对其它“垂直的”数轴对、坐标切换和对称变换进行分类

Other "orthogonal" pairs of axes

Intuition: for any nonzero vector $(x,y)$, each two "neighboring" pairs among $(x,y)$, $(-y,x)$, $(-x,y)$, $(-y,-x)$ should be orthogonal. (Draw a picture to gain intuition!)

So by the parametric definition of a line, we see that the "$(x,y)$-axes" (the line $\{(tx,ty):t\in\mathbb{R}\}$) and "$(-y,x)$-axes" (the line $\{(-ty,tx):t\in\mathbb{R}\}$) should also be orthogonal axes. (Note that we have not formally defined orthogonality yet; that will take place in the next section.)

Change of coordinates, and reflections over general lines

We can re-write $(a,b) = a(1,0)+b(0,1)$ in the form $c(x,y)+d(-y,x)$ for some reals $c,d$ depending linearly on $a,b$. It's just a system of linear equations: $cx - dy = a$ and $cy + dx = b$ is equivalent to $c = \frac{ax+by}{x^2+y^2}$ and $d = \frac{-ay+bx}{(-y)^2+x^2}$.

In other words, the point $(a,b)$ in the $(1,0),(0,1)$-axes system is the same as $\frac{ax+by}{x^2+y^2}(x,y) + \frac{-ay+bx}{x^2+y^2}(-y,x)$ in the $(x,y),(-y,x)$-axes system.

This also allows us to define reflections easily, as in the case of $x,y$-axes before: the reflection of $(a,b)$ over the line $\{(tx,ty)\}$ is just $\frac{ax+by}{x^2+y^2}(x,y) - \frac{-ay+bx}{x^2+y^2}(-y,x)$, where we've just negated the $(-y,x)$ coordinate.