G25 - Vector geometry proofs (Higher) | AQA GCSE Mathematics

Use vectors to prove that lines are parallel, points lie on one straight line, or an intersection divides a line in a particular ratio. You will replace measurements from a picture with routes that work for every shape satisfying the conditions. Clear arrows and a final geometric reason turn vector algebra into a proof.

Routes with labelled vectors (Higher)

Draw any convex quadrilateral and join the midpoints of its four sides in order. Must the new shape be a parallelogram, even when the original has no parallel sides? It looks like a job for a ruler. Vectors will let you put the ruler away.

You already know

Vectors add head to tail. Reversing an arrow negates the vector, and a scalar multiplies its length, reversing its direction when negative. A proof needs an argument covering every allowed case, rather than a convincing example.

First practise choosing a route. If $O=(0,0)$ , $A=(4,0)$ and $B=(1,6)$ , the route from $A$ to $B$ can go backwards to $O$ , then forwards to $B$ :

\overrightarrow{AB} =\begin{pmatrix}-4\\0\end{pmatrix}+\begin{pmatrix}1\\6\end{pmatrix} =\begin{pmatrix}-3\\6\end{pmatrix}

You reach the same endpoint as the direct arrow. A detour changes the path, but not the displacement.

Position vector

A position vector describes a point relative to a chosen origin $O$ . If $\overrightarrow{OA}=\mathbf{a}$ , then $\mathbf{a}$ is the position vector of $A$ . The origin need not be the bottom-left corner of a diagram or lie on coordinate axes.

Letters let one calculation cover many triangles. Keep the arrow direction visible before replacing it with a letter.

[DIAGRAM: asset_name: Two routes between the same points; asset_slug: aqa8300_g29_triangle_routes; caption: The route from A to B goes backwards along OA, then forwards along OB; description: A scalene triangle OAB with O at0comma0, A at6comma1 and B at2comma5. Blue arrows run O to A labelled2a and O to B labelled3b. A sage arrow runs A to B labelled3b minus2a. A small route statement below says A to O to B. Not drawn accurately.]
The route from A to B goes backwards along OA, then forwards along OB

The route rule

For any points $A$ , $B$ and $C$ ,

\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}

In particular, if $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$ ,

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

The arrows in a route must join head to tail.

Not on the formulae sheet: learn it.

Write a route as letters first. It is much easier to catch a reversed arrow before the diagram has become a page of algebra.

Worked example 1: a route through the origin

Question. In triangle $OAB$ , $\overrightarrow{OA}=2\mathbf{a}$ and $\overrightarrow{OB}=3\mathbf{b}$ . Work out $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ . Non-calculator.

Plan. Take the route $A$ to $O$ to $B$ .

\begin{aligned} \overrightarrow{AB}&=\overrightarrow{AO}+\overrightarrow{OB} &&\text{join the route}\\ &=-2\mathbf{a}+3\mathbf{b} &&\text{reverse }OA\\ &=3\mathbf{b}-2\mathbf{a} &&\text{collect the terms} \end{aligned}

Why this step? The known arrow is $O$ to $A$ , but our route travels $A$ to $O$ . That is where the minus sign comes from.

Answer. $\overrightarrow{AB}=3\mathbf{b}-2\mathbf{a}$ .

Check. Add $\overrightarrow{OA}$ : $2\mathbf{a}+(3\mathbf{b}-2\mathbf{a})=3\mathbf{b}=\overrightarrow{OB}$ .

Change only the requested direction. The triangle and its given vectors stay the same.

Worked example 2: reversing the route

Question. In the same triangle, work out $\overrightarrow{BA}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ . Non-calculator.

Plan. Follow $B$ to $O$ to $A$ , reversing $OB$ .

\begin{aligned} \overrightarrow{BA}&=\overrightarrow{BO}+\overrightarrow{OA} &&\text{head-to-tail route}\\ &=-3\mathbf{b}+2\mathbf{a} &&\text{reverse }OB\\ &=2\mathbf{a}-3\mathbf{b} &&\text{simplify} \end{aligned}

Why this step? Changing the start and finish reverses the entire vector, so both terms change sign.

Answer. $\overrightarrow{BA}=2\mathbf{a}-3\mathbf{b}$ .

Check. $\overrightarrow{AB}+\overrightarrow{BA}=(3\mathbf{b}-2\mathbf{a})+(2\mathbf{a}-3\mathbf{b})=\mathbf{0}$ , as a complete return journey should.

The symbol $\mathbf{0}$ means the zero vector. Giving a vector as just a length, or writing $AB$ when you mean $\overrightarrow{AB}$ , hides the direction that makes the argument work.

