# Collinear points

In Geometry, a set of points are said to be **collinear** if they all lie on a single line. Because there is a line between any two points, every pair of points is collinear. Demonstrating that certain points are collinear is a particularly common problem in olympiads, owing to the vast number of proof methods.

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## Slope-based collinearity test

Collinearity tests are primarily focused on determining whether a given 3 points \(A, B\), and \(C\) are collinear. This is because it is easily extensible: e.g. showing 4 points \(A, B, C, D\) are collinear can be done by first showing \(A, B, C\) are collinear then showing \(B, C, D\) are collinear as well.

When the coordinates of the points are given, this problem is relatively simple, if sometimes computationally challenging. One simple test simply finds the equation of the line between \(A\) and \(B\) using standard methods, then finding the equation of the line between \(B\) and \(C\), and comparing.

Let \(A = (1, 2), B = (2, 4), C = (3, 6)\). Are \(A, B, C\) collinear?

The slope of the line between \(A\) and \(B\) is \(\frac{4 - 2}{2 - 1} = 2\), making the equation of the line between \(A\) and \(B\) \(y = 2x + b\), and setting \(x = 1, y = 2\) we determine \(b = 0\). Thus, the equation of the line becomes \(y = 2x\). Similarly, the slope between \(B\) and \(C\) is \(\frac{6 - 4}{3 - 2} = 2\), making the equation of the line between \(B\) and \(C\) also \(y = 2x\). This is the same line, so \(A, B, C\) are collinear.

This can be slightly improved:

## Linear algebra-based collinearity test

Another, more advanced, method utilizes the fact that the polygon formed by \(A\), \(B\), and \(C\) is necessarily degenerate, and therefore has area 0. The shoelace formula, which uses some linear algebra, may thus be used to find the area of "triangle" \(ABC\), and if it is 0 the points are necessarily collinear. This is a popular technique when using barycentric coordinates, where explicitly finding equations of lines is computationally difficult.

Let \(A = (1, 2), B = (2, 4), C = (3, 6)\). Are \(A, B, C\) collinear?

We find that the determinant of the matrix

\[\begin{pmatrix}1&1&2\\1&2&4\\1&3&6\end{pmatrix}\]

is 0. Thus, by the shoelace formula, the area of \(\triangle ABC\) is 0 and so \(A, B, C\) are collinear.

## Other collinearity tests

Collinearity is a particularly nice property when not explicitly constructed; in particular, there are a number of sets of interesting points that turn out to be collinear, and thus one method of proving collinearity is to demonstrate that the conditions for these theorems hold. For example, to show \(A, B, C\) are collinear, one might show that one can construct a configuration so that \(A, B, C\) are the intersections specified by Pascal's theorem (which states these intersections are collinear).

- For a given triangle, orthocenter, circumcenter, and the center of the nine point circle are
*always*collinear (regardless of the triangle). Together, they form the Euler line. - Menelaus' theorem gives a criterion for points on the sides of a triangle to be collinear.
- Given 6 points on a conic, typically a circle, Pascal's theorem states that certain intersections are collinear.
- Monge's theorem gives a criterion for three circles to induce a set of collinear points. This is a major win for projective geometry, and Desargues' theorem in particular.

## See Also

**Cite as:**Collinear points.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/collinear-points/