The collinearity test based on coordinates

The practical criterion

Finding a coefficient k by hand can sometimes be tedious. A quicker method is used instead. Let u(x;y) and v(x';y') be given. The vectors u and v are collinear if and only if:

xy' - yx' = 0

This quantity is called the determinant of the two vectors.

Example

Let u(3;-2) and v(-6;4). We calculate: xy' - yx' = 34 - (-2)(-6) = 12 - 12 = 0. Therefore, u and v are collinear (we can verify that v = -2*u).

Another example: u(5;1) and v(2;3). Determinant = 53 - 12 = 15 - 2 = 13, which is not equal to 0, so u and v are not collinear.

Why this criterion works

If v = ku, then x' = kx and y' = ky, so xy' - yx' = x(ky) - y(kx) = kxy - kx*y = 0. The converse can be shown in a similar way, provided that u is non-zero.

Method

1. Identify the coordinates of the two vectors.
2. Calculate xy' – yx'.
3. If the result is 0: the vectors are collinear. Otherwise, they are not.

Common pitfall

Be careful with the order of the terms: it is xy' - yx', and definitely not xx' - yy'. A simple error in order or sign completely changes the conclusion. Always check your calculations by rewriting the coordinates in the correct order.