Q.Show, using determinants, that the points $(2,3)$, $(4,7)$ and $(6,11)$ are collinear.

Three points are collinear exactly when the area-of-triangle determinant built from their coordinates equals zero.

The three points are collinear if and only if $\begin{vmatrix} 2 & 3 & 1 \\ 4 & 7 & 1 \\ 6 & 11 & 1 \end{vmatrix} = 0.$

Apply the row operations $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$ (Property 5 — the determinant's value is unchanged): $R_2-R_1 = (4-2,\ 7-3,\ 1-1) = (2,4,0), \qquad R_3-R_1 = (6-2,\ 11-3,\ 1-1) = (4,8,0).$ The determinant becomes $\begin{vmatrix} 2 & 3 & 1 \\ 2 & 4 & 0 \\ 4 & 8 & 0 \end{vmatrix}.$

Expanding along the third column, whose only non-zero entry is in row 1: $= 1\begin{vmatrix} 2 & 4 \\ 4 & 8 \end{vmatrix} = 1\big[(2)(8)-(4)(4)\big] = 1(16-16) = 0.$

Since the determinant equals $0$, the area of the triangle formed by these three points is zero, so the three points do not form a genuine triangle at all — they lie on a single straight line, i.e. they are collinear. …

The same conclusion can be reached, as shown in the verification, by comparing the slope between each pair of points — collin …

A common error is concluding the points are NOT collinear purely from a visual guess without actually computing the determinant (or the slopes) — the determinant test must always be carried out numerically, since three p …