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Formulas/maths/Straight Lines/Condition for Collinearity

Condition for Collinearity

Three points (x₁,y₁), (x₂,y₂), (x₃,y₃) are collinear if and only if the slopes of any two pairs are equal.
Derivation

Three points A(x1,y1)A(x_1,y_1), B(x2,y2)B(x_2,y_2), C(x3,y3)C(x_3,y_3) are collinear if and only if they all lie on one line — equivalently, the slope from AA to BB equals the slope from AA to CC :

y2−y1x2−x1=y3−y1x3−x1\frac{y_2 - y_1}{x_2 - x_1} = \frac{y_3 - y_1}{x_3 - x_1}

Cross-multiplying:

(y2−y1)(x3−x1)=(y3−y1)(x2−x1)(y_2 - y_1)(x_3 - x_1) = (y_3 - y_1)(x_2 - x_1)

Expanding and rearranging:

x1(y2−y3)+x2(y3−y1)+x3(y1−y2)=0x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0

This is exactly the expansion of the determinant:

 ∣x1y11x2y21x3y31∣=0 \boxed{\,\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0\,}

The determinant form is symmetric in all three points and is preferred when no natural "base" point exists.

Key Idea
Vector form. In vector language, $A$, $B$, $C$ are collinear if and only if $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel — i.e. their cross product vanishes:
> > $$ > \overrightarrow{AB} \times \overrightarrow{AC} = 0 > $$ > > With $\overrightarrow{AB} = (x_2-x_1,\, y_2-y_1)$ and $\overrightarrow{AC} = (x_3-x_1,\, y_3-y_1)$, the cross product (z-component) is: > > $$ > (x_2-x_1)(y_3-y_1) - (x_3-x_1)(y_2-y_1) = 0 > $$ > > This is exactly the determinant condition above expanded. The determinant **is** the cross product.