How to Determine Number of Line Segments Going Through Points
A common problem in geometry classes is the determination of how many lines can be drawn through a set of points in a plane, two points at a time. No three points in the set are allowed to lie in a straight line. A simple example is if you have three points on a circle. Clearly they don’t form a line; no single line will pass through all three. But three lines can be drawn that pass through two points at a time. A simple formula solves the problem for you.
Draw, or suppose you have, n points in a plane 1. No three points lie in a straight line. You want to know how many lines can be drawn through two points at a time.
For example, you may have a circle with eight points, denoted A through H.
Pick one point and determine how many pairs of points it can be in. If there are n points, the answer is n-1. This is how many lines can pass through that first point and another point at the same time.
Continuing with the above example, A can be matched up with B or C or D or E or F or G or H. That’s seven possible matches.
Pick the next point over. Its pairing with the first point has already been counted, but its pairing with the n-2 other points hasn’t. Add n-2 to your earlier number, n-1, as possible lines through the points.
Continuing with the above example, B can have a line going through it and C through H. You don’t count a line going through B and A, since you already did that in Step 2. So the possible lines through B are six.
Continue with the pattern, adding n-3, then n-4, and so on. So the total sum of possible lines is n-1 + n-2 + n-3 + … + 1. This is the same as summing up 1 + 2 + 3 + … + n-1. It can be shown that the formula for 1 + 2 + 3 + … + n-1 is n(n-1)/2.
Continuing with the above example, there were eight points, so n=8 gives a total number of possible lines through the points of n(n-1)/2 = 8_7/2 = 28. You can verify this yourself by adding the 7 found in Step 2 to the 6 found in Step 3 to 5, 4, 3, 2 and 1 to get 28. It also matches the result discussed in the introduction where the number of points was n=3: n(n-1)/2 = 3_2/2 = 3 possible lines.