Vector geometry — Paper 4 · IGCSE Mathematics 0580

11 (a) A = 1 4 Find (i) A2, [2] f p (ii) A–1, the inverse of A. [2] f p - 1 0 (b) Describe fully the single transformation represented by the matrix . c 0 1 m … … [2] (c) Find the matrix that represents a clockwise rotation of 90º about the origin. [2] f p (d) C A NOT TO P a SCALE O B b In the diagram, O is the origin and P lies on AB such that AP : PB = 3 : 4. OA = a and OB = b . (i) Find OP, in terms of a and b, in its simplest form. OP = … [3] (ii) The line OP is extended to C such that OC = m OP and BC = ka. Find the value of m and the value of k. m = … k = … [2]

13 marks

Mark scheme: 11(a)(i)  1 −18  2   M1 for two or three correct elements  6 13  11(a)(ii) 1  4 3  2  4 3    or better isw M1 for det = 11 or [k]   isw 11  −1 2   −1 2  11(b) Reflection 1 y-axis oe 1 11(c)  0 1  2   B1 for one correct column or row  −1 0  11(d)(i) 1 4 3 3 M2 for correct unsimplified answer seen (4a + 3b) or a + b JJJG 3 JJJG 4 7 7 7 or AP = (b – a) oe or BP = (a – b) oe 7 7 JJJG JJJG or M1 for AB = b – a or BA = a – b or correct JJJG route for OP 11(d)(ii) 7 2 B1 for each value [m =] 3 m or M1 for (4a + 3b) = b + ka oe 4 7 [k =] 3

This question in 0580/41 Oct/Nov 2017

11 (a) OA = AB = AC = 3 - 7 6 Find (i) OB , OB = … [3] (ii) BC. BC = [2] f p (b) S R NOT TO SCALE b X P Q a PQRS is a parallelogram with diagonals PR and SQ intersecting at X. PQ = a and PS = b . Find QX in terms of a and b. Give your answer in its simplest form. QX = … [2] 2 5 (c) M = c 1 8 m Calculate (i) M2, M2 = [2] f p (ii) M -1 . M -1 = [2] f p

11 marks

Mark scheme: 11(a)(i) 12.6 or 12.64 to 12.65 3 2 2 M2 for 12 + ( − 4 ) OR  12  B1 for    − 4  M1 for (their12)2 + ( their − 4 ) 2 11(a)(ii)  −11  2  −11   k    B1 for   or   or for  13   k   13  JJJG  − 8  [ BA = ]    7  11(b) 1 2 M1 for correct route or correct (b − a) oe unsimplified answer 2 JJJG or B1 for QS = b – a oe 11(c)(i)  9 50  2 B1 for 2 correct elements    10 69  11(c)(ii) 1  8 − 5  2  8 −5  1  a b    oe isw B1 for k   or   11  − 1 2   − 1 2  11  c d  or det = 11 soi

This question in 0580/41 May/June 2018

11 B A C Q NOT TO P SCALE O OAB is a triangle and ABC and PQC are straight lines. P is the midpoint of OA, Q is the midpoint of PC and OQ : QB = 3 : 1. OA = 4a and OB = 8b . (a) Find, in terms of a and/or b, in its simplest form (i) AB, AB = … [1] (ii) OQ, OQ = … [1] (iii) PQ. PQ = … [1] (b) By using vectors, find the ratio AB : BC. … : … [3]

6 marks

Mark scheme: 11(a)(i) 8b – 4a oe 1 11(a)(ii) 6b 1 11(a)(iii) 6b – 2a or 2(3b – a) 1 FT –2a + their (a)(ii) JJJG JJJG 11(b) 2 : 1 oe final answer 3 Dep on correct BC or correct AC seen JJJG B2 for BC = 4b–2a JJJG or M1 for a correct route for BC in terms of a and b JJJG or for a correct route for AC in terms of a and b If no/incorrect working seen then SC1 for final answer of 2 : 1 (oe)

This question in 0580/43 Oct/Nov 2019

2 (a) p = q = 5 7 (i) Find 2 p + q . [2] f p (ii) Find p . … [2] - 3 (b) A is the point (4, 1) and AB = e 1o. Find the coordinates of B. ( … , … ) [1] (c) The line y = 3 x - 2 crosses the y-axis at G. Write down the coordinates of G. ( … , … ) [1] (d) D NOT TO T SCALE M O C In the diagram, O is the origin, OT = 2TD and M is the midpoint of TC. OC = c and OD = d . Find the position vector of M. Give your answer in terms of c and d in its simplest form. … [3]

