Condition for 3 Points in Plane to be Collinear

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Theorem

Let $A = \tuple {x_1, y_1}, B = \tuple {x_2, y_2}, C = \tuple {x_3, y_3}$ be points in the Cartesian plane.

Then:

$A$, $B$ and $C$ are collinear

if and only if the determinant:

$\begin {vmatrix}

x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix}$ equals zero.

Proof

We have that:

$A$, $B$ and $C$ are collinear

if and only if:

the area of $\triangle ABC = 0$

if and only if:

$\dfrac 1 2 \size {\paren {\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} } } = 0$ (from Area of Triangle in Determinant Form)

if and only if:

$\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} = 0$

$\blacksquare$

Sources

• 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text I$. Coordinates: $9$. Collinear points
• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 10$: Formulas from Plane Analytic Geometry: $10.10$: Area of Triangle with Vertices at $\tuple {x_1, y_1}$, $\tuple {x_2, y_2}$, $\tuple {x_3, y_3}$