Question

6. Prove: Through a point P there are exactly three lines parallel to p, the polar of P (i.e., the three lines have no points in common with line p)

Answer 1

If points A,B and C are on one line and A', B' and C' are on another line then the points of intersection of the lines AC' and CA', AB' and BA', and BC' and CB' lie on a common line called the Pappus line of the configuration.

Axioms

1. There exists at least one line.

2. Every line has exactly three points.

3. Not all lines are on the same point.

4. If a point is not on a given line, then there exists exactly one line on the point that is parallel to the given line.

5. If P is a point not on a line, there exists exactly one point P' on the line such that no line joins P and P'.

6. With the exception in Axiom 5, if P and Q are distinct points, then exactly one line contains both of them.

Let P be any point. By corrected axiom 3, there is a line not containing P. This line contains points A,B,C [Axiom 2]. P lies on lines meeting two of these points, say B and C [Axiom 5]. There is exactly one line through P parallel to BC [Axiom 4].

Axiom 6

With the exception in Axiom 5, if P and Q are distinct points, then exactly one line contains both of them.

There can be no other line through X since by Axiom 4 it would have to meet BC at a point other than A, B or C [Axioms 6 and 5], and this would contradict Axiom 2.

The Pappus geometry has 9 points and 9 lines.

1. There exists at least one line.

2. Every line has exactly three points.

3. Not all lines are on the same point.

4. If a point is not on a given line, then there exists exactly one line on the point that is parallel to the given line.

5. If P is a point not on a line, there exists exactly one point P' on the line such that no line joins P and P'.

6. With the exception in Axiom 5, if P and Q are distinct points, then exactly one line contains both of them.