Solution: The sum of the lengths of any two line segments is equal to the length of the third line segment then all three points are collinear.
Consider, $A=(1,5) B=(2,3)$ and $C=(-2,-11)$
Find the distance between points; say $A B, B C$ and $C A$
$$
\begin{array}{l}
A B=\sqrt{(2-1)^{2}+(3-5)^{2}}=\sqrt{(1)^{2}+(-2)^{2}}=\sqrt{1+4}=\sqrt{5} \\
B C=\sqrt{(-2-2)^{2}+(-11-3)^{2}}=\sqrt{(-4)^{2}+(-14)^{2}}=\sqrt{16+196}=\sqrt{212} \\
C A=\sqrt{(-2-1)^{2}+(-11-5)^{2}}=\sqrt{(-3)^{2}+(-16)^{2}}=\sqrt{9+256}=\sqrt{265}
\end{array}
$$
Since $A B+B C \neq C A$
Therefore, the points $(1,5),(2,3)$, and $(-2,-11)$ are not collinear.