Date | November 2009 | Marks available | 2 | Reference code | 09N.1.sl.TZ0.10 |
Level | SL only | Paper | 1 | Time zone | TZ0 |
Command term | Find | Question number | 10 | Adapted from | N/A |
Question
A line joins the points A(2, 1) and B(4, 5).
Find the gradient of the line AB.
Let M be the midpoint of the line segment AB.
Write down the coordinates of M.
Let M be the midpoint of the line segment AB.
Find the equation of the line perpendicular to AB and passing through M.
Markscheme
\({\text{Gradient}} = \frac{{(5 - 1)}}{{(4 - 2)}}\) (M1)
Note: Award (M1) for correct substitution in the gradient formula.
\(= 2\) (A1) (C2)
[2 marks]
Midpoint = (3, 3) (accept x = 3, y = 3 ) (A1) (C1)
[1 mark]
\({\text{Gradient of perpendicular}} = -\frac{{1}}{{2}}\) (A1)(ft)
\(y = - \frac{{1}}{{2}} x + c\) (M1)
\(3 = - \frac{{1}}{{2}} \times 3 + c\)
\(c = 4.5\)
\(y = -0.5x + 4.5\) (A1)(ft)
OR
\(y - 3 = -0.5(x - 3)\) (A1)(A1)(ft)
Note: Award (A1) for –0.5, (A1) for both threes.
OR
\(2y + x = 9\) (A1)(A1)(ft) (C3)
Note: Award (A1) for 2, (A1) for 9.
[3 marks]
Examiners report
While parts (a) and (b) were answered or at least attempted with various success, few candidates made progress in part (c). Some candidates used the coordinates of point A or B rather than M and others could not find the gradient of the perpendicular line.
While parts (a) and (b) were answered or at least attempted with various success, few candidates made progress in part (c). Some candidates used the coordinates of point A or B rather than M and others could not find the gradient of the perpendicular line.
While parts (a) and (b) were answered or at least attempted with various success, few candidates made progress in part (c). Some candidates used the coordinates of point A or B rather than M and others could not find the gradient of the perpendicular line.