Indirect proof for Line-Intersection theorem My geom teacher gave us this question about this specific theorem.
Line-Intersection Theorem.
If two lines intersect then their intersections have exactly one point.
She wanted us to negate the statement above and then provide a contradiction for it. 
In my mind it's actually pretty easy but I'm not sure of my answer. I want to cross reference my answer with other's
 A: You can also build it on the Axiom that there is one and only one line between any two distinct points in space.
Suppose two distinct lines $l_1$ and $l_2$ intersect at more than one point. Pick any two of these points. Both $l_1$ and $l_2$ are lines which join these two points. But thiis contradicts the fact the two lines are disjoint proving there can only be one point of intersection between any two distinct lines. 
As the other answer implies it is also important to assert the fact that the two lines must be distinct since the result is not true for overlapping lines. 
A: Hint: If you know analytic geometry (I refer only to the $\mathbb R^2$-plane), you could try to study the system of equations
$$l_1:   y=mx+q $$
$$l_2:  y=nx+r $$
in the case where the pairs $(m,q)$ and $(n,r)$ are not equal (i.e. the lines $l_1$ and $l_2$ are distinct) and $m\neq n$, i.e. the lines are not parallel. 
A: Suppose lines m and l intersect in more than 1 point or 0 points.  1) They cannot intersect in 0 points, otherwise they would not intersect.  
2) Now suppose m and l intersect in 2 or more points.  Then by postulates, and if you consider any 2 of the points, m and l must both be the same line.  But this contradicts our hypothesis that different, distinct lines are intersecting.  Therefore lines m and l intersect in exactly 1 pt.  QED