Watch out

Wrong: $\overrightarrow{AB}=\overrightarrow{OA}+\overrightarrow{OB}$ . Those arrows both start at $O$ , so they do not form the required route. Right: $\overrightarrow{AB}=-\overrightarrow{OA}+\overrightarrow{OB}$ .

Put the same route into a new triangle.

Points dividing a line in a ratio (Higher)

If $M$ lies on $AB$ and $AM:MB=1:2$ , the whole segment has three equal shares. The movement from $A$ to $M$ is one third of the movement from $A$ to $B$ , not one half.

[DIAGRAM: asset_name: A ratio gives a fraction of the whole vector; asset_slug: aqa8300_g29_ratio_point; caption: AM is one of the three shares making the whole vector AB; description: Triangle OAB has O at0comma0, A at6comma1, B at2comma5. M lies one third of the way from A to B at14over3 comma7over3. Arrows O to A and O to B are labelled a and b. A to M is sage, M to B grey, with AM to MB equals1to2 shown above the side. A sage dashed arrow O to M is labelled OM. Not drawn accurately.]
AM is one of the three shares making the whole vector AB

To find $\overrightarrow{OM}$ , first reach $A$ , then travel the required fraction of $AB$ . That keeps the starting point in the calculation.

A point inside a segment

If $M$ lies between $A$ and $B$ , with $AM:MB=m:n$ , where $m,n>0$ , then

\overrightarrow{AM}=\frac{m}{m+n}\overrightarrow{AB}

If $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$ ,

\overrightarrow{OM} =\mathbf{a}+\frac{m}{m+n}(\mathbf{b}-\mathbf{a})

Not on the formulae sheet: learn it.

You do not need to memorise a separate position-vector fraction. “Reach $A$ , then take the correct fraction of $AB$ ” builds it each time.

Worked example 3: one share out of three

Question. $\overrightarrow{OA}=\mathbf{a}$ , $\overrightarrow{OB}=\mathbf{b}$ , and $M$ lies on $AB$ with $AM:MB=1:2$ . Work out $\overrightarrow{OM}$ . Non-calculator.

Plan. Go from $O$ to $A$ , then one third of the way along $AB$ .

\begin{aligned} \overrightarrow{AB}&=\mathbf{b}-\mathbf{a} &&\text{finish minus start}\\ \overrightarrow{OM}&=\mathbf{a}+\frac13(\mathbf{b}-\mathbf{a}) &&\text{one of three shares}\\ &=\frac23\mathbf{a}+\frac13\mathbf{b} &&\text{collect matching vectors} \end{aligned}

Why this step? The fraction multiplies $\mathbf{b}-\mathbf{a}$ , which starts at $A$ . Multiplying $\mathbf{b}$ alone would put the point on $OB$ .

Answer. $\overrightarrow{OM}=\frac23\mathbf{a}+\frac13\mathbf{b}$ .

Check. $\overrightarrow{AM}=\frac13(\mathbf{b}-\mathbf{a})$ and $\overrightarrow{MB}=\frac23(\mathbf{b}-\mathbf{a})$ . Their lengths are in ratio $1:2$ .

Change only the ratio to $2:1$ . The new point lies nearer $B$ , so you take a larger fraction of the same arrow.

Worked example 4: two shares out of three

Question. Keep $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$ . Point $N$ lies on $AB$ , with $AN:NB=2:1$ . Work out $\overrightarrow{ON}$ . Non-calculator.

Plan. Go to $A$ , then take two thirds of $AB$ .

\begin{aligned} \overrightarrow{AN}&=\frac23(\mathbf{b}-\mathbf{a}) &&\text{two of three shares}\\ \overrightarrow{ON}&=\mathbf{a}+\frac23(\mathbf{b}-\mathbf{a}) &&\text{start at }O\\ &=\frac13\mathbf{a}+\frac23\mathbf{b} &&\text{collect terms} \end{aligned}

Why this step? The order $AN:NB$ tells you that the first piece has two shares. Reversing the ratio would locate the earlier point $M$ .

Answer. $\overrightarrow{ON}=\frac13\mathbf{a}+\frac23\mathbf{b}$ .

Check. $NB$ is the remaining one third of $AB$ , half the length of $AN$ , so $AN:NB=2:1$ .

