Date | May 2011 | Marks available | 2 | Reference code | 11M.1.sl.TZ1.10 |
Level | SL only | Paper | 1 | Time zone | TZ1 |
Command term | Find | Question number | 10 | Adapted from | N/A |
Question
The diagram shows the straight lines \({L_1}\) and \({L_2}\) . The equation of \({L_2}\) is \(y = x\) .
Find
(i) the gradient of \({L_1}\) ;
(ii) the equation of \({L_1}\) .
Find the area of the shaded triangle.
Markscheme
(i) \(\frac{{0 - 2}}{{6 - 0}}\) (M1)
\( = - \frac{1}{3}{\text{ }}\left( { - \frac{2}{6}{\text{, }} - 0.333} \right)\) (A1) (C2)
(ii) \(y = - \frac{1}{3}x + 2\) (A1)(ft) (C1)
Notes: Follow through from their gradient in part (a)(i). Accept equivalent forms for the equation of a line.
[3 marks]
\({\text{area}} = \frac{{6 \times 1.5}}{2}\) (A1)(M1)
Note: Award (A1) for \(1.5\) seen, (M1) for use of triangle formula with \(6\) seen.
\( = 4.5\) (A1) (C3)
[2 marks]
Examiners report
In this question, many candidates did not use the \(x\) and \(y\) intercepts to find the slope and attempted to read ordered pairs from the graph.
Part b proved difficult for many candidates, often using trigonometry rather than the more straight forward area of the triangle.