... of course there's no single line that all the points lie on. They've been defined to be non-collinear.
Edit: can't reply because of HN's stupid rate-limit mechanism, but to this:
>So the theorem proves that no matter which way you arrange any finite set of points, except for all on the same line, then you can always find a line with exactly two points.
Of course you can. It's absolutely implied by the problem definition. My 9 year old could do this, given a ruler and a pencil, with 100% success rate. I absolutely do not believe this is a novel "theorem"
So the theorem proves that no matter which way you arrange any finite set of points, except for all on the same line, then you can always find a line with exactly two points.
You may think "I'm sure I can arrange these points in a way where EVERY line will cross three or more points" but you will fail if you try unless ALL points are colinear.
For all arbitrarily sized (but finite) sets of not collinear points, there's always a line that passes through exactly two points in the set.
Given N points, N > 2, can you arrange them in a Euclidean plane so that (1) they are not all on the same line, and (2) every line that goes through two of the points must also go through at least one more of the points?
The theorem says that you cannot do this.