So I"ve got a homework question I"m stuck on. It"s asking me to develop a formula that when given $n$ points, it gives the number of straight lines that can be drawn through those points.

For example, the first two questions were "How many lines can be drawn through 3 points?" Which is 3, and "How many lines can be drawn through 4 points?" Which is 6. Now, it says "Develop a formula that gives the number of lines that can be drawn through $n$ points." I understand the relationship, and i"ve developed a formula, but it relies on having a predetermined list of answers.

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I"ve found that where $a_n$ is the number of points you have, you can find the number of lines with $a_n = a_{n-1} + (n-1)$ where $a_{n-1}$ is the previous item in the list. However, I don"t think that"s the formula that my teacher is looking for. Is there a less complicated way to solve this problem?

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edited Jul 30 "14 at 8:32 zarathustra
asked Aug 22 "12 at 23:11 SomekidwithHTMLSomekidwithHTML
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It says "drawn through $n$ points", but it really means "drawn through any two of $n$ points", where we assume no three points are collinear. So the number of lines is the same as the number of ways to choose two points out of $n$, where order doesn"t count. Do you know about permutations, combinations, binomial coefficients?

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answered Aug 22 "12 at 23:16 Robert IsraelRobert Israel
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We will assume that our $n$ points are in "general position." This means that no three of our points are on the same line.

Let our points be $P_1, P_2, P_3, \dots, P_n$.

First draw all the lines through $P_1$ and every other point. There are $n-1$ of these.

Now draw all the lines through $P_2$, and every other point, except for the line through $P_1$ and $P_2$, since that has already been drawn. There are $n-2$ of these, through $P_3$, $P_4$, and so on up to $P_n$.

Now draw all the lines through $P_3$ and every other point that have not already been drawn. We have already taken care of the line through $P_3$ and $P_1$, and also of the line through $P_3$ and $P_2$, so there are $n-3$ of these.

Continue. At the end, all we are drawing is the line through $P_{n-1}$ and $P_n$, just $1$ line.

Thus the total number of lines is $$(n-1)+(n-2)+(n-3)+\cdots+2+1,$$which will look more familiar as $$1+2+3+\cdots +(n-1).$$You probably know a formula for the sum of the first $k$ positive integers.

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Another way: Sit in turn on every one of our $n$ points. Draw the lines through that point and every other point. So for each of our $n$ points, you draw $n-1$ lines, for a total of $n(n-1)$.

However, this means that you have drawn every line twice. The line through $P_i$ and $P_j$ has been drawn once when you were sitting on $P_i$, and once again when you were sitting on $P_j$. So $n(n-1)$ overcounts our lines by a factor of $2$. That means that the actual number of lines is $$\frac{n(n-1)}{2}.$$