# Points following an axiom [closed]

Mark any finite number of points on a plane. It should meet this axiom.

Axiom:

Original way of stating it: If a line (infinite) is drawn passing through exactly n (n>0) points, any line drawn parallel to it will pass through exactly n or 0 points.

New way of stating it: If any 2 parallel lines are selected, either one (or both) of them passes through zero points, or they pass through the same number ($n$) of points.

Don't my original and new way of stating the axiom mean the same thing?

## closed as off-topic by Ian MacDonald, Len, leoll2, Mike Earnest, mdc32Apr 27 '15 at 1:47

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Ian MacDonald, Len, leoll2, Mike Earnest, mdc32
If this question can be reworded to fit the rules in the help center, please edit the question.

• Infinitely many points are OK? – xnor Apr 25 '15 at 15:15
• This appears to be a plain old math problem. Consider whether text-book style problems are appropriate meta.puzzling.stackexchange.com/questions/2783/… – Bob Apr 25 '15 at 15:27
• this is any square $n$X$n$ plan permeated by $k$<$n²$ perpendicular lines to each other and to plan axes , where all the points taking part of these lines are removed from this plan . – Abr001am Apr 25 '15 at 15:36
• I'm not sure how to classify this problem. It looks like an elementary geometry problem (not even a problem, maybe); if so, should be moved to another SE or deleted. If it is a lateral-thinking puzzle, then the tags are wrong, also I can't see anything outside the box here. I'm very confused. – leoll2 Apr 25 '15 at 15:41
• in my view , it is too easy for a puzzle , only if i did interept it properly – Abr001am Apr 25 '15 at 15:50

• Also any set of $n > 1$ collinear points would work.. – Ben Frankel Apr 25 '15 at 16:22
The parallel line will always pass through exactly $0$ points of the other line, else they wouldn't be parallel!
What if the two lines coincide? They both pass through the $n$ points, but they're called coincident and not parallel