Collinear points

Collinear points

Points are collinear when they lie on one straight line. To prove that \(A\), \(B\) and \(C\) are collinear, show two facts: a vector joining two of the points is a scalar multiple of a vector joining another pair, for example \(\overrightarrow{AC}=k\overrightarrow{AB}\), and the two vectors share a point, here \(A\). Parallel lines through a common point are the same line.

Method

1. Choose two vectors that share a point, such as \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\), or \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\).
2. Express both in the same terms: columns, or the given vectors \(\mathbf{a}\) and \(\mathbf{b}\).
3. Show that one is a number times the other.
4. Write both facts and the conclusion in words, then read off the ratio of lengths if it is asked for.

Example: \(A(2,\ -1)\), \(B(4,\ 0)\), \(C(10,\ 3)\). \(\overrightarrow{AB}=\begin{pmatrix}2\\ 1\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}8\\ 4\end{pmatrix}=4\overrightarrow{AB}\). So \(AC\) is parallel to \(AB\), and they share the point \(A\); hence \(A\), \(B\) and \(C\) are collinear. As \(AC=4AB\), \(BC=3AB\), so \(AB:BC=1:3\).

With letters: if \(\overrightarrow{OP}=2\mathbf{a}+\mathbf{b}\) and \(\overrightarrow{OQ}=6\mathbf{a}+3\mathbf{b}\), then \(\overrightarrow{OQ}=3\overrightarrow{OP}\), so \(O\), \(P\) and \(Q\) are collinear, with \(OP:PQ=1:2\).

Check

A scalar multiple on its own is not enough. In the right-hand figure \(\overrightarrow{CD}=\overrightarrow{AB}\), so \(AB\) and \(CD\) are parallel, but they lie on two different lines, so \(A\), \(B\), \(C\) are not collinear. Always name the common point.