| Date | November 2007 | Marks available | 2 | Reference code | 07N.1.sl.TZ0.13 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Sketch | Question number | 13 | Adapted from | N/A |
Question
The mid-point, M, of the line joining A(s , 8) to B(−2, t) has coordinates M(2, 3).
Calculate the values of s and t.
Find the equation of the straight line perpendicular to AB, passing through the point M.
Markscheme
\(s = 6\) (A1)
\(t = - 2\) (A1) (C2)
[2 marks]
\({\text{gradient of AB}} = \frac{{ - 2 - 8}}{{ - 2 - 6}} = \frac{{ - 10}}{{ - 8}} = \frac{5}{4}\) (A1)(ft)
(A1) for gradient of AM or BM \( = \frac{5}{4}\)
\({\text{Perpendicular gradient}} = - \frac{4}{5}\) (A1)(ft)
Equation of perpendicular bisector is
\(y = - \frac{4}{5}x + c\)
\(3 = - \frac{4}{5}(2) + c\) (M1)
\(c = 4.6\)
\(y = -0.8x + 4.6\)
or \(5y = -4x + 23\) (A1)(ft) (C4)
[4 marks]
Examiners report
(a) It was surprising how many errors were made in finding the values for s and t
(b) The candidates had difficulty in finding the equation of a straight line. They could find the gradient of the line AB and a number could give the gradient of the perpendicular line but most did not substitute the correct midpoint to find the equation of the line.
