# Existence of lines not containing given points in general position

I remember seeing something like the following problem in the past and would like to know if it has a solution (or if I can find a source for it).

Problem Given a finite set of points in the plane in general position (no three are collinear), does there exist a line that intersects no points?

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As Qiaochu alludes to in his answer, no general position assumption is necessary for this statement. – Matt E Oct 28 '10 at 16:15
The question seems to be a partial recollection of the Sylvester-Gallai problem, whose general position requirement is "not all collinear". For the problem as it was posted, any line far away from the points is a solution. – T.. Oct 28 '10 at 17:33
Yes, I think Qiaochu Yuan and T.. have the right idea, I was definitely mis-remembering the Sylvester-Gallai problem. – AnonymousCoward Oct 29 '10 at 4:36

Yes. Choose any direction in the plane. There are continuum many lines parallel to that direction. Only finitely many hit one of the points. So continuum many of these lines do not hit any points. This argument works as well for countably many points.

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Ah thanks, I don't know why this didn't occur to me. – AnonymousCoward Oct 28 '10 at 14:34

What is this general position stuff? Take any line outside the convex hull.

Maybe you're thinking of the Sylvester-Gallai theorem.

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Such statements are often easier to see in the projectively dual version. The dual statement would be that given any finite set of lines in the projective plane (which secretely we are thinking of the dual plan of lines in the projective plane where our original points lay), and a point $P$ not on them (this is the point in the dual plane corresponding to the line at infinity in our original plan), there is a point $Q$ distinct from $P$, and also not lying on any of the lines.

Just stating it again without the parenthetical explanations:

Given any finite set of lines in the projective plane, and a point $P$ not lying on them any of them, there exists a point $Q$ distinct from $P$ and not lying on any of them.

This will (pretty clearly) be true just provided we are working over an infinite field. (Thus we don't need that $\mathbb R$ is uncountable; the fact that it is infinite is enough. This is made clear in Agusti Roig's answer, of which the above is slighty more theoretical reformulation.)

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Surprisingly, the Sylvester-Gallai problem (where the line has to pass through two, and only two, of the given points) is a projective statement about points and lines that cannot be proved using only projective geometry. This is because it is false over some fields. e.g., the inflection points of a nonsingular cubic over the complex numbers are a counterexample. So a projective proof would need some special axiom ruling out one or more finite configurations that don't occur in RP^2, whose presence is equivalent to a square root of a negative number in the field of coordinates. – T.. Oct 28 '10 at 17:10

Ross' argument with equations says that, if you choose a Cartesian coordinate system in the plane, every straight line can be written as

$$y = ax + b \ , \qquad \text{or} \qquad x = 0 \ ,$$

for some real numbers $a, b$. Now, choosing a direction means picking up some $a \in \mathbb{R}$. Everyone of your finite (or countable) number of points $(x_n, y_n)_{n=1,2, \dots}$ determines a straight line: you just solve the equation for $b$ in terms of $y_n, a, x_n$

$$y_n = a x_n + b \qquad \Longleftrightarrow \qquad b = y_n - ax_n \ .$$

And then you just need to choose any $b \neq y_n - ax_n, \ n = 1,2,\dots$, which you can do, since there is just a finite (or countable) number of them, but an uncountable quantity of real numbers.

Alternatively, you could fix the height $b$ instead of the slope $a$; for instance, $b=0$, and the points $(x_n, y_n)_{n=1,2,\dots}$ would give you a finite (or countable) number of slopes of straight lines from the origin $(0,0)$ to each of your points, using the same equation as above:

$$a = \dfrac{y_n}{x_n} \ .$$

(In case you have $x_n = 0$, you just have the straight line $x=0$.)

Now you pick up any $a\neq\dfrac{y_n}{x_n}$, which you can do for the same reason as before, and you are done.

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For a very nice treatment of the combinatorics of point and line configurations and the symmetry of such configurations, look at Branko Grünbaum's book: Configurations of Points and Lines, American Mathematical Society, 2009.

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Thanks for the suggestion. – AnonymousCoward Oct 28 '10 at 22:50