9 marks

Mark scheme: 2(a)(i)  6  2 B1 for each    17  2(a)(ii) 6.4[0] or 6.403... 2 M1 for 42 + 52 2(b) (1, 2) 1 2(c) (0, –2) 1 2(d) 1 1 3 B2 for correct unsimplified answer c + d  2 2 3 or M1 for CT = – c + d oe 3  2 or TC = c – d oe 3 or for correct route

This question in 0580/42 May/June 2020

5 (a) a = b = 8 - 5 (i) Find (a) b - a , [1] f p (b) 2a + b , [2] f p (c) b . … [2] 13 (ii) a + kb = , where k and m are integers. e mo Find the value of k and the value of m. k = … m = … [3] (b) C B NOT TO q M N SCALE O A p OABC is a parallelogram and O is the origin. M is the midpoint of OB. N is the point on AB such that AN : NB = 3 : 2. OA = p and OC = q. (i) Find, in terms of p and q, in its simplest form. (a) OB OB = … [1] (b) CM CM = … [2] (c) MN MN = … [2] (ii) CB and ON are extended to meet at D. Find the position vector of D in terms of p and q. Give your answer in its simplest form. … [3]

16 marks

Mark scheme: 5(a)(i)(a)  5  1   final answer  −13  5(a)(i)(b)  − 4  2  − 4   k   − 6    final answer B1 for answer   or   or    11   k   11   16  seen 5(a)(i)(c) 5.39 or 5.385… 2 M1 for 22 + ([–]5)2 5(a)(ii) [k =] 8 3 B2 for k = 8 or m = –32 [m =] – 32 or M1 for – 3 + 2k = 13 oe or for m = –5 × their k + 8 correctly evaluated 5(b)(i)(a) p + q final answer 1 5(b)(i)(b) 1 1 1 p – q 2 M1 for unsimplified answer or any correct p – q or (p – q) or final  2 2 2 2 vector route for CM , e.g. answer 1 – q + their (b)(i)(a) 2 5(b)(i)(c) 1 1 5p + q 2 M1 for unsimplified answer or any correct p + q or final answer  2 10 10 vector route for MN 5(b)(ii) 5 5p + 3q 3 B2 for unsimplified correct answer p + q or final answer OR 3 3 3 M1 for p + q seen 5 B1 for final answer of form kp + q (k > 1) 5 or final answer p + jq oe (any j) 3

This question in 0580/42 May/June 2021

4 (a) A is the point (1, 5) and B is the point (3, 9). M is the midpoint of AB. (i) Find the coordinates of M. ( … , … ) [2] (ii) Find the equation of the line that is perpendicular to AB and passes through M. Give your answer in the form y = mx + c . y = … [4] - 2 - 2 (b) The position vector of P is and the position vector of Q is e 3o e 5o. (i) Find the vector PQ. [2] f p (ii) R is the point such that PR = 3PQ . Find the position vector of R. [2] f p (c) U NOT TO SCALE u Y T O t OT = t , OU = u and UY = 2YT. (i) Find OY in terms of t and u. Give your answer in its simplest form. OY = … [2] (ii) Z is on OT and YZ is parallel to UO. Find OZ in terms of t and/or u. Give your answer in its simplest form. OZ = … [1]

13 marks

Mark scheme: 4(a)(i) (2, 7) 2 B1 for each coordinate 4(a)(ii) 1 4 Correct equivalent in different form − x + 8 oe scores 3 marks. 2 9 − 5 4 M1 for gradient of AB = or or 2 3 − 1 2 M1 dep for gradient 1 p = −their grad of AB M1 (dep on previous M1) for substitution of their midpoint into y = (their p)x + c oe where their p ≠ 0 4(b)(i) 0 2 0 k  B1 for  or  2 k 2  4(b)(ii)  − 2  2 FT their PQ    9  0 B1FT for  6 4(c)(i) 2 1 1 2  2 t + u or (2t + u) final answer M1 for UY = ( t –u) oe 3 3 3 3  1 or TY = (u – t) oe 3 or correct route soi 4(c)(ii) 2 1 t cao 3

