IGCSE Maths 0580 Vector Geometry Past Paper Questions and Worked Solutions

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IGCSE Maths (0580) · Transformations and Vectors · 7.4

Every vector geometry question from 9 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 7.4. 9 Extended-tier questions. Each question is labelled with its question paper code (for example 0580/21/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Transformations and Vectors past paper questions.

0580 Vector Geometry past paper questions: Extended (Papers 2 and 4)

17 marksExtended

May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q25

The diagram shows triangle ABC and straight line ACD. P lies on AB and AP : PB = 4 : 1. Q lies on BC and BQ = 2⁄5BC. AB = m and AC = 5n.

0580/21 May/June 2026 Q25 – triangle ABC with P on AB, Q on BC and line ACDm5nABCDPQNOT TOSCALE

(a)(i) Find BC. (ii) Find PQ, in terms of m and n in its simplest form.

(b) CD = 3n. Show that P, Q and D lie on a straight line.

Show solution
Part (a)(i)
  1. Go from B to A, then from A to C: BC = BA + AC.
  2. BA is the reverse of AB, so BA = −m. BC = −m + 5n
Answer −m + 5n
Part (a)(ii)
  1. AP : PB = 4 : 1, so PB is 1⁄5 of AB: PB = 1⁄5m.
  2. BQ = 2⁄5BC = 2⁄5(5n − m) = 2n − 2⁄5m.
  3. PQ = PB + BQ = 1⁄5m + 2n − 2⁄5m.
  4. PQ = 2n − 1⁄5m
Answer 2n − 1⁄5m
Part (b)
  1. Find PD. AD = AC + CD = 5n + 3n = 8n, and PA = −4⁄5m.
  2. PD = PA + AD = −4⁄5m + 8n.
  3. Factorise out 4: PD = 4(2n − 1⁄5m) = 4PQ.
  4. PD is a multiple of PQ, so they are parallel. They also share the point P.
  5. P, Q and D are collinear
Answer P, Q and D are collinear
25 marksExtended

May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q22

OABC is a parallelogram. OA = a and OC = c. P is a point on AC and AP : PC = 1 : 4.

0580/22 May/June 2026 Q22 – parallelogram OABC with P on diagonal ACacOCABPNOT TOSCALE

(a) Find, in terms of a and c, in its simplest form (i) AP (ii) OP.

(b) Q is a point on AB and AQ = 1⁄4c. Show that O, P and Q lie in a straight line.

Show solution
Part (a)(i)
  1. Go from A to C: AC = AO + OC = −a + c = c − a.
  2. AP : PC = 1 : 4, so AP is 1⁄5 of AC.
  3. AP = 1⁄5(c − a)
Answer AP = 1⁄5(c − a)
Part (a)(ii)
  1. OP = OA + AP = a + 1⁄5c − 1⁄5a.
  2. OP = 4⁄5a + 1⁄5c
Answer OP = 4⁄5a + 1⁄5c
Part (b)
  1. Write OP with a common factor: OP = 1⁄5(4a + c).
  2. OQ = OA + AQ = a + 1⁄4c = 1⁄4(4a + c).
  3. Both are multiples of (4a + c), so OQ = 5⁄4OP. They are parallel and both start at O.
  4. O, P and Q are collinear
Answer O, P and Q are collinear
33 marksExtended

May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q28

O is the origin, OR = r and OS = s. P is a point on RS and RP : PS = 2 : 5. Find the position vector of P. Give your answer, in terms of r and s, in its simplest form.

0580/23 May/June 2026 Q28 – triangle ORS with P on RSrsORSPNOT TOSCALE
Show solution
  1. Find the vector from R to S: RS = RO + OS = −r + s = s − r.
  2. RP : PS = 2 : 5, so RP is 2⁄7 of RS.
  3. OP = OR + RP = r + 2⁄7(s − r).
  4. Expand and collect: r − 2⁄7r + 2⁄7s = 5⁄7r + 2⁄7s.
  5. 5⁄7r + 2⁄7s
Answer 5⁄7r + 2⁄7s
46 marksExtended

Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q19

OABCDE is a regular hexagon. OA = p and AB = q.

0580/42 February/March 2026 Q19 – regular hexagon OABCDE with OA = p and AB = qpqOABCDENOT TOSCALE

(a) Find, in terms of p and/or q, in its simplest form (i) DE (ii) BE (iii) OE.

