Three points A, B and C are collinear if the vector between two of them is a scalar multiple of the vector between two others that share a point. Find AB and AC (or AB and BC), show one equals k times the other for some number k, then state that since the vectors are parallel and share the common point, A, B and C lie on a straight line. The common point is the part students forget to write down.
Showing that AB = k × AC only proves that line AB and line AC point in the same direction, meaning they are parallel lines. Two parallel lines that never touch are not the same line. What turns "parallel" into "collinear" is that AB and AC share the point A. Because they start from the same point and point in the same direction, they must lie along the same straight line.
This is why every collinearity proof needs two things: the scalar multiple relationship, and a sentence naming the shared point. Markers specifically look for the second part, and it is the most commonly dropped line in a rushed answer.
Step 1
Pick the point common to both vectors, usually the middle point
Step 2
Form two vectors sharing that point, e.g. AB and AC
Step 3
Show one vector = k × the other (find k)
Step 4
State: parallel + common point ⇒ same straight line
A, B and C have coordinates A(1, 2), B(4, 6) and C(10, 14). Prove that A, B and C are collinear. If turning two coordinates into a vector is still shaky, revise column vectors and magnitude first, because every collinearity proof starts with that subtraction.
Solution
AB = B − A = (4−1, 6−2) = (3, 4).BC = C − B = (10−4, 14−6) = (6, 8).(6, 8) = 2 × (3, 4), so BC = 2AB. The vectors are parallel.Quick check: if two column vectors are scalar multiples of each other, their components are always in the same ratio. Here 6 ÷ 3 = 2 and 8 ÷ 4 = 2, both give the same scalar, confirming the vectors are parallel before you even write out the full working.
O is the origin. OA = 6a − 2b, OB = 9a + kb and OC = 15a − 5b, where a and b are non-parallel vectors. Given that A, B and C are collinear, find the value of k.
Solution
AB = OB − OA = (9a + kb) − (6a − 2b) = 3a + (k + 2)b.AC = OC − OA = (15a − 5b) − (6a − 2b) = 9a − 3b.AB = λ × AC for some number λ. Compare the a-components: 3 = 9λ, so λ = 1/3.k + 2 = λ × (−3) = 1/3 × (−3) = −1. So k = −1 − 2 = −3.k = −3, AB and AC share the common point A and AB = ⅓AC, so A, B and C are collinear.The step students get wrong
When a and b are non-parallel vectors, you must compare their coefficients separately, one equation for the a-terms and one for the b-terms. Do not add the two vectors' components together as if a and b were ordinary numbers. Solve the a-equation first to find λ, then substitute that λ into the b-equation to find the unknown constant. Also do not forget the final sentence naming the shared point, it is the part that turns "parallel" into "collinear" and is often the only mark lost.
If the question already gives you that AB and AC both come out as multiples of the same vector, you can sometimes read off the ratio without fully solving for λ. In worked example 1, once you see that BC's components are exactly double AB's components, you already know the scalar is 2, no equation needed. This shortcut only works when both vectors are already given in the same fully worked-out form, so use it to check your answer rather than to skip the working in an exam.
Comparing coefficients of a and b separately in worked example 2 is the same skill you use when changing the subject of a formula: isolate one unknown at a time and substitute back in. The scale-factor idea behind "AB is k times AC" also shows up in similar triangle tests, where corresponding sides are in the same fixed ratio.
Does it matter whether I use AB and AC, or AB and BC?
No, either pair works, as long as the two vectors you choose share a common point. AB and AC share point A. AB and BC share point B. Pick whichever pair the question's given information makes easiest to calculate.
What if the scalar multiple comes out negative?
A negative scalar still proves the vectors are parallel, it just means they point in opposite directions along the same line. The collinearity conclusion still holds: state that the vectors are parallel (even with opposite direction) and share a common point, so the three points lie on one straight line.
Can I use gradients instead of vectors to prove collinearity?
If you have coordinates, showing that the gradient of AB equals the gradient of BC (and that B is common to both) proves the same thing. But when the question gives position vectors in terms of a and b rather than coordinates, the scalar multiple method is the only one that works, since there is no gradient to calculate. The same "compare a ratio, then confirm with a shared reference point" logic appears in the congruence tests, where matching ratios alone are not enough without a shared side or angle.
Is collinearity the same as saying B is the midpoint of AC?
No. Collinearity only means the three points lie on the same straight line, B could be anywhere along it. B is the midpoint only if AB and BC are equal in both direction and magnitude, which is a stronger, more specific condition than simply being parallel.
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Mr. Gan works with students who want a repeatable method for vector geometry, not a guess at which pair of points to compare.
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