| Date | November 2017 | Marks available | 2 | Reference code | 17N.1.sl.TZ0.2 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 2 | Adapted from | N/A |
Question
The coordinates of point A are \((6,{\text{ }} - 7)\) and the coordinates of point B are \(( - 6,{\text{ }}2)\). Point M is the midpoint of AB.
\({L_1}\) is the line through A and B.
The line \({L_2}\) is perpendicular to \({L_1}\) and passes through M.
Find the coordinates of M.
Find the gradient of \({L_1}\).
Write down the gradient of \({L_2}\).
Write down, in the form \(y = mx + c\), the equation of \({L_2}\).
Markscheme
\((0,{\text{ }}2.5)\)\(\,\,\,\)OR\(\,\,\,\)\(\left( {0,{\text{ }} - \frac{5}{2}} \right)\) (A1)(A1) (C2)
Note: Award (A1) for 0 and (A1) for –2.5 written as a coordinate pair. Award at most (A1)(A0) if brackets are missing. Accept “\(x = 0\) and \(y = - 2.5\)”.
[2 marks]
\(\frac{{2 - ( - 7)}}{{ - 6 - 6}}\) (M1)
Note: Award (M1) for correct substitution into gradient formula.
\( = - \frac{3}{4}{\text{ }}( - 0.75)\) (A1) (C2)
[2 marks]
\(\frac{4}{3}{\text{ }}(1.33333 \ldots )\) (A1)(ft) (C1)
Note: Award (A0) for \(\frac{1}{{0.75}}\). Follow through from part (b).
[1 mark]
\(y = \frac{4}{3}x - \frac{5}{2}{\text{ }}(y = 1.33 \ldots x - 2.5)\) (A1)(ft) (C1)
Note: Follow through from parts (c)(i) and (a). Award (A0) if final answer is not written in the form \(y = mx + c\).
[1 mark]
