Linear fractional transformations look like \(f(x) = \frac{ax+b}{cx+d}\) for some real constants \(a,b,c,d\). Perhaps I can convince you they are fractional, but how are they linear? Considering the simplest example, how could something like \(f(x) = 1/x\) be linear? We’d try to show that to some extent, it is linear. The main idea is that we embed the real line into the cartesian plane and cheat a bit by using projections.

Consider the transformation \(r: (x,y) \to (y,x)\). This is nothing more but reflecting along \(y=x\), and it’s certainly linear in the sense that \(r(u+v) = r(u) + r(v)\) for whatever coordinates \(u, v\) we throw at \(r\). We now embed the real line inside the plane as the set \(\{(x, 1) \forall x \in \mathbb{R}\}\), i.e. the line \(y=1\). Under the reflection \(r\), we would map some point \((c,1)\) to \((1,c)\). Graphically, we are mapping the red points to the blue points as you could see below.

How does this have anything to do with \(f(x) = \frac{1}{x}\)? The trick is that for each blue point \((1,c)\), we draw a green line connecting it and the origin, and see where the green line intersects with the red line. We can see that that intersection point is exactly \((1/c, 1)\)!

Secretly, we’re using two similar triangles. One has vertices \((0,0), (0,1), (\frac{1}{c}, 1)\) and the other has vertices \((0,0), (0,c), (1,c)\).

As such we have sort of been able to map \(x\) to \(\frac{1}{x}\), but we cheated a little by going up one dimension and doing some projection at the end. As an exercise, could we make a similar construction for \(f(x) = 2/x\) by choosing a different transformation \(r\)? How about \(f(x) = x + 1\)? How about \(f(x) = 1 + \frac{1}{x}\)?

There’s another way of finding inverse points, often used in straighedge and compass construction, that uses circles and a pair of similar right-angled triangles. How is that method different from the above?

To make all of this formal, one needs to study a bit of projective geometry. You can read more about linear fractional transformations from its wiki.