Proof: [br]Consider any line [i]l[/i] in an affine plane. By Theorem 7.1, we know that there are k distinct points that lie on [i]l.[/i] Now, consider the point P not on [i]l. [/i]Axiom 2 tells us that any two distinct points have exactly one line in common. There are k points on line l distinct from P. Thus, there are at least k distinct lines through point P. By Axiom 3, we also know that there is a line through point P that does not intersect line [i]l[/i]. Thus, there are at least k+1 distinct lines through point P. Now, consider that there are k+2 lines through point P. By Axiom 1, this would imply that there is a point O distinct from those that lie on line [i]l, [/i]and that does not lie on the line through P parallel to [i]l. [/i]We then know that the line through point P and O must also intersect line [i]l [/i]at some point R because Axiom 3 tells us that there is exactly one line through point P not intersecting [i]l. [/i]Either this point R is distinct from the other k distinct points on [i]l[/i] or point R is one of those k distinct points. If R is one of the k distinct points then we still have k+1 distinct lines through point P. If R is not one of the k distinct points then there are now k+1 points on line [i]l [/i]and we no longer have order k. Thus, there must be exactly k+1 lines through point P. Without loss of generality, we can then say that given any point in an affine geometry of order k there are exactly k+1 lines through that point. Or, each point in an affine geometry of order k lines on k+1 lines.