This question in 0580/43 May/June 2021

7 (a) y 8 7 6 5 4 3 A 2 1 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 7 x – 1 B – 2 – 3 – 4 (i) On the grid, draw the image of (a) shape A after an enlargement, scale factor 2, centre (0, 1), [2] (b) shape A after a reflection in the line y = x - 1. [3] (ii) Describe fully the single transformation that maps shape A onto shape B. … … [3] (b) C B NOT TO q M SCALE O A p OABC is a trapezium and O is the origin. M is the midpoint of AB. OA = p , OC = q and OA = 2CB. Find, in terms of p and q, the position vector of M. Give your answer in its simplest form. … [3]

11 marks

Mark scheme: 7(a)(i)(a) Shape at (–2, 1) ( –4, 1) ( –4, 7) (0, 7) 2 B1 for 3 correct points or for enlargement SF2 from any centre 7(a)(i)(b) Shape at (2, –2) (2, –3) (5, –1) (5, –3) 3 B2 for correct orientation but wrong position or for 3 correct points or B1 for y = x – 1 drawn 7(a)(ii) Rotation 3 B1 for each 90 [anticlockwise] oe (0, 0) oe 7(b) 3 1 1 3 p + 2q 3  1  1  p + q or ( 3p + 2q ) or M2 for AM = AM =  - p + q + p  oe 4 2 4 4 2  2   final answer or M1 for correct route for AB oe soi by –½ p + q  or for OM soi

This question in 0580/41 Oct/Nov 2021

9 (a) F is the point (5, - 2 ) and FG = . 3 Find (i) the coordinates of point G, ( … , … ) [1] (ii) 5 FG , [1] f p (iii) FG . … [2] (b) Q A B a P NOT TO SCALE O c C OABC is a parallelogram. P is a point on AC and Q is the midpoint of AB. OA = a and O C = c . (i) Find, in terms of a and/or c (a) AQ, AQ = … [1] (b) O Q . O Q = … [1] 2 1 (ii) OP = a + c 3 3 (a) Show that O, P and Q lie on a straight line. [2] (b) Write down the ratio OP : OQ. Give your answer in the form 1 : n. 1 : … [1]

9 marks

Mark scheme: 9(a)(i) (3, 1) 1 9(a)(ii)  −10  1    15  9(a)(iii) 3.61 or 3.605 to 3.606 2 M1 for (–2)2 + 32 oe 9(b)(i)(a) 1 1 c 2 9(b)(i)(b) 1 1 FT a + their (b)(i)(a) a + c oe 2 9(b)(ii)(a)  1 2  O P = + c) oe B1 for O P or PQ factorised 3(2a  1 and OQ = (2a + c) oe or for correct multiplicative statement on 2 relationship without factorised vectors   OR 2 e.g. OQ = 1.5 O,P OQ = O,P 2 1 O=P (a + c) 3  3 2 2 PQ = O,P OR  2 1  1 1 1.5 a +  c = a + 1 c PQ = + c)  3 3  2 3(a 2 and correct comment e.g. have the same base vector or that they are multiples of one another and they share a common point OR    e.g. OQ = 1.5 O,P 2 PQ = O P 9(b)(ii)(b) 1.5 oe 1

This question in 0580/42 Oct/Nov 2021

6 (a) p = q = 3 1 Find (i) 3q, [1] f p (ii) p - q , [1] f p (iii) p . … [2] - 4 (b) B is the point (2, 7) and AB = e 6o. Find the coordinates of A. ( … , … ) [2] (c) G NOT TO SCALE K O H M In triangle OGH, M is the midpoint of OH and K divides GH in the ratio 5 : 2. OG = g and OH = h . Find MK in terms of g and h. Give your answer in its simplest form. MK = … [4]

10 marks

Mark scheme: 6(a)(i)  −3  1    3  6(a)(ii) 3 1  2 6(a)(iii) 3.61 or 3.605 to 3.606 2 M1 for 22 + 32 oe 6(b) (6, 1) 2 B1 for each 6(c) 2 3 4 B3 for correct unsimplified expression for g + h 7 14 MK 2 or B2 for [ MK =] g + kh 7 3 or [ MK =] kg + h 14 2 or HK = (g – h) oe 7 5 or GK = (h – g) oe 7 or M1 for correct route for MK

