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Date May 2014 Marks available 2 Reference code 14M.1.sl.TZ2.4
Level SL only Paper 1 Time zone TZ2
Command term Write down 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 \(L\).

[2]
a.

Find the equation of the line \(L\) in the form \(y = ax + b\).

[3]
b.

Write down a vector equation for the line \(L\).

[2]
c.

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]

a.

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]

b.

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]

c.

Examiners report

[N/A]
a.
[N/A]
b.
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c.

Syllabus sections

Topic 4 - Vectors » 4.3 » Vector equation of a line in two and three dimensions: \(r = a + tb\) .
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