Date | May 2008 | Marks available | 2 | Reference code | 08M.1.sl.TZ2.6 |
Level | SL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 6 | Adapted from | N/A |
Question
Write down the gradient of the line \(y = 3x + 4\).
Find the gradient of the line which is perpendicular to the line \(y = 3x + 4\).
Find the equation of the line which is perpendicular to \(y = 3x + 4\) and which passes through the point \((6{\text{, }}7)\).
Find the coordinates of the point of intersection of these two lines.
Markscheme
\(3\) (A1) (C1)
[1 mark]
\( - 1/3\) (ft) from (a) (A1)(ft) (C1)
[1 mark]
Substituting \((6{\text{, }}7)\) in \(y ={\text{their }}mx + c\) or equivalent to find \(c\). (M1)
\(y = \frac{{ - 1}}{3}x + 9\) or equivalent (A1)(ft) (C2)
[2 marks]
\((1.5{\text{, }}8.5)\) (A1)(A1)(ft) (C2)
Note: Award (A1) for \(1.5\), (A1) for \(8.5\). (ft) from (c), brackets not required.
[2 marks]
Examiners report
This question was well answered by some candidates and poorly answered by others. It seemed to be part of the syllabus that might have been fully taught by some schools and not by others. It was surprising to see how many candidates could not find the gradient of a perpendicular line when this has been tested for many years.
This question was well answered by some candidates and poorly answered by others. It seemed to be part of the syllabus that might have been fully taught by some schools and not by others. It was surprising to see how many candidates could not find the gradient of a perpendicular line when this has been tested for many years.
This question was well answered by some candidates and poorly answered by others. It seemed to be part of the syllabus that might have been fully taught by some schools and not by others. It was surprising to see how many candidates could not find the gradient of a perpendicular line when this has been tested for many years.
This question was well answered by some candidates and poorly answered by others. It seemed to be part of the syllabus that might have been fully taught by some schools and not by others. It was surprising to see how many candidates could not find the gradient of a perpendicular line when this has been tested for many years.