01 · Internal ratio
A(−3,2), B(5,6). Find P if AP : PB = 3 : 1.
Hint
P is three quarters of the way from A to B.
Worked solution
p = (−3,2) + ¾(8,4) = (3,5)
Understand · explore · practise
Vector geometry
Divide lines in a ratio and use vectors to prove parallelism, collinearity and geometric results. Includes midpoint theorems, intersections, regular hexagons and diagonal ratios.
Before you startPosition vectors, vector arithmetic and simultaneous equations
01 / Divide a segment
Travel the required fraction of the whole displacement.
If P divides AB internally in the ratio AP : PB = m : n, with m,n > 0, then AP is m/(m + n) of AB.
p = a + [m/(m + n)](b − a)
= (na + mb)/(m + n)
For A(−2,1), B(4,7) and AP : PB = 1 : 2, p = a + (b − a)/3 = (0,3). Notice that b gets the weight 1/3, not 2/3: P is only one third of the way from A to B.
More generally p = a + t(b − a). For 0 < t < 1 the point is inside the segment; t = 0 and 1 are its endpoints; t outside [0,1] puts it on the extended line.
02 / Parallel and collinear
A scalar multiple gives a direction relationship.
Two non-zero vectors are parallel when one is a non-zero scalar multiple of the other. A positive multiple gives the same direction; a negative multiple gives the opposite direction.
For A(1,2), B(4,4), C(10,8), AB = 3i + 2j and AC = 9i + 6j = 3AB. Since these vectors share A, the three distinct points are collinear. Also BC = 2AB, so AB : BC = 1 : 2 in lengths.
Parallel arrows with unrelated starts do not put all their endpoints on one line. To prove three points are collinear, compare two non-zero displacements that connect those points, such as AB and AC.
Unknown coefficients
For non-parallel a,b, suppose 3a + kb is parallel to 6a − 4b. Write 3a + kb = λ(6a − 4b). Comparing the a coefficient gives λ = 1/2, then k = −2. The independence of a,b is essential.
03 / Compare coefficients
Two non-parallel directions cannot cancel accidentally.
If a and b are non-zero and non-parallel, and pa + qb = ra + sb, then p = r and q = s.
Rearrange to (p − r)a = (s − q)b. If one coefficient were non-zero, this would make a a scalar multiple of b, contradicting their non-parallel directions. Both coefficients must therefore be zero.
Without that condition, coefficient comparison can fail. If b = 2a, then 2a + b = 4a, although the displayed coefficients of a and b differ.
04 / The midpoint theorem
A midpoint connection is half a third side.
In triangle ABC, let P be the midpoint of AB and Q the midpoint of AC. Using position vectors:
p = (a + b)/2
q = (a + c)/2
PQ = q − p = (c − b)/2
Thus PQ is parallel to BC and half its length. Triangles APQ and ABC are similar: the two sides from A have the same 1/2 scale factor and include the same angle. Their areas have ratio 1 : 4.
05 / Parallelogram diagonals
Describe the same intersection along two routes.
In a non-degenerate parallelogram OABC, let OA = a and OC = c. Then OB = a + c. If P lies on both diagonals:
p = t(a + c)
p = a + u(c − a)
Equate coefficients of the non-parallel vectors a,c: t = 1 − u and t = u. Hence t = u = 1/2. The intersection is the midpoint of both diagonals, proving that the diagonals bisect each other.
06 / Find an intersection
Each line needs its own parameter.
In triangle OAB, M is the midpoint of AB and N is the midpoint of OB. Let the medians OM and AN meet at G.
g = t(a + b)/2
g = a + u(b/2 − a)
t/2 = 1 − u, t/2 = u/2
The second equation gives t = u; then t/2 = 1 − t gives t = u = 2/3. Thus g = (a + b)/3. G divides each median in the ratio 2 : 1 measured from its vertex.
Translate to a general triangle
For vertices A,B,C with position vectors a,b,c, the same argument gives the centroid position (a + b + c)/3. Each median contains it: for example a + (2/3)((b + c)/2 − a) = (a + b + c)/3.
07 / A line parallel to a side
Use a parameter before choosing a numerical ratio.
In triangle OAB, M lies on OA with OM = t a, where 0 < t < 1. A line through M parallel to OB meets AB at N. Write N in two ways:
n = ta + λb
n = a + μ(b − a)
Comparing coefficients gives t = 1 − μ and λ = μ, so μ = λ = 1 − t. Therefore:
n = ta + (1 − t)b
MN = (1 − t)b
AN : NB = (1 − t) : t
For t = 1/4, N is three quarters of the way from A to B and AN : NB = 3 : 1. The algebra also explains how the ratio changes when M moves.
