Mathematics · May / June 2026 · Paper 43 · Question 14
0580
As printed in the paper
The question as text
A is the point (5, 2) and Ó is the point (Ç3, 6).
Read automatically from the paper; the printed page above is exact.
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14(a)[3]
8.94 or 8.944 to 8.945 [3]
Mark at most accurate. Accept 80 or 4 5 as final answer for 3 marks. Allow recovery from –82 or –42. Accept 5 + 3 and 2 – 6
M2 for (–3 – 5)2 + (6 – 2)2 oe − 8
or M1 for (–3 – 5) and (6 – 2) oe M1 implied by AB = ± oe 4
14(b)[4]
[y =] 2x + 9 final answer [5]
− For 1st M1 look for 6 = –3m + c and 2 = 5m + c 4
M1 for [gradient of AB =] 6 2 oe −− rearranged to form am = b. M1 is awarded at this 3 5 stage provided either a or b is correct. If correct it
M1 for gradient of perpendicular, should lead to 8m = –4. 1 p = –1 ÷ their − oe 2
M1 for substituting (–2 ,5) into y = their px + c oe where their p is not 0 or their gradient of AB
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