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?