Vectors in two dimensions — Paper 4 · IGCSE Mathematics 0580

1 (a) y 7 6 5 4 3 T 2 1 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 x –1 –2 –3 –4 P –5 –6 –7 (i) Describe fully the single transformation that maps triangle T onto triangle P. … … [2] - 2 (ii) Translate triangle T by the vector [2] e - 5o. (iii) Rotate triangle T through 90° anticlockwise about (0, 0). [2] 1 (iv) Enlarge triangle T by scale factor - with centre (0, 0). [2] 2 (b) y B (5, 6) NOT TO SCALE A (3, 2) O x (i) Find the column vector AB. AB = [1] f p (ii) Find AB . AB = … [2] (iii) B is the mid-point of the line AC. Find the co-ordinates of C. ( … , … ) [2] (iv) Find the equation of the straight line that passes through A and B. … [3] (v) The straight line that passes through A and B cuts the y-axis at D. Write down the co-ordinates of D. ( … , … ) [1]

17 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) Reflection 2 B1 for each y = –1 1(a)(ii) Triangle at 2 − 2   k  (0, –3), (4, –1), (4, –3) B1 for translation   or    k   −5  or for three correct vertices 1(a)(iii) Triangle at 2 B1 for rotation about (0, 0) 90° clockwise (–2, 2), (–2, 6), (–4, 6) or 90° anticlockwise with wrong centre or for three correct vertices 1(a)(iv) Triangle at (–3, –1), (–3, –2), 2 1 B1 for scale factor − with wrong centre (–1, –1) 2 1 or scale factor with centre (0, 0) 2 or for three correct vertices 1(b)(i)  2  1   cao  4  1(b)(ii) 4.47 or 4.472… 2 M1 for (their 2) 2 + (their 4) 2 1(b)(iii) (7, 10) 2 B1 for each 1(b)(iv) y = 2 x − 4 oe 3 6 − 2 M1 for gradient = oe or answer y = mx – 4 5 − 3 M1 for substituting (3, 2) or (5, 6) into y = their mx + c or into y – k = their m(x – h) or into their y = mx – 4 1(b)(v) (0, –4) 1 FT their (b)(iv)

This question in 0580/43 Oct/Nov 2018

8 (a) Make p the subject of (i) 5p + 7 = m , p = … [2] (ii) y 2 - 2p 2 = h . p = … [3] (b) y A (0, 5) NOT TO SCALE B (-3, 4) x O (i) Write OA as a column vector. OA = [1] f p (ii) Write AB as a column vector. AB = [1] f p (iii) A and B lie on a circle, centre O. Calculate the length of the arc AB. … [6]

13 marks

Mark scheme: 8(a)(i) m − 7 2 7 m oe final answer M1 for 5p = m – 7 or p + = 5 5 5 8(a)(ii) 2 2 3 M1 for first correct step isolate term in p or [± ] y − h or [ ± ] h − y oe divide by ±2 2 −2 M1 for second correct step FT their first step final answer 8(b)(i) 0 1  5 8(b)(ii)  − 3  1    − 1  8(b)(iii) 3.22 or 3.216... to 3.220... 6 B3 for [angle AOB =] 36.8 or 36.9 or 36.84 to 36.87 or M2 for tan[AOB] = 34 oe or for [AOB = ]2 × sin-1  2 2  (5 − 4) + (0 −−3)   oe  10    or for cos [AOB =] 2 5 2 + 5 2 − (5 − 4) 2 + (0 −−3) 2 ( ) oe 2 × 5 × 5 or M1 for recognition of right-angle with perpendicular from B to OA or x-axis or for [AB2 = ] (5 − 4) 2 + (0 −− 3) 2 or better oe or (their AB)2 = 52 + 52 – 2 × 5 × 5 × cosOAB oe their angle AOB M2 for × 2 × π × 5 oe 360 or M1 for radius = 5 soi

This question in 0580/42 Oct/Nov 2019

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

8 (a) AB = BC = DC = - 1 5 - 3 Find (i) AC, AC = [2] f p (ii) BD, BD = [2] f p (iii) BC . … [2] (b) C NOT TO D a SCALE E b O A B In the diagram, OAB and OED are straight lines. O is the origin, A is the midpoint of OB and E is the midpoint of AC. AC = a and CB = b . Find, in terms of a and b, in its simplest form (i) AB, AB = … [1] (ii) OE, OE = … [2] (iii) the position vector of D. … [3]