A midpoint is the case $1:1$ : its position vector is $\frac12(\mathbf{a}+\mathbf{b})$ . If a point is beyond $B$ , use the stated direction and whole length instead: $\overrightarrow{AX}=\frac32\overrightarrow{AB}$ gives $\overrightarrow{OX}=\mathbf{a}+\frac32(\mathbf{b}-\mathbf{a})$ . The multiplier exceeds $1$ because you pass $B$ .

Watch out

Wrong: $AM:MB=1:2$ , so $\overrightarrow{AM}=\frac12\overrightarrow{AB}$ . The denominator must count all three shares. Right: $\overrightarrow{AM}=\frac13\overrightarrow{AB}$ .

Use five total shares in the next question.

Proving that lines are parallel (Higher)

A non-zero vector and a non-zero scalar multiple point along the same direction or the reverse direction. Their supporting lines are parallel, or they are the very same line. That small distinction will matter when proving collinearity.

Recognising parallel directions

For non-zero vectors $\mathbf{u}$ and $\mathbf{v}$ , if

\mathbf{u}=k\mathbf{v}\qquad(k\ne0)

then they have parallel directions. A positive $k$ gives the same direction and a negative $k$ the opposite direction. Their lengths are in ratio $|k|:1$ .

Not on the formulae sheet: learn it.

Look for a common factor in the whole expression. Matching only the coefficient of $\mathbf{a}$ , while ignoring the coefficient of $\mathbf{b}$ , proves nothing about the direction.

Worked example 5: the midpoint line in a triangle

Question. In triangle $OAB$ , $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$ . Points $M$ and $N$ are the midpoints of $OA$ and $OB$ . Prove that $MN\parallel AB$ . Non-calculator.

Plan. Express both arrows from their endpoints, then compare them.

\begin{aligned} \overrightarrow{MN}&=-\frac12\mathbf{a}+\frac12\mathbf{b} &&\text{route }M\text{ to }O\text{ to }N\\ &=\frac12(\mathbf{b}-\mathbf{a}) &&\text{factor out }\frac12\\ &=\frac12\overrightarrow{AB} &&\text{identify the whole arrow} \end{aligned}

Why this step? The factor must multiply both terms. It then scales one complete arrow into the other.

Answer. $\overrightarrow{MN}=\frac12\overrightarrow{AB}$ , so $MN$ is parallel to $AB$ .

Check. The factor also says $MN$ is half as long as $AB$ , agreeing with similar triangles $OMN$ and $OAB$ .

Change only the fraction along each side, from one half to two thirds. The same comparison works.

Worked example 6: a different parallel cross-section

Question. In the same triangle, $P$ lies on $OA$ with $OP=\frac23 OA$ , and $Q$ lies on $OB$ with $OQ=\frac23 OB$ . Prove that $PQ\parallel AB$ . Non-calculator.

Plan. Subtract the position vector of $P$ from that of $Q$ , then factor.

\begin{aligned} \overrightarrow{PQ}&=\frac23\mathbf{b}-\frac23\mathbf{a} &&\text{finish minus start}\\ &=\frac23(\mathbf{b}-\mathbf{a}) &&\text{one factor for both terms}\\ &=\frac23\overrightarrow{AB} &&\text{recognise the direction} \end{aligned}

Why this step? Both points are the same fraction of the way along their sides, so the same fraction factors out.

Answer. $\overrightarrow{PQ}=\frac23\overrightarrow{AB}$ , so $PQ\parallel AB$ .

Check. $OP:OA=OQ:OB=PQ:AB=2:3$ , consistent with similar triangles.

Now the opening puzzle pays off. Let $P,Q,R,S$ be consecutive midpoints of the sides of any convex quadrilateral $ABCD$ , with position vectors $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$ . Write the midpoint positions, then subtract:

\begin{aligned} \overrightarrow{PQ} &=\frac{\mathbf{b}+\mathbf{c}}2-\frac{\mathbf{a}+\mathbf{b}}2 =\frac12(\mathbf{c}-\mathbf{a})\\ \overrightarrow{SR} &=\frac{\mathbf{c}+\mathbf{d}}2-\frac{\mathbf{a}+\mathbf{d}}2 =\frac12(\mathbf{c}-\mathbf{a}) \end{aligned}

So $PQ$ and $SR$ are equal vectors. The same argument gives $\overrightarrow{QR}=\overrightarrow{PS}=\frac12(\mathbf{d}-\mathbf{b})$ . Both pairs of opposite sides are equal and parallel: the midpoint shape is a parallelogram, however untidy the original quadrilateral looks.

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G25 - Vector geometry proofs (Higher)

Routes with labelled vectors (Higher)

Points dividing a line in a ratio (Higher)

Proving that lines are parallel (Higher)

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