Collinear vectors: the 3-second determinant test

The test

Two vectors $\vec{u}(x_1, y_1)$ and $\vec{v}(x_2, y_2)$ are collinear iff:

$\det(\vec{u}, \vec{v}) = x_1 \cdot y_2 - x_2 \cdot y_1 = 0$

The mnemonic

It’s a cross product:

$x_1$	$x_2$
$y_1$	$y_2$

$\det = (x_1 \times y_2) - (x_2 \times y_1)$

Down-diagonal minus up-diagonal.

Why is it better than “same slope”?

• No division: no issue if $x_1 = 0$ or $x_2 = 0$.
• Works for vertical vectors: $\vec{u}(0, 3)$ and $\vec{v}(0, 5)$ → $\det = 0 \times 5 - 0 \times 3 = 0$. Collinear. ✓

Example

Are $\vec{u}(2, 3)$ and $\vec{v}(4, 6)$ collinear?

$\det = 2 \times 6 - 4 \times 3 = 12 - 12 = 0$.

Yes, they’re collinear.

Classic application: 3 collinear points

Three points $A$, $B$, $C$ are collinear iff $\vec{AB}$ and $\vec{AC}$ are collinear.

?À toi de jouer

Are u(3, −2) and v(6, 4) collinear?

A.No (det = 24)B.YesC.Cannot tellD.Only if multiplied by a scalar

Valider