Date | May 2011 | Marks available | 4 | Reference code | 11M.1.sl.TZ1.13 |
Level | SL only | Paper | 1 | Time zone | TZ1 |
Command term | Find | Question number | 13 | Adapted from | N/A |
Question
The equation of the line R1 is 2x+y−8=0 . The line R2 is perpendicular to R1 .
Calculate the gradient of R2 .
The point of intersection of R1 and R2 is (4, k) .
Find
(i) the value of k ;
(ii) the equation of R2 .
Markscheme
y=−2x+8 (M1)
Note: Award (M1) for rearrangement of equation or for −2 seen.
m(perp)=12 (A1) (C2)
[2 marks]
(i) 2(4)+k−8=0 (M1)
Note: Award (M1) for evidence of substituting x=4 into R1 .
k=0 (A1) (C2)
(ii) y=12x+c (can be implied) (M1)
Note: Award (M1) for substitution of 12 into equation of the line.
0=12(4)+c
y=12x−2 (A1)(ft) (C2)
Notes: Follow through from parts (a) and (b)(i). Accept equivalent forms for the equation of a line.
OR
y−y1=12(x−x1) (M1)
Note: Award (M1) for substitution of 12 into equation of the line.
y=12(x−4) (A1)(ft) (C2)
Notes: Follow through from parts (a) and (b)(i). Accept equivalent forms for the equation of a line.
[4 marks]
Examiners report
Parts a and bi of Question 13 appeared to be accessible to most candidates, but part bii was not well attempted. Many candidates did not show their working and lost method marks due to their incorrect answers.
Parts a and bi of Question 13 appeared to be accessible to most candidates, but part bii was not well attempted. Many candidates did not show their working and lost method marks due to their incorrect answers.