This question in 0580/41 Oct/Nov 2022

11 (a) = 40m 2 Find the two possible values of m. m = … or … [3] (b) A B P NOT TO a SCALE O c C OABC is a parallelogram. OA = a and OC = c . P is the point on CB such that CP : PB = 3 : 1. (i) Find, in terms of a and/or c, in their simplest form, (a) AC, AC = … [1] (b) CP, CP = … [1] (c) OP. OP = … [1] (ii) OP and AB are extended to meet at Q. Find the position vector of Q. … [2]

8 marks

Mark scheme: 11(a) 2.5 and – 2.5 oe 3 42025 M2 for 1681m2 = oe 4 or M1 for (9m)2 + (40m)2 oe 11(b)(i)(a) c – a final answer 1 11(b)(i)(b) 3 1 a final answer 4 11(b)(i)(c) 3 1 FT c + their (b)(i)(b), must be a vector in terms of a and/or c in its c + a final answer simplest form 4 11(b)(ii) 4 2 1 4 a + c oe B1 for [ BQ = ] c or [ AQ = ] c 3 3 3 or M1 for a correct route or for answer a + kc oe, where k > 1

This question in 0580/42 Oct/Nov 2022

10 (a) a = b = 2 5 (i) On the grid, draw and label vector 2a. [1] (ii) On the grid, draw and label vector ( a - b ) . [2] (b) M C B NOT TO q SCALE N A O p OABC is a trapezium with OA parallel to CB. M is the midpoint of CB and N is the point on AB such that AN : NB = 1 : 2 . 3 O is the origin, OA = p , OC = q and CB = p . 4 (i) Find, in terms of p and/or q, in its simplest form (a) OB OB = … [1] (b) AB AB = … [2] (c) MN . MN = … [3] (ii) OA and MN are extended to meet at G. Find the position vector of G in terms of p. … [2]

11 marks

Mark scheme: 10(a)(i) 2a drawn correctly with direction arrow 1 10(a)(ii) a − b drawn correctly with direction arrow 2  4  B1 for   seen or implied  −3  or M1 for correctly drawing their a – b with an arrow 10(b)(i)(a) 3 1 q + p final answer 4 10(b)(i)(b) 1 2 M1 for a correct route q – p final answer 4 10(b)(i)(c) 13 2 3 3 2 p – q final answer M2 for p – (their (b)(i)(b)) oe 24 3 8 3 3 1 or for – p – q + p + (their (b)(i)(b)) oe 8 3 or M1 for a correct route or for 2 [BN =] – (their (b)(i)(b)) 3 1 or [AN = ] (their (b)(i)(b)) 3 2 13 or final answer kp – q oe or p –kq oe 3 24 10(b)(ii) 19 2 3 p oe final answer M1 for AG = p ÷ 2 soi 16 8 or for answer kp oe

This question in 0580/43 Oct/Nov 2022

10 (a) ABC is a triangle. - 11 B is the point ( 1, - 10) , A is the point (4, 14) and CA = . e 8o (i) Find the coordinates of C. ( … , … ) [2] (ii) Find BA. BA = [1] f p (iii) Find CA . … [2] (b) M T NOT TO a R SCALE O N b OMN is a triangle. OM = a and ON = b . R is a point on MN such that MR : RN = 3 : 2. ORT is a straight line. 2 3 (i) Show that OR = a + b . 5 5 [3] (ii) (a) NT = 4a + kb and OT = c OR . Find the value of k and the value of c. k = … c = … [4] (b) Find MT . MT = … [1]

13 marks

Mark scheme: 10(a)(i) (15, 6) 2 B1 for each 10(a)(ii)  3  1    24  10(a)(iii) 13.6 or 13.60… 2 M1 for (–11)2 + 82 oe 10(b)(i) 3 M3 a + (b – a) 5 2 or b + (a – b) 5 3 2 3 M2 for [ MR =] (b – a) oe leading to a + b with no 5 5 5 2 errors or [ NR = ] (a – b) oe 5 or M1 for MN = b – a or NM = a – b or a correct route for OR 10(b)(ii)(a) k = 5, c = 10 4 B2 for c = 10 2 3 or M1 for c( a + b) = b + 4a + kb oe 5 5 2 or for c = 4 5 and 3 M1 for  their c = k + 1 5 10(b)(ii)(b) 3a + 6b final answer 1 FT 3a + ( their k + 1) b

