C1 January 2010 Q3
3. The line \(l_1\) has equation \(3x + 5y - 2 = 0\)
The line \(l_2\) is perpendicular to \(l_1\) and passes through the point \((3, 1)\).
| Scheme | Marks |
|---|---|
| Putting the equation in the form \(y = mx\ (+c)\) and attempting to extract the \(m\) or \(mx\) (not the \(c\)), or finding 2 points on the line and using the correct gradient formula. | M1 |
| Gradient \(= -\dfrac{3}{5}\) (or equivalent) | A1 |
| (2) |
Notes
Condone sign errors and ignore the \(c\) for the M mark, so...
both marks can be scored even if \(c\) is wrong (e.g. \(c = -\dfrac{2}{5}\)) or omitted.
Answer only: \(-\dfrac{3}{5}\) scores M1 A1. Any other answer only scores M0 A0.
\(y = -\dfrac{3}{5}x + \dfrac{2}{5}\) with no further progress scores M0 A0 (\(m\) or \(mx\) not extracted).
| Scheme | Marks |
|---|---|
| Gradient of perp. line \(= \dfrac{-1}{\text{"}\left(-\frac{3}{5}\right)\text{"}}\) (Using \(-\dfrac{1}{m}\) with the \(m\) from part (a)) | M1 |
| \(y - 1 = \text{"}\left(\dfrac{5}{3}\right)\text{"}(x - 3)\) | M1 |
| \(y = \dfrac{5}{3}x - 4\) (Must be in this form... allow \(y = \dfrac{5}{3}x - \dfrac{12}{3}\) but not \(y = \dfrac{5x - 12}{3}\)) This A mark is dependent upon both M marks. | A1 |
| (3) | |
| (5 marks) |
Notes
2nd M: For the equation, in any form, of a straight line through \((3, 1)\) with any numerical gradient (except 0 or \(\infty\)).
(Alternative is to use \((3, 1)\) in \(y = mx + c\) to find a value for \(c\), in which case \(y = \dfrac{5}{3}x + c\) leading to \(c = -4\) is sufficient for the A1).
(See general principles for straight line equations at the end of the scheme).