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 L1 and L2 . The equation of L2 is y=x .
Find
(i) the gradient of L1 ;
(ii) the equation of L1 .
Find the area of the shaded triangle.
Markscheme
(i) 0−26−0 (M1)
=−13 (−26, −0.333) (A1) (C2)
(ii) y=−13x+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]
area=6×1.52 (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.