Prove collinearity with a common point

Prove collinearity with a common point

Choose two displacements sharing a point, then test whether one is a scalar multiple of the other. The same scalar must work in every coordinate. State the common point when turning the vector equation into a collinearity conclusion.

Three distinct points A,B,C are collinear iff $\overrightarrow{AB}=k\overrightarrow{BC}$ for a nonzero real k. They share B. The picture shows one possible ordering; the scalar can be positive or negative for other orderings.

Justify both directions

For the forward implication, points on the same line give parallel, nonzero AB and BC, hence a scalar multiple. Conversely, the scalar relation gives parallel segments AB and BC; because they share B, their supporting lines coincide, so A,B,C lie on one straight line.

The source gives two equivalent forms: $\overrightarrow{AB}=\lambda\overrightarrow{AC}$ with A common, and $\overrightarrow{AC}=\mu\overrightarrow{BC}$ with C common, for nonzero real λ,μ. The distinct-point condition is essential and also excludes coincident endpoints implicit in some scalar choices.

Use the linked examples to factor a common displacement, compare every coordinate, and justify the common point. To disprove collinearity, show that no scalar satisfies the vector equation. Merely finding one scalar that fails is insufficient. Self-Practice 1 also identifies the nondegenerate parallelogram O,A,B,D from its directed opposite sides.