math problem! anyone know how to prove this?

Question

if n points lie in a plane and no three are collinear, prove that there are n(n-1)/2 lines joining these points?

Answer 1

Two points determine any line, extending for as long as may be wished/wanted.. For convenience I'll call them "end points," since you want lines "joining these points."

Given n points in a plane, you can pick the "first" end point in 'n' ways. Given that, you can pick the "second" end point from the remaining points in a further (n - 1) ways for each of your first choices. Therefore there are altogether n (n - 1) ways of picking the endpoints, DISTINGUISHING the order in which those endpoints are picked.

But either of the ends (A and B, say) could have been the "first" point picked, the other then being the "second" point picked. Yet the line (point A -----> point B) is exactly the same as the line (point B -----> point A). This is true for EVERY one of the n (n - 1) pairs that you first picked out. So the number of lines (otherwise, the number of independent ways of picking out pairs of points WITHOUT REGARD TO ORDER) is half of that n (n - 1) ways first found, that is:

n (n - 1) / 2. QED.

CHECK : You can CHECK that this is right by considering just 2 or 3 points. n = 2 gives you 2(1)/2, or just 1 line, and n = 3 gives you 3(2)/2 = 3 lines. Obviously thes two results are correct: from 2 points you indeed get just 1 line joining them, and from 3 points you get the 3 sides of the triangle they define.

(It's always a good idea to do a simple check like this, of the correctness of an argument made for general 'n.' If it works, that's fine; but if it doesn't, at least you know that there must be a flaw somewhere in your argument.)

Live long and prosper.

Answer 2

2 points are enough to determine a line. So pick two points from the n, and you've got a line. There are n choose 2 or n(n-1)/2 ways to choose 2 points from the n.

Conversely, If you have a line that goes through points in the plane then you know that there are exactly two points on that line since no three are collinear.