4MA0 Higher January 2016 Paper 3 Q18
18 \(A\) is the point with coordinates (1, 3)
\(B\) is the point with coordinates (−2, −1)
The line L has equation \(3y = 4 - 2x\)
Is line L parallel to \(AB\)?
Show your working clearly.
(3)
| Scheme | Marks |
|---|---|
| eg. \(\dfrac{3 - -1}{1 - -2} \left(= \dfrac{4}{3}\right)\) | M1 |
\(y = \dfrac{4}{3} - \dfrac{2}{3}x\) or \(y = \dfrac{4 - 2x}{3}\) or (\(m\) = ) \(-\dfrac{2}{3}\) | M1 |
| No with reason | A1 |
| (3) |
Notes
M1: for gradient of line \(AB\)
A1: accept no with \(-\dfrac{2}{3}\) (or – 0.666…) and \(\dfrac{4}{3}\) (or 1.333…)
\(-\dfrac{2}{3}\) and \(\dfrac{4}{3}\) may be seen as coefficients of \(x\) in \(y = mx + c\) OR
shows that \(3y = 4 - 2x\) and line through \(AB\) \(\left(y = \dfrac{4}{3}x + \dfrac{5}{3}\right)\) intersect at \(x = -\dfrac{1}{6}\) or \(y = \dfrac{13}{9}\) OR
L has a negative gradient (with evidence), line \(AB\) has a positive gradient (with evidence)
NB: Any value given for a gradient must be correct
Alternatively: M1 for \(y = \dfrac{4}{3}x + \dfrac{5}{3}\)
M1 for \(\dfrac{4}{3}x + \dfrac{5}{3} = \dfrac{4}{3} - \dfrac{2}{3}x\) oe
M2 for sketch of L with \(\left(0, \dfrac{4}{3}\right)\) and (2, 0) marked on axes AND (1,3) and (-2 , -1) joined