at 12:20
• Is it sufficient to show the quantity is invariant under just those two?
– bxw
Aug 18 '20 at 12:24
• In analytical geometry If you want to keep orientation, then yes, but if not, then additionally to these 2 are considered reflections. Aug 18 '20 at 12:58
• What's meant by the phrase "keep orientation"? I was under the impression that translations and rotations could be represented as a composition of reflections.
– bxw
Aug 19 '20 at 12:34

## 2 Answers

Since gradients are invariant under translations, we may assume without loss of generality that the two systems of Cartesian coordinates have the same origin, and each line passes through that common origin. The transformation from coordinates $$x,\,y$$ to coordinates $$X,\,Y$$ satisfies$$X=x\cos\theta-y\sin\theta,\,Y=x\sin\theta+y\cos\theta$$for some $$\theta\in\Bbb R$$. If $$y=mx$$ and $$Y=nX$$,$$0=x\sin\theta+mx\cos\theta-nx\cos\theta+nmx\sin\theta\implies n=\frac{m+\tan\theta}{1-m\tan\theta}.$$Finally,$$\frac{\frac{m_2+\tan\theta}{1-m_2\tan\theta}-\frac{m_1+\tan\theta}{1-m_1\tan\theta}}{1+\frac{m_1+\tan\theta}{1-m_1\tan\theta}\frac{m_2+\tan\theta}{1-m_2\tan\theta}}=\frac{\left(m_{2}-m_{1}\right)\left(1+\tan^{2}\theta\right)}{\left(1+m_{1}m_{2}\right)\left(1+\tan^{2}\theta\right)}=\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}.$$In closing, it's worth noting that Boothby's request to use a change of Cartesian coordinates not only gives us more work to do than necessary, it makes the final result look like an accident. It is not. Writing $$m_1=\tan\theta_1$$ etc., $$\frac{m_2-m_1}{1+m_1m_2}=\tan(\theta_2-\theta_1)$$, so the result follows from the rotational invariance of angles in the plane.

• Why do we need to know gradients are invariant under translations in this case? If we were working with a different expression, would this still be a condition we needed?
– bxw
Aug 24 '20 at 13:58
• @bxw If we didn't use that fact, the coordinate systems might have a more general relationship, & their expressions for the line might be more general too. It just removes from the problem what are actually spurious degrees of freedom.
– J.G.
Aug 24 '20 at 14:10
• I understand, thank you.
– bxw
Aug 24 '20 at 14:15

If you have two cartesian coordinate systems, $$Oxy$$ and $$\Omega\xi\eta$$, then the equation relating them is $$\begin{pmatrix}\xi \\ \eta \end{pmatrix} = \begin{pmatrix}a & b \\ c & d \end{pmatrix} \begin{pmatrix}x \\ y \end{pmatrix} + \begin{pmatrix}\xi(O) \\ \eta(O) \end{pmatrix},$$ where

1. the matrix $$\begin{pmatrix}a & b \\ c & d \end{pmatrix}$$ in invertible and
2. $$\xi(O)$$ and $$\eta(O)$$ are the coordinates of $$O$$ in the second coordinate system.