(b) OB and DC are extended to meet at G. Find the position vector of G in terms of p and q. Give your answer in its simplest form.

Show solution
  1. In a regular hexagon, opposite sides are parallel and equal. So ED is parallel to OA and DE is parallel to AB.
Part (a)(i)
  1. DE has the same length as AB but points the opposite way.
  2. DE = −q
Answer DE = −q
Part (a)(ii)
  1. The long diagonal BE is parallel to OA and twice as long, pointing the opposite way.
  2. BE = −2p
Answer BE = −2p
Part (a)(iii)
  1. Go round the hexagon from O to E: OE = OA + AB + BE.
  2. OE = p + q − 2p = q − p
Answer q − p
Part (b)
  1. OB = OA + AB = p + q. G lies on OB extended, so OG = k(p + q) for some number k.
  2. Find D by going O → A → B → C → D. BC = q − p and CD = −p, so OD = p + q + (q − p) − p = 2q − p.
  3. DC is parallel to OA, so points on DC extended are 2q − p + tp.
  4. Match the q parts: k = 2. Match the p parts: −1 + t = 2, so t = 3.
  5. OG = 2(p + q) = 2p + 2q
Answer 2p + 2q
54 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q24

OBD and ACD are straight lines. O is the origin, the position vector of A is a and the position vector of B is b. BC = 1⁄3OA. M is the midpoint of CD. Find the position vector of M, in terms of a and b, in its simplest form.

0580/21 October/November 2025 Q24 – triangle OAD with B on OD, C on AD and M the midpoint of CDabOADBCMNOT TOSCALE
Show solution
  1. BC is parallel to OA and one third of its length, so triangle DBC is similar to triangle DOA with scale factor 1⁄3.
  2. That means DB = 1⁄3DO, so OB is the other 2⁄3 of OD.
  3. OB = b = 2⁄3OD, so OD = 3⁄2b.
  4. OC = OB + BC = b + 1⁄3a.
  5. M is the midpoint of CD, so OM is the average of OC and OD: OM = 1⁄2(OC + OD).
  6. OM = 1⁄2(b + 1⁄3a + 3⁄2b) = 1⁄2(1⁄3a + 5⁄2b)
  7. OM = 1⁄6a + 5⁄4b
Answer 1⁄6a + 5⁄4b
64 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/42/O/N/25 · Q18

OA is parallel to CB and OA : CB = 4 : 3. OA = a and OB = b.

0580/42 October/November 2025 Q18 – quadrilateral OABC with OA parallel to CBabOABCMNOT TOSCALE

(a) Find AB in terms of a and b. (b) M is the midpoint of OC. Find AM in terms of a and b, in its simplest form.

Show solution
Part (a)
  1. Go from A back to O, then on to B: AB = AO + OB = −a + b.
  2. AB = b − a
Answer AB = b − a
Part (b)
  1. CB is parallel to OA and CB : OA = 3 : 4, so CB = 3⁄4a. Going the other way, BC = −3⁄4a.
  2. OC = OB + BC = b − 3⁄4a.
  3. M is the midpoint of OC, so OM = 1⁄2OC = 1⁄2b − 3⁄8a.
  4. AM = AO + OM = −a + 1⁄2b − 3⁄8a.
  5. Collect the a terms: −a − 3⁄8a = −11⁄8a.
  6. AM = −11⁄8a + 1⁄2b
Answer AM = −11⁄8a + 1⁄2b
75 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q26

ABCD is a parallelogram. AB = m and AD = p. F is a point on BC with BF = 4FC. E is a point on AD with AE : ED = 1 : 2.

0580/43 October/November 2025 Q26 – parallelogram ABCD with F on BC and E on ADmpADCBEFNOT TOSCALE

(a) Find EF, in terms of m and p, in its simplest form.

(b) EF and DC are extended to meet at the point G. Find CG, in terms of m and/or p, in its simplest form.

