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Could someone explain to me what this simple problem is asking me to do?

Draw three straight lines intersecting by twos?

Does this mean that each line in the set of three has to be intersected by the other two in the set thus forming a triangle?

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It is admittedly not entirely clear, but presumably what is meant is that no two of the lines are parallel, and that the three are not concurrent. So yes, they are the lines through pairs of vertices of a triangle. –  André Nicolas Oct 25 '11 at 4:01
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Essentially, yes: you get a triangle with its sides extended indefinitely. –  Brian M. Scott Oct 25 '11 at 4:26
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up vote 3 down vote accepted

It is more common in mathematics to say that lines intersect "pairwise" or "in pairs", which means that any pair of lines from the arrangement has a point of intersection. For three distinct lines this condition is satisfied if the lines are extensions of the sides of a triangle, or if all the lines intersect at one point. It could be the purpose of the exercise to notice that there is more than one type of configuration that meets the requirement.

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When you say there is more than one type of configuration that meets the requirement, is this the same thing as saying that there are a countably infinite number of triangles that can be formed? –  Matthew Hoggan Oct 26 '11 at 1:34
    
I was thinking of a configuration as a collection of distinct lines and distinct points plus the information of which lines contain which points, so with three distinct lines there are only two possibilities up to a relabelling of the points and lines. One is the triangle and the other is three lines through a point. If you consider the distances or angles in the configuration as additional distinguishing marks, then as you say there will be infinitely many different shapes of triangle and infinitely many different triples of lines through a single point. Uncountably many, though. –  zyx Oct 26 '11 at 1:58
    
thank you for the clarification!!! –  Matthew Hoggan Oct 26 '11 at 3:33
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