Is $1$ the halfway point between $0$ and infinity? I am a math newbie, but I stumbled across this while I was working on a game in python. This is the proof I have, I believe it shows that the halfway point between $0$ and infinity is $1$:



If it is true, I'm probably not the first to figure this out, but even still, I just want some kind of confirmation.
This is very mind blowing for me right now :D
While I personally would not use the word "halfway", I don't find it totally unreasonable to say that 1 is in some sort of special position between $0$ and $\infty$. The sense I'm thinking of is that in the projective line (Wikipedia), over $\mathbb{R}$ or $\mathbb{C}$, the map $z\mapsto\frac{1}{z}$ flips all the numbers "below" $1$ to all the numbers "above" $1$ and vice versa (in particular, swapping $0$ and $\infty$), while fixing $1$.

Added: You can think of the projective line over $\mathbb{R}$ as a way of describing or capturing all the possible "slopes of lines" in the plane as a single mathematical object, and this object has a "symmetry" $z\mapsto \frac{1}{z}$ which makes $1$ stand out (though note that $-1$ stands out just as much). This directly relates to the idea behind your diagram, since you labeled your diagram with the slopes of the lines (the equations you wrote are incorrect though, as others have already pointed out.)

With regards to the projective line over $\mathbb{C}$, a.k.a. the Riemann sphere (Wikipedia), this is often manifested in illustrations as choosing the unit circle to be the "equator" between the "north pole" $\infty$ and "south pole" $0$. Here's the picture from Wikipedia:



(Of course, one can question to what extent the map $z\mapsto \frac{1}{z}$ is special.)
Short answer:

No, $1$ is the halfway point between $0$ and $2$.
The halfway point between $0$ and infinity is infinity.
(Unless you want to model a non-Euclidean world)

Long answer:

No, the image gives the values for $y'(x) = m$ of some lines $y(x) = m x$,
not the values for $y(x)$.

This value $m$ is not linear dependent on the angle $\alpha$ of the line $y(x)$ with the positive $x$-axis:

$$
m = \tan \alpha
$$


Your pick is the value at the mid angle $\pi/4$ (green line). 

It is indeed in the middle of the $\alpha$ interval $[0,\pi/2]$ (the blue line is $x=\pi/2$), but not a good pick for a middle value of $y'$ between $\alpha=0, y'=m=0$ and $\alpha=\pi/2, y'=m=\infty$.
Almost all value of $y'$ is in the right half of the interval, in $[\pi/4,\pi/2]$.

Here the average is better suited.

In fact over the angles from $0$ to $\pi/2$ the average is
$$
\overline{y'} = \frac{\int\limits_0^{\pi/2}\tan \alpha\,d\alpha}{\int\limits_0^{\pi/2} d\alpha}
= \frac 2 \pi \left[ -\ln \lvert \cos \alpha \rvert \vphantom{\frac11} \right]_{\alpha=0}^{\alpha\to \pi/2} \to \infty
$$
as expected.