Date | November 2008 | Marks available | 2 | Reference code | 08N.1.sl.TZ0.7 |
Level | SL only | Paper | 1 | Time zone | TZ0 |
Command term | Find | Question number | 7 | Adapted from | N/A |
Question
A straight line, L1 , has equation x+4y+34=0 .
Find the gradient of L1 .
The equation of line L2 is y=mx . L2 is perpendicular to L1 .
Find the value of m.
The equation of line L2 is y=mx . L2 is perpendicular to L1 .
Find the coordinates of the point of intersection of the lines L1 and L2 .
Markscheme
4y=−x−34 or similar rearrangement (M1)
Gradient=−14 (A1) (C2)
m=4 (A1)(ft)(A1)(ft) (C2)
Note: (A1) Change of sign
(A1) Use of reciprocal
[2 marks]
Reasonable attempt to solve equations simultaneously (M1)
(−2, −8) (A1)(ft) (C2)
Note: Accept x=−2 y=−8. Award (A0) if brackets not included.
[2 marks]
Examiners report
The omission of the negative sign was a common fault.
Most candidates managed to answer this correctly from their (a).
This part proved challenging for the majority. Once again, the use of the GDC was expected.