Is there a way to prove that two points $P, Q$ are collinear using a method similar to that if points $A, B, C$ are collinear then $$\operatorname{area} \triangle ABC = \frac{1}{2}\begin{vmatrix}x_A & y_A & 1 \\ x_B & y_B & 1 \\ x_C & y_C & 1\end{vmatrix} = 0$$
I know that it is obvious that two points are collinear but I would just like to know if it can be proven in a way similar to this.
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$\begingroup$No. Not in Euclidean geometry. I can't comment on non-Euclidean geometry.
The method you mention for three points cannot be applied to two because with three points one can define an enclosed region of the plane and therefore find its area. If the area is zero then the points are collinear. This will not work with two points because two points are not enough to define an enclosed region of the plane. So the concept of area won't make sense with only two points
There's another way. We can show three points $P, Q, R$ are collinear by showing the vectors $\vec{PQ}$ and $\vec{PR}$ are parallel. Note that this works in 2D and 3D. This... also can't be carried over to two points $P$ and $Q$. I suppose the best we could is to say that $P$ and $Q$ are collinear if and only if the vectors $\vec{PQ}$ and $\vec{PQ}$ are parallel. But of course they are.. they're the same vector. Or we could say if and only if $\vec{PQ}$ and $\vec{QP}$ are parallel. And of course they are. Each is $-1$ times the other. I only bring this other method up to further illustrate how the notion of "proving collinearity" for two points doesn't really make sense.
And here's why - as mentioned in the comments on the question, the fact that two points are collinear is part of the axioms. Specifically, axiom #1. Axioms are statements and properties that are assumed to be true. They're taken for granted. There's nothing to prove.
$\endgroup$ 1 $\begingroup$If think the best analoguous argument you make when you pass from three points to two points is the following:
Three points A,B,C are collinear iff the area of the triangle $\Delta_{ABC}$ is $0$.
Two points A,B coincide iff the length of the segment $AB$ is $0$.
And of course:
Four points A,B,C,D are coplanar iff the volume of the tetraedron $\Delta_{ABCD}$ is $0$.
$\endgroup$ $\begingroup$The straight two-point analogue of the property would be in 1D,
$$\begin{vmatrix}x_A & 1 \\ x_B & 1 \end{vmatrix} = 0$$
and expresses that the two points are coincident (i.e. they don't define a line, just like three collinear points don't define a plane, four coplanar points don't define space...)
If you want a truly 2D expression, you can
- add any point along $AB$ (such as $A$, $B$ or the midpoint),
$$\begin{vmatrix}x_A & y_A & 1 \\ x_B & y_B & 1 \\ \frac{x_A+x_B}2 & \frac{y_A+y_B}2 & 1\end{vmatrix} = 0$$
- add a point of the line at infinity, $$\begin{vmatrix}x_A & y_A & 1 \\ x_B & y_B & 1 \\ 0 & 0 & 0\end{vmatrix} = 0.$$
Two points are always collinear, I presumed that you are asking the same question for 3 points
Then there are the following ways to prove it
Prove that the line joining 3 points make 180° angle
Prove that the area of triangle joining three points is 0
Let the points are A,B,C then prove that AB+BC=AC
All the mentioned things are actually same , but stated in different ways
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