| Date | May 2012 | Marks available | 2 | Reference code | 12M.1.sl.TZ1.2 |
| Level | SL only | Paper | 1 | Time zone | TZ1 |
| Command term | Calculate | Question number | 2 | Adapted from | N/A |
Question
The coordinates of point A are (−4, p) and the coordinates of point B are (2, −3) .
The mid-point of the line segment AB, has coordinates (q, 1) .
Find the value of
(i) q ;
(ii) p .
Calculate the distance AB.
Markscheme
(i) \(\frac{{ - 4 + 2}}{2} = q\) (M1)
Note: Award (M1) for correct substitution in the correct formula.
q = –1 (A1)
(ii) \(\frac{{p + ( - 3)}}{2} = 1\) (M1)
Note: Award (M1) for correct substitution into the correct formula or consistent with their equation in (i).
p = 5 (A1) (C4)
Notes: Award A marks for integer values. Penalise if answers left as a fraction the first time a fraction is seen.
[4 marks]
\({\text{AB}} = \sqrt {{{(2 + 4)}^2} + {{( - 3 - 5)}^2}} \) (M1)
Note: Award (M1) for the correct substitution of their coordinates for A and B in the correct formula.
AB = 10 (A1)(ft) (C2)
Note: Follow through from their answer to part (a)(ii).
[2 marks]
Examiners report
(a) Despite some good answers in this part of the question, sign errors in setting up one or both of the equations meant that marks were lost by some candidates. This error was particularly prevalent in finding the value of p and the equation \(\frac{{(p + 3)}}{2} = 1\) was seen quite often. In part (b), there was a requirement for a correct substitution of their coordinates for A and B into the correct formula for Pythagoras and, while (2 + 4)2 was often seen, sign errors in the second component of (–3 – 5)2 proved to be the downfall of a significant number of candidates. As a consequence, the final two marks were lost. Given these errors, there were still a significant number of full mark responses to this question.
(a) Despite some good answers in this part of the question, sign errors in setting up one or both of the equations meant that marks were lost by some candidates. This error was particularly prevalent in finding the value of p and the equation \(\frac{{(p + 3)}}{2} = 1\) was seen quite often. In part (b), there was a requirement for a correct substitution of their coordinates for A and B into the correct formula for Pythagoras and, while (2 + 4)2 was often seen, sign errors in the second component of (–3 – 5)2 proved to be the downfall of a significant number of candidates. As a consequence, the final two marks were lost. Given these errors, there were still a significant number of full mark responses to this question.