08 / Paths around polygons
Build positions by following known edges.
In a regular hexagon OABCDE, let OA = a and OE = c. Adjacent sides have the same length and turn by 60°. The intermediate direction from A to B is a + c.
OB = 2a + c
OC = 2a + 2c
OD = a + 2c
OE = c
The centre has position a + c. Check it as the midpoint of OC, AD or BE. Equivalent routes must produce the same vector.
A general trapezium path
Let ABCD be a trapezium with AB = a, BC = b and DC = k a, where k > 0 and the two base directions agree. Taking A as origin gives C = a + b and D = (1 − k)a + b. If M is a fraction r of the way from D to C, then:
AM = [1 − k(1 − r)]a + b
For k = 2 and r = 3/4 this is a/2 + b. The connected path prevents a sign guess on the parallel base.
09 / Extension: diagonal ratios
Midpoints produce a special three-part division.
In parallelogram OABC, use a = OA and c = OC. Pick E on AB with E = a + t c, and F on BC with F = t a + c, where 0 < t < 1. Join O to E and F. The model changes t while keeping the parallelogram fixed.
The diagonal from A to C has points a + s(c − a). For the intersection with OE, compare this with λ(a + t c). This gives λ = 1 − s and λt = s, so s = t/(1 + t). The intersection with OF similarly gives s = 1/(1 + t).
Three diagonal fractions:
t/(1 + t), (1 − t)/(1 + t), t/(1 + t)
All three are equal precisely when t = 1/2. Then E and F are midpoints and the two lines trisect AC. At other t-values, the outer pieces remain equal but the middle piece changes.
Choose where the two side points lieMove at your pace
t = 0.5. OE and OF meet AC at fractions 0.3333 and 0.6667 of the way from A. The three diagonal pieces have fractions 0.3333, 0.3333, 0.3333. At t = 1/2, all three are exactly 1/3. Decimal fractions are rounded to four places.
Watch midpoint connections divide a diagonal into equal thirds
Pause, replay or seek freely. The notes explain the same idea and stay in view.
10 / Your turn
State the vector relationship that proves the geometry.
Assume named triangles and parallelograms are non-degenerate.
02 · Midpoint check
A has position 2a − b and B has position −4a + 5b. Find the midpoint position.
Hint
Average the two positions.
Worked solution
−a + 2b
03 · Collinearity
Show that A(−1,1), B(2,3), C(8,7) are collinear and find AB : BC.
Hint
Compare AB with AC.
Worked solution
AB = (3,2), AC = (9,6) = 3AB
BC = 2AB ⇒ AB : BC = 1 : 2
04 · Opposite directions
a and b are non-parallel. Are 2a − 3b and −6a + 9b parallel? Do they point the same way?
Hint
Find the scalar multiplier.
Worked solution
−6a + 9b = −3(2a − 3b)
They are parallel and point in opposite directions.
05 · A parameter
For non-parallel a,b, find k if 4a + kb is parallel to 2a − 5b.
Hint
The scalar multiplier must be 2.
Worked solution
k = −10
06 · Midpoint areas
P,Q are midpoints of two sides of triangle ABC meeting at A. If triangle ABC has area 28, find the area of APQ.
Hint
Length scale 1/2 means area scale 1/4.
Worked solution
Area APQ = 7
07 · A centroid
Find the centroid of a triangle with vertices (−2,0), (4,1), (1,8).
Hint
Average all three position vectors.
Worked solution
G = (1,3)
08 · A moving parallel cut
In the parallel-cut construction, OM = (2/5)OA. Find AN : NB and MN in terms of b = OB.
Hint
Use t = 2/5 in the general result.
Worked solution
AN : NB = 3 : 2
MN = (3/5)b
09 · Not always thirds
In the diagonal model, t = 1/3. Find the three fractions of AC and check that they total 1.
Hint
Substitute into the three fraction formulas.
Worked solution
1/4, 1/2, 1/4
1/4 + 1/2 + 1/4 = 1
10 · A polygon route
In the regular hexagon above, find the vector from B to D.
Hint
Subtract the position of B from the position of D.
Worked solution
BD = (a + 2c) − (2a + c) = c − a
11 / Recap
Choose routes that describe the same point.
- For an internal ratio m:n, travel m/(m+n) of the whole vector.
- A non-zero scalar multiple proves parallelism.
- For collinearity, connect the three points with vectors sharing a point.
- Compare coefficients only in independent directions.
- Use separate parameters for separate lines, then equate their point positions.
- A geometric proof must include the relationship, not just a final numerical answer.
Section 1 of 11 · Divide a segment