12 marks

Mark scheme: 8(a)(i) 4 2  4   k   B1 for   or   4  k   4  8(a)(ii)  −4  2  −4   k    B1 for   or    8   k   8  8(a)(iii) 5.39 or 5.385.. 2 M1 for (–2)2 + 52 oe 8(b)(i) a + b 1   8(b)(ii) 3 2 M1 for a correct route, e.g. OA + AE a + b 2 8(b)(iii) 4 3  4 2a + b M2 for unsimplified OD or for b 3 3  or M1 for OD attempted in terms of a and b  1  2 or for CD = b or DB = b seen 3 3

This question in 0580/43 Oct/Nov 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

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

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

4 (a) y 6 5 4 3 2 T 1 0 x 1 2 3 4 5 6 7 8 9 10 – 1 – 2 (i) Enlarge triangle T by scale factor 3, centre (0, 2). [2] (ii) (a) Rotate triangle T about (4, 2) by 90˚ clockwise. Label the image P. [2] (b) Reflect triangle T in the line x + y = 6 . Label the image Q. [3] (c) Describe fully the single transformation that maps triangle P onto triangle Q. … … [2] (b) a H O Z NOT TO SCALE b K The diagram shows triangle OHK, where O is the origin. The position vector of H is a and the position vector of K is b. Z is the point on HK such that HZ : ZK = 2 : 5. Find the position vector of Z, in terms of a and b. Give your answer in its simplest form. … [3]

12 marks

Mark scheme: 4(a)(i) Triangle at (3, –1), (9, –1), (9, 2) 2 B1 for correct shape, size and orientation or for correct plots but no triangle 4(a)(ii)(a) Triangle at (3, 3), (4, 3), (3, 5) 2 B1 for correct shape size and orientation or for rotation about (4, 2) 90˚ anticlockwise or for correct plots but no triangle 4(a)(ii)(b) Triangle at (4, 3), (5, 3), (5, 5) 3 B2 for correct shape size and orientation or for correct plots but no triangle or M1 for x + y = 6 drawn 4(a)(ii)(c) Reflection 2 B1 for each x = 4 4(b) 5 2 3 a + b final answer 7 7 B2 for correct unsimplified answer OR 2 5 M2 for HZ = ( b − a ) or KZ = ( a − b ) oe 7 7 or M1 for HK = –a + b or KH = –b + a or for a correct route

This question in 0580/42 Feb/March 2023

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

12 y B NOT TO SCALE A x O O is the origin (0, 0), A is the point (8, 1) and B is the point (2, 5). (a) Write as column vectors. (i) OB OB = [1] f p (ii) AB AB = [1] f p (b) Find the equation of the line AB. Give your answer in the form y = mx + c . y = … [3] (c) Find the equation of the perpendicular bisector of AB. Give your answer in the form y = mx + c . y = … [4] (d) The line AB meets the y-axis at P. The perpendicular bisector of AB meets the y-axis at Q. Find the length of PQ. … [2]

11 marks

Mark scheme: 12(a)(i) 2 1  5 12(a)(ii)  −6  1    4  12(b) 2 19 3 1 − 5 [ y =] − x + oe M1 for gradient = oe 3 3 8 − 2 M1 for substituting (8, 1) or (2, 5) into y = their mx + c 12(c) 3 9 4 B1 for (5, 3) oe [ y = x − oe 1 ]2 2 M1 for gradient = −their gradient of AB M1 substituting their midpoint into y = their mx + c 12(d) 65 2 19 9 oe M1 for their – their − oe 6 3 2

This question in 0580/42 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

10 A is the point (2, 1). 2 AB = e o 4 Find the coordinates of B. ( … , … ) [2]

2 marks

Mark scheme: 10 (4, 5) 2 B1 for each

This question in 0580/41 Oct/Nov 2025

18 C B NOT TO SCALE b O a A In the diagram, OA is parallel to CB. OA | CB = 4 | 3 OA = a and OB = b . (a) Find AB in terms of a and b. AB = … [1] (b) M is the midpoint of OC. Find AM in terms of a and b. Give your answer in its simplest form. AM = … [3]

4 marks

Mark scheme: 18(a) b – a 1  18(b) 11 1 3 B2 for a correct unsimplified vector for AM seen − a + b oe simplified 8 2  1 3 or for OM = b − a oe soi final answer 2 8  3 or B1 for OC = b − a oe soi 4  or for a correct vector route for AM along lines on the diagram

This question in 0580/42 Oct/Nov 2025