**Question Corner and Discussion Area**

Suppose I am given thatThe following hints should help you answer this question.How do I prove that, if one line contains exactly

- If
PandQare two points, there is exactly one line containingPandQ- If
Lis any line, there is a pointPwhich does not lie onL- There are at least three points on every line
- Any two distinct lines intersect at exactly one point
- There exists at least one line.
npoints, then

- Every line contains exactly
npoints?- Every point lies on exactly
nlines?- The space contains points and lines?

First, try proving that, for any two lines *L* and *M*, there is at
least one point *R* not on either of them. Do this using the fact that
*L* and *M* each have at least three points, so you can find a point
*P* on *L* which isn't the intersection point, and a point *Q* on *M*
which isn't the intersection point. The line *PQ* will have a third point
*R* on it (because every line has at least three points). *R* is not
on *L* (if it were, *L* and *PQ* would both be lines containing *R* and
*P*. Since there is a unique line joining any two points, *L* would have
to equal *PQ*, contradicting the fact that *Q* is not on *L*). Similarly,
*R* is not on *M*.

Now you can prove that any two lines *L* and *M* have the same number of points.
They have their intersection point *X* in common. For each remaining
point *P* on *L*, the line *RP* intersects *M* in a point *f*(*P*). Show that
*f* establishes a 1-1 correspondence between the points on *L* (other than
*X*) and the points on *M* (other than *X*).

To prove part (2), for any point *P*, show that there must be at least
one line *L*
not containing *P*. Every line through *P* intersects *L* in a point.
Show that this process sets up a 1-1 correspondence between lines through
*P* and points on *L*, proving that *P* is on exactly *n* lines.

Now pick a point *P*. Every other point must lie on a line through *P*.
There are *n* such lines, with *n*-1 points on each, so there are
*n*(*n*-1) points not including *P*. Thus there are *n*(*n*-1)+1 points in
total. A similar argument gives you the number of lines.

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