How to Determine Number of Line Segments Going Through Points

By Paul Dohrman

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. 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.

About the Author

Paul Dohrman's academic background is in physics and economics. He has professional experience as an educator, mortgage consultant, and casualty actuary. His interests include development economics, technology-based charities, and angel investing.