| 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, \({L_1}\) , has equation \(x + 4y + 34 = 0\) .
Find the gradient of \({L_1}\) .
The equation of line \({L_2}\) is \(y = mx\) . \({L_2}\) is perpendicular to \({L_1}\) .
Find the value of \(m\).
The equation of line \({L_2}\) is \(y = mx\) . \({L_2}\) is perpendicular to \({L_1}\) .
Find the coordinates of the point of intersection of the lines \({L_1}\) and \({L_2}\) .
Markscheme
\(4y = - x - 34\) or similar rearrangement (M1)
\({\text{Gradient}} = - \frac{1}{4}\) (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)
\(( - 2{\text{, }} - 8)\) (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.
