Q.Show, using determinants, that the points $(1,4)$, $(3,10)$ and $(-1,-2)$ 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 $\begin{vmatrix} 1 & 4 & 1 \\ 3 & 10 & 1 \\ -1 & -2 & 1 \end{vmatrix} = 0.$

Apply the row operations $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$ (Property 5, value unchanged): $R_2-R_1=(3-1,\ 10-4,\ 1-1)=(2,6,0),\qquad R_3-R_1=(-1-1,\ -2-4,\ 1-1)=(-2,-6,0).$ The determinant becomes $\begin{vmatrix} 1 & 4 & 1 \\ 2 & 6 & 0 \\ -2 & -6 & 0 \end{vmatrix}.$

Expand along the third column, whose only non-zero entry is the $1$ in row 1 (sign $+$): $= 1\begin{vmatrix} 2 & 6 \\ -2 & -6 \end{vmatrix} = (2)(-6)-(6)(-2) = -12+12 = 0.$

Since the determinant is $0$, the triangle these points would form has zero area — they lie on one straight line, i.e. they are collinear. …

The same conclusion follows, as in the verification, by checking that the slope between each pair of points is …

A common error is to guess collinearity from a rough sketch instead of computing. Points that look roughly in line may not be exactly collinear, so …