TITLE: Methods for showing three points in $\mathbb{R}^2$ are colinear (or not)
QUESTION [10 upvotes]: A common question is to prove/disprove that three points in $\mathbb{R}^2$ are colinear.  For, example

Show that $(-1, 8)$, $(1, -2)$ and $(2, 1)$ lie on a common line. 

What are some methods one could use to answer questions of this nature?

REPLY [9 votes]: Here are three tests of collinearity.
Distance Test
This is the first test I learned to check if three points were collinear. It actually tests if one point is between two others. Given three points $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$, the point $(x_2,y_2)$ is between $(x_1,y_1)$ and $(x_3,y_3)$ when
$$
|(x_1,y_1)-(x_3,y_3)|=|(x_1,y_1)-(x_2,y_2)|+|(x_2,y_2)-(x_3,y_3)|
$$
If any of the three points is between the other two, they are collinear; that is, if any of the distances is the sum of the other two.
$\hspace{3cm}$
Area Test
The area of a triangle is given by half the determinant of the vectors of two of the sides. That is, given
$\hspace{4cm}$
The area of the triangle is
$$
\frac12\det\begin{bmatrix}x_2-x_1&y_2-y_1\\\\x_3-x_1&y_3-y_1\end{bmatrix}
$$
The three points are collinear exactly when this area is $0$.
Volume Test
Given three points in $\mathbb{R}^2$, first we move them into $\mathbb{R}^3$, $z=1$ to be precise. Next the volume of the tetrahedron formed by three points and the origin is one sixth the determinant of the three points.
That is, the volume of
$\hspace{3cm}$
is
$$
\frac16\det\begin{bmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{bmatrix}
$$
The volume of a pyramid is one third of the base times the altitude. In the case above, the altitude is $1$. Thus, the points are collinear exactly when this volume is $0$.