Date | May 2014 | Marks available | 2 | Reference code | 14M.1.sl.TZ2.4 |
Level | SL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 4 | Adapted from | N/A |
Question
The line \(L\) is parallel to the vector \(\left( \begin{array}{l}3\\2\end{array} \right)\).
The line \(L\) passes through the point \((9, 4)\).
Find the gradient of the line \(L\).
Find the equation of the line \(L\) in the form \(y = ax + b\).
Write down a vector equation for the line \(L\).
Markscheme
attempt to find gradient (M1)
eg reference to change in \(x\) is \(3\) and/or \(y\) is \(2\), \(\frac{3}{2}\)
gradient \( = \frac{2}{3}\) A1 N2
[2 marks]
attempt to substitute coordinates and/or gradient into Cartesian equation
for a line (M1)
eg \(y - 4 = m(x - 9),{\text{ }}y = \frac{2}{3}x + b,{\text{ }}9 = a(4) + c\)
correct substitution (A1)
eg \(4 = \frac{2}{3}(9) + c,{\text{ }}y - 4 = \frac{2}{3}(x - 9)\)
\(y = \frac{2}{3}x - 2{\text{ }}\left( {{\text{accept }}a = \frac{2}{3},{\text{ }}b = - 2} \right)\) A1 N2
[3 marks]
any correct equation in the form r = a + tb (any parameter for t), where a indicates position eg \(\left( \begin{array}{l}9\\4\end{array} \right)\) or \(\left( \begin{array}{c}0\\ - 2\end{array} \right)\), and b is a scalar multiple of \(\left( \begin{array}{l}3\\2\end{array} \right)\)
eg r = \(\left( \begin{array}{c}9\\4\end{array} \right) + t\left( \begin{array}{c}3\\2\end{array} \right),\left( \begin{array}{c}x\\y\end{array} \right) = \left( \begin{array}{c}3t + 9\\2t + 4\end{array} \right)\), r = 0i − 2 j + s(3i + 2 j) A2 N2
Note: Award A1 for a + tb, A1 for L = a + tb, A0 for r = b + ta.
[2 marks]