This question in 0580/41 Oct/Nov 2023

7 (a) p = q = - 5 5 (i) Find 3q. [1] f p (ii) (a) Find p - q . [1] f p (b) Find p - q . … [2] (b) M NOT TO SCALE a S O N b In triangle OMN, O is the origin, OM = a and ON = b . S is a point on MN such that MS : SN = 5 : 3 . Find, in terms of a and/or b, the position vector of S. Give your answer in its simplest form. … [3]

7 marks

Mark scheme: 7(a)(i)  −12  1    15  7(a)(ii)(a) 1  12     −10  7(a)(ii)(b) 15.6 or 15.62… 2 M1dep for their122 + ( their [ −]10 ) 2 oe, dep their 12 ≠ 0 and their –10 ≠ 0 7(b) 3 5 3 a + b final answer 8 8 B2 for an unsimplified correct answer 5 or MS = ( b − a ) soi 8 3 or NS = ( −+b a ) soi 8 or B1 for correct route for OS or for MN = b – a or NM = a – b

This question in 0580/42 Feb/March 2024

6 (a) Work out 2 e o - e o. - 5 - 7 f p [2] - 6 (b) MN = e o. 4 (i) M is the point (2, -5). Find the coordinates of N. ( … , … ) [1] (ii) Find MN . … [2] (c) A Q C NOT TO SCALE a P O B 2c OACB is a trapezium with OB = 2AC. OA = a and OB = 2c . 4 AP : PB = 4 : 1 and AQ = AC . 5 (i) Write each of the following in terms of a and c. Give each answer in its simplest form. (a) AB … [1] (b) CB … [1] (c) OP … [2] (d) QP … [2] (ii) Use your answers to make two statements about the relationship between lines QP and CB. … … [2]

13 marks

Mark scheme: 6(a)  4  2  6  4  k    B1 for   or answer  or    −3   −10  k  −3  6(b)(i) (–4, –1) 1 6(b)(ii) 7.21 or 7.211… 2 M1 for (–6)2 + 42 6(c)(i)(a) 2c – a 1 6(c)(i)(b) c – a 1 6(c)(i)(c) 1 2 4 (a + 8c) final answer M1 for [ AP =]  their(2c – a) 5 5 1 or [ BP = ]  – their (2c – a) 5 or for a correct vector route using the lines on the diagram 6(c)(i)(d) 4 2 4 4 (– a + c) final answer M1 for [QP = ] – c +  their(2c – a) 5 5 5 or for a correct vector route 6(c)(ii) [QP is] parallel [to CB ] 2 Dep both statements consistent with 4 their (c)(i)(b) and their (c)(i)(d) and both vectors QP = CB oe in terms of a and c 5 B1 for each dep on statement consistent with their (c)(i)(b) and their (c)(i)(d) and both vectors in terms of a and c

This question in 0580/42 Oct/Nov 2024

21 B E NOT TO n A SCALE C m O OAB is a triangle. C is the midpoint of OA. OC = m and CB = n . E lies on AB and AE : EB = 4 : 5 . Find, in terms of m and n, the position vector of E. Give your answer in its simplest form. … [4]

4 marks

Mark scheme: 21 14 4 4 4 m + noe simplified final answer B3 for 2m + ( −m + n ) oe 9 9 9 or M2 for 4 4 AE = ( −m + n ) or EA = (m − n ) 9 9 5 5 or BE = (m − n ) or EB = (n − m ) 9 9 M1 for OB = m + n or BO = –m – n or AB = n – m or BA = m – n or for a correct vector route along the lines on the diagram

This question in 0580/42 Feb/March 2025

26 F B C m NOT TO SCALE A E D p ABCD is a parallelogram. AB = m and AD = p . F is a point on BC and BF = 4FC. E is a point on AD and AE : ED = 1 : 2. (a) Find EF , in terms of m and p, in its simplest form. … [3] (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. … [2]

5 marks

Mark scheme: 26(a) 7 3 M2 for correct unsimplified expression in m + p terms of m and p 15 or M1 for a correct route 26(b) 3 2 2 1 m M1 for using the ratio p : p 7 3 5

This question in 0580/43 Oct/Nov 2025