Show solution
Part (a)
  1. AE : ED = 1 : 2, so AE = 1⁄3AD = 1⁄3p.
  2. BF = 4FC means BF is 4⁄5 of BC. BC = AD = p, so BF = 4⁄5p.
  3. Go from E to F the long way: EF = EA + AB + BF = −1⁄3p + m + 4⁄5p.
  4. Collect the p terms: −1⁄3 + 4⁄5 = −5⁄15 + 12⁄15 = 7⁄15.
  5. EF = m + 7⁄15p
Answer EF = m + 7⁄15p
Part (b)
  1. FC and ED are both parallel to p, so triangles GCF and GDE are similar.
  2. FC = 1⁄5p and ED = 2⁄3p, so the ratio FC : ED = 1⁄5 : 2⁄3 = 3 : 10.
  3. So GC : GD = 3 : 10. Since GD = GC + CD, the length CD is 10 − 3 = 7 of these parts.
  4. CG is 3 parts out of CD’s 7 parts, and it points the same way as DC (which is m).
  5. CG = 3⁄7m
Answer CG = 3⁄7m
85 marksExtended

May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q21

OABC is a rhombus and O is the origin. The diagonals of the rhombus intersect at P. OP = 2m and AP = n (as vectors).

0580/21 May/June 2025 Q21 – rhombus OABC with diagonals meeting at P, OP = 2m and AP = n2mnOABCPNOT TOSCALE

(a) Find, in terms of m and n, in its simplest form (i) OA (ii) OC.

(b) D is the point such that AD = 10m − 3n. Show that OADC is a trapezium.

Show solution
Part (a)(i)
  1. Go from O to P, then from P back to A. PA is the reverse of AP, so PA = −n.
  2. OA = OP + PA = 2m − n, so OA = 2m − n
Answer OA = 2m − n
Part (a)(ii)
  1. The diagonals of a rhombus bisect each other, so P is the midpoint of AC and PC = AP = n.
  2. OC = OP + PC = 2m + n
  3. OC = 2m + n
Answer OC = 2m + n
Part (b)
  1. A trapezium has one pair of parallel sides. Show that CD is parallel to OA.
  2. Go from C to D using known vectors: CD = CO + OA + AD.
  3. CD = −(2m + n) + (2m − n) + (10m − 3n) = 10m − 5n.
  4. Factorise: CD = 5(2m − n) = 5 × OA.
  5. CD is a multiple of OA, so CD is parallel to OA; OADC has a pair of parallel sides, so it is a trapezium.
Answer CD is a multiple of OA, so CD is parallel to OA; OADC has a pair of parallel sides, so it is a trapezium.
93 marksExtended

May/June 2025 · Paper 2 · 0580/22/M/J/25 · Q23

In the diagram, OA is parallel to BC. BC = 3OA. M is the midpoint of AC. The position vector of A is a and the position vector of B is b. Find the position vector of M. Give your answer in terms of a and b, in its simplest form.

0580/22 May/June 2025 Q23 – quadrilateral OACB with OA parallel to BC and M the midpoint of ACabOABCMNOT TOSCALE
Show solution
  1. BC is parallel to OA and three times as long, so BC = 3a.
  2. OC = OB + BC = b + 3a.
  3. AC = OC − OA = (b + 3a) − a = 2a + b.
  4. M is halfway along AC: OM = OA + 1⁄2AC = a + 1⁄2(2a + b).
  5. OM = a + a + 1⁄2b = 2a + 1⁄2b
Answer 2a + 1⁄2b

Question papers used on this page

All vector geometry questions above, listed by 0580 question paper code.

QP codeFile namePaperQuestions
0580/21/M/J/260580_s26_qp_21May/June 2026 · Paper 2 (Extended)Q25
0580/22/M/J/260580_s26_qp_22May/June 2026 · Paper 2 (Extended)Q22
0580/23/M/J/260580_s26_qp_23May/June 2026 · Paper 2 (Extended)Q28
0580/42/F/M/260580_m26_qp_42Feb/Mar 2026 · Paper 4 (Extended)Q19
0580/21/O/N/250580_w25_qp_21Oct/Nov 2025 · Paper 2 (Extended)Q24
0580/42/O/N/250580_w25_qp_42Oct/Nov 2025 · Paper 4 (Extended)Q18
0580/43/O/N/250580_w25_qp_43Oct/Nov 2025 · Paper 4 (Extended)Q26
0580/21/M/J/250580_s25_qp_21May/June 2025 · Paper 2 (Extended)Q21
0580/22/M/J/250580_s25_qp_22May/June 2025 · Paper 2 (Extended)Q23

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