Mathematics · May / June 2026 · Paper 21 · Question 25

5n – m [1]

1 m or ( 1 10 ) − − 2n n m 5 5 [3]

M2 for 1 2 4 3 + (5n – m) oe − + − (5n – m) m their m 5 n their 5 5 5 5

or M1 for a correct route e.g. PB + BQ Condone a mix of notation &&&& &&&& &&& BQ = 2 5 their (5n – m) or for or QC or QB correct for M1. &&&& 3 − their (5n – m) or for CQ = 5

&&&& 3 − QD = 6n m oe 5 &&&& 4 − or PD = 8n m oe 5 &&&& &&&& For showing any two of QD , PD and &&&& PQ share a common factor, e.g. &&&& 1 − QD = 3(2n m) 5 &&&& 1 − or PD = 4(2n m) 5 [2]

B1 Allow unsimplified B1 Other pairs with common factors are acceptable &&&& &&&& 1 1 = − − e.g. PQ 2 5 n m and QD = 3(2n m) 5 &&&& &&&& PD = 4 1 10 −m) and = ( − ) 5 (10n PQ n m 5 If factorisation not shown, allow correct vectors in terms of m and n followed by a correct relationship, e.g. &&&& &&&& &&&& &&&& &&& &&&& QD = 3 4 PD , 4 QD = 3 PD , QD : PD = 3 : 4, &&&& QD 3 &&&& = PD 4 &&&& &&&& &&&& &&& &&&& &&&& PQ = 1 4 PD , 4 PQ = PD , PQ : PD = 1 : 4, &&&& PQ 1 &&&& = PD 4 &&&& &&&& &&&& &&&& &&&& &&&& PQ , 1 QD =3 QD = PQ , QD : PQ = 3 : 1, 3 &&&& QD &&&& = 3 PQ &&&& &&&& A correct conclusion, e.g. &&&& &&&& QD is a multiple of/is parallel to PQ and they have a common point Q, so they are collinear oe [2]

B1 e.g. with common point P if PQ and PD used, &&&& &&&& or with common point D if QD and PD used. Allow without vector arrows for all B1 marks in 25(b).