Given five coplanar points such that no line passes through exactly two of them, prove that they are all collinear. The problem: 

There are five points on a plane. There is no line that passes through exactly two points. Prove that these five points are collinear.

My attempt: The method I am trying to use is proof by contradiction. Let $P_1$, $P_2$, $P_3$, $P_4$, and $P_5$ be the points. Let's assume that only points $P_1$, $P_2$, $P_3$, and $P_4$ lie on a same line. Let $P_5$ be a distinct point. The line that passes through $P_1\ldots P_4$ has a slope $k_1$. Since $P_5$ is a distinct point, we can define another line with different slope that passes trough points $P_1$ and $P_5$. That is a line that passes through two points. That is a contradiction, since such a line should not exist. So, point $P_5$ has to lie on a same line that passes through other points. So, all of the five points are collinear. $\square$

Is my proof correct? 

 A: If there are five points, and no line goes through exactly two of them, and the five points are not collinear, then exactly three or four points must be collinear.
First, suppose exactly three points are collinear, as $A$, $B$, $C$ below. Then $AD$, $AE$, $BD$, $BE$, $CD$, $CE$, and $DE$ pass through exactly two points, which contradicts the supposition. 
And even if two sets of three points are collinear, as $F$, $G$, $H$ and $F$, $J$, $K$, still $GJ$, $GK$, $HJ$, and $HK$ pass through exactly two points, against the supposition.
Finally, if exactly four points are collinear, as $L$, $M$, $N$, $O$, then $LP$, $MP$, $NP$, and $OP$ pass through exactly two points, again a contradiction.
Therefore, the five points must be collinear.
A: If not all five points are collinear, then some three of the points must form a triangle.  Each edge of this triangle defines a line on which are two of the points.  In order to satisfy the three-point-per-line requirement, the other two points must be distributed upon each of these three edge-lines - not possible!  Each pair of lines shares one point, so they cannot also share a second, distinct point, so each of the two remaining points must be placed on only one of the three lines, but this can only be done for two of the lines, leaving the third passing through only two points.
A: Assume first that three points are collinear.  Then, two of the five points may be connected with a distinct line, a contradiction.  (break into subcases here)
Next, assume that four points are collinear.  Then, the remaining point may be connected with one of the four collinear points, a contradiction.  The remaining possibility is that all five are collinear.
I don't think you even need to mention different slopes.  Just that in Euclidean plane, between any two points there is a unique line connecting them.
