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Date	November 2019	Marks available	2	Reference code	19N.2.SL.TZ0.S_2
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Write down	Question number	S_2	Adapted from	N/A

Question

Consider the lines $L_{1}$ ${L_1}$ and $L_{2}$ ${L_2}$ with respective equations

$L_{1} : y = - \frac{2}{3} x + 9$ ${L_1}\,{\text{:}}\,y = - \frac{2}{3}x + 9$ and $L_{2} : y = \frac{2}{5} x - \frac{19}{5}$ ${L_2}{\text{:}}\,y = \frac{2}{5}x - \frac{{19}}{5}$ .

A third line, $L_{3}$ ${L_3}$ , has gradient $- \frac{3}{4}$ $- \frac{3}{4}$ .

Find the point of intersection of $L_{1}$ ${L_1}$ and $L_{2}$ ${L_2}$ .

[2]

Write down a direction vector for $L_{3}$ ${L_3}$ .

[1]

$L_{3}$ ${L_3}$ passes through the intersection of $L_{1}$ ${L_1}$ and $L_{2}$ ${L_2}$ .

Write down a vector equation for $L_{3}$ ${L_3}$ .

[2]

Markscheme

valid approach (M1)

eg $L_{1} = L_{2}$ ${L_1} = {L_2}$ , $x = 12$ $x = 12$ , $y = 1$ $y = 1$

$(12, 1)$ $\left( {12{\text{, }}1} \right)$ (exact) A1 N2

[2 marks]

$(\begin{matrix} - 4 3 \end{matrix})$ $\left( {\begin{array}{*{20}{c}} { - 4} \\ 3 \end{array}} \right)$ (or any multiple of $(\begin{matrix} - 4 3 \end{matrix})$ $\left( {\begin{array}{*{20}{c}} { - 4} \\ 3 \end{array}} \right)$ ) A1 N1

[1 mark]

any correct equation in the form r = a + $t$ $t$ b (accept any parameter for $t$ $t$ ) where
a is a position vector for a point on $L_{1}$ ${L_1}$ , and b is a scalar multiple of $(\begin{matrix} - 4 3 \end{matrix})$ $\left( {\begin{array}{*{20}{c}} { - 4} \\ 3 \end{array}} \right)$ A2 N2

eg r $= (\begin{matrix} 121 \end{matrix}) + t (\begin{matrix} - 4 3 \end{matrix})$ $= \left( {\begin{array}{*{20}{c}} {12} \\ 1 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { - 4} \\ 3 \end{array}} \right)$

Note: Award A1 for the form a + $t$ b, A1 for the form $L$ = a + $t$ b, A0 for the form r = b + $t$ a.

[2 marks]

Examiners report

[N/A]

Syllabus sections

Topic 2—Functions » SL 2.10—Solving equations graphically and analytically

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