Mathematics · May / June 2026 · Paper 22 · Question 22

1 5 (c – a) or 1 5 c – 1 5 a [1]

AC = c – a or CA = a – c 2

M1 for Allow e.g. AO + OC or AC for AC &&& &&&& &&&&&&&&&&&& or for 1 4 This is for a correct route for AP , dealing AO + OC + CA AC or 5 5 correctly with the ratio;

condone a mix of notation, 4 −+ + e.g. a c 5 CA

1 5 (4a + c) or 4 5 a + 1 5 c &&&& &&& [1]

FT their (a)(i) + a simplified Their (a)(i) must be in terms of a and/or c for this mark.

For showing any two of OQ , OP and &&&& PQ share a common factor, e.g. &&&& 1 4 ( ) = + OP a c and 5 &&&& OQ = 1 4 (4a + c) [2]

B1 Other pairs with common factors are acceptable, e.g. &&&& &&& PQ = 1 5 (a + 1 OP = 4 1 4 c) and a + c 5 4 &&&& &&&& OQ = 1 4 ( ) PQ = 1 + 4 a c and 20 (4a + c) If factorisation not shown, allow two correct vectors in terms of a and c from &&&& &&&& 4 1 OQ = a + 1 = + OP a c , 4 c 5 5 &&&& PQ = 1 5 a + 1 20 c (which don’t have to be simplified), followed by a correct relationship, e.g. &&&& &&&& &&&& &&&& &&& &&& OQ = 5 4 OP , 4 OQ = 5 OP , OQ : OP = 5 : 4, &&&& OQ 5 &&&& = OP 4 &&&& &&&& &&&& &&& &&&& &&& PQ = 1 4 OP , 4 PQ = OP, PQ : OP = 1 : 4, &&&& PQ 1 &&&& = OP 4 &&&& &&& &&&& &&&& &&&& &&&& PQ , 1 OQ = 5 OQ = PQ , OQ : PQ = 5 : 1, 5 &&&& OQ &&&& = 5 PQ Allow without vector arrows for both B1 marks in 22(b). &&& &&&& A correct conclusion e.g. &&&& &&& OQ is a multiple of/is parallel to OP and they have a common point O, so collinear oe [1]

B1 e.g. with common point P if OP and PQ used, or &&&& &&&& with common point Q if OQ and PQ used.