Vector Geometry Proofs GCSE: Collinearity and Parallel Lines

What tools do you use in vector proofs?

You need three building blocks:

1. Vector path between two points

$$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}$$

where O is the origin and a, b are the position vectors of A and B.

2. Parallel vectors

Two vectors are parallel if one is a scalar multiple of the other: p = kq for some scalar k. Parallel vectors point in the same or opposite direction (k negative means opposite).

3. Midpoint

The position vector of the midpoint M of AB is:

$$\overrightarrow{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$$

How do you prove two lines are parallel?

Express a vector along each line. If one is a scalar multiple of the other, the lines are parallel.

Worked example:

In triangle OAB, let $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. Point M is the midpoint of OA and point N is the midpoint of OB. Prove that MN is parallel to AB.

Step 1: Find the position vectors of M and N.

$$\overrightarrow{OM} = \frac{1}{2}\mathbf{a}, \quad \overrightarrow{ON} = \frac{1}{2}\mathbf{b}$$

Step 2: Express $\overrightarrow{MN}$.

$$\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = \frac{1}{2}\mathbf{b} - \frac{1}{2}\mathbf{a} = \frac{1}{2}(\mathbf{b} - \mathbf{a})$$

Step 3: Express $\overrightarrow{AB}$.

$$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$$

Step 4: Compare.

$$\overrightarrow{MN} = \frac{1}{2}\overrightarrow{AB}$$

Since $\overrightarrow{MN}$ is a scalar multiple of $\overrightarrow{AB}$, the lines MN and AB are parallel. Also, MN = ½AB, confirming the midpoint theorem. □

How do you prove three points are collinear?

Three points A, B, C are collinear (lie on the same straight line) if $\overrightarrow{AC}$ is a scalar multiple of $\overrightarrow{AB}$ AND the vectors share a common point.

Worked example:

$\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$. Point C has position vector $\frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}$. Point D has position vector $\frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{b}$.

Show that O, C, D, and the midpoint M of AB are collinear.

Find each position vector:

• $\overrightarrow{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{b}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$

Express vectors from O:

• $\overrightarrow{OC} = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}$
• $\overrightarrow{OD} = \frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{b}$
• $\overrightarrow{OM} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$

Observe: $\overrightarrow{OC} = \frac{2}{3}(\mathbf{a} + \frac{1}{2}\mathbf{b})$... let's try a cleaner comparison.

$\overrightarrow{OC} = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}$. Is this a scalar multiple of $\mathbf{a} + \mathbf{b}$? Check: $k(\mathbf{a}+\mathbf{b})$ would need k = 2/3 and k = 1/3 simultaneously — contradiction. So C, D, and M are not simply multiples of a + b.

Instead, check whether $\overrightarrow{CD}$ is parallel to $\overrightarrow{OM}$:

$$\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC} = \left(\frac{1}{3} - \frac{2}{3}\right)\mathbf{a} + \left(\frac{2}{3} - \frac{1}{3}\right)\mathbf{b} = -\frac{1}{3}\mathbf{a} + \frac{1}{3}\mathbf{b} = \frac{1}{3}(\mathbf{b} - \mathbf{a})$$

$$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$$

So $\overrightarrow{CD} = \frac{1}{3}\overrightarrow{AB}$ — CD is parallel to AB (and one-third of its length). This shows CD is parallel to AB, not collinearity with O.

The key collinearity technique: to show A, B, C are collinear, compute $\overrightarrow{AB}$ and $\overrightarrow{AC}$. If $\overrightarrow{AC} = k\overrightarrow{AB}$ for some scalar k, all three points share point A and lie on the same line.

Common worked structure for exam proofs

Frequently asked questions

What is the difference between a vector and a position vector?

A vector describes a displacement (a direction and distance) with no fixed starting point. A position vector gives the position of a point relative to a fixed origin O: $\overrightarrow{OA}$ is the position vector of A. In proofs, you usually work with position vectors and compute displacement vectors like $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$.

If two vectors are parallel, does that mean the lines they lie on are parallel?

Not necessarily — parallel vectors could lie on the same line (collinear) or on separate parallel lines. To confirm the lines are separate (not collinear), you need to check that the lines do not share a common point.

How do I find the position vector of a point that divides a line in a ratio?

If M divides AB in the ratio m : n, then: $$\overrightarrow{OM} = \frac{n\mathbf{a} + m\mathbf{b}}{m + n}$$

For the midpoint (1:1 ratio), this gives ½(a + b), as expected.

What algebra mistakes are most common in vector proofs?

The most frequent error is dropping a negative sign when computing b − a or reversing the direction of a vector path. Always write $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ (destination minus source), never the other way round. Then simplify carefully, collecting terms in a and terms in b separately.

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