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Date	May 2022	Marks available	4	Reference code	22M.1.AHL.TZ2.15
Level	Additional Higher Level	Paper	Paper 1	Time zone	Time zone 2
Command term	Show that	Question number	15	Adapted from	N/A

Question

The equation of the line $y = m x + c$ can be expressed in vector form $r = a + λ b$ .

The matrix $M$ is defined by $(\begin{matrix} 6 & 3 \\ 4 & 2 \end{matrix})$ .

The line $y = m x + c$ (where $m \neq - 2$ ) is transformed into a new line using the transformation described by matrix $M$ .

Find the vectors $a$ and $b$ in terms of $m$ and/or $c$ .

[2]

Find the value of $det M$ .

[1]

Show that the equation of the resulting line does not depend on $m$ or $c$ .

[4]

Markscheme

(one vector to the line is $(\begin{matrix} 0 \\ c \end{matrix})$ therefore) $a = (\begin{matrix} 0 \\ c \end{matrix})$ A1

the line goes $m$ up for every $1$ across

(so the direction vector is) $b = (\begin{matrix} 1 \\ m \end{matrix})$ A1

Note: Although these are the most likely answers, many others are possible.

[2 marks]

(from GDC OR $6 \times 2 - 4 \times 3$ ) $|M| = 0$ A1

[1 mark]

METHOD 1

$(\begin{matrix} X \\ Y \end{matrix}) = (\begin{matrix} 6 & 3 \\ 4 & 2 \end{matrix}) (\begin{matrix} x \\ m x + c \end{matrix}) = (\begin{matrix} 6 x + 3 m x + 3 c \\ 4 x + 2 m x + 2 c \end{matrix})$ M1A1

$= (\begin{matrix} 3 (2 x + m x + c) \\ 2 (2 x + m x + c) \end{matrix})$ A1

therefore the new line has equation $3 Y = 2 X$ A1

which is independent of $m$ or $c$ AG

Note: The AG line (or equivalent) must be seen for the final A1 line to be awarded.

METHOD 2

take two points on the line, e.g $(0, c)$ and $(1, m + c)$ M1

these map to $(\begin{matrix} 6 & 3 \\ 4 & 2 \end{matrix}) (\begin{matrix} 0 \\ c \end{matrix}) = (\begin{matrix} 3 c \\ 2 c \end{matrix})$

and $(\begin{matrix} 6 & 3 \\ 4 & 2 \end{matrix}) (\begin{matrix} 1 \\ m + c \end{matrix}) = (\begin{matrix} 6 + 3 m + 3 c \\ 4 + 2 m + 2 c \end{matrix})$ A1

therefore a direction vector is $(\begin{matrix} 6 + 3 m \\ 4 + 2 m \end{matrix}) = (2 + m) (\begin{matrix} 3 \\ 2 \end{matrix})$

(since $m \neq - 2$ ) a direction vector is $(\begin{matrix} 3 \\ 2 \end{matrix})$

the line passes through $(\begin{matrix} 3 c \\ 2 c \end{matrix}) - c (\begin{matrix} 3 \\ 2 \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$ therefore it always has the origin as a jump-on vector A1

the vector equation is therefore $r = μ (\begin{matrix} 3 \\ 2 \end{matrix})$ A1

which is independent of $m$ or $c$ AG

Note: The AG line (or equivalent) must be seen for the final A1 line to be awarded.

METHOD 3

$r = (\begin{matrix} 6 & 3 \\ 4 & 2 \end{matrix}) ((\begin{matrix} 0 \\ c \end{matrix}) + λ (\begin{matrix} 1 \\ m \end{matrix})) = (\begin{matrix} 3 c \\ 2 c \end{matrix}) + λ (\begin{matrix} 6 + 3 m \\ 4 + 2 m \end{matrix})$ M1A1

$= c (\begin{matrix} 3 \\ 2 \end{matrix}) + (2 + m) λ (\begin{matrix} 3 \\ 2 \end{matrix})$ A1

$= μ (\begin{matrix} 3 \\ 2 \end{matrix})$

where $μ = c + (2 + m) λ$ is an arbitrary parameter. A1

which is independent of $m$ or $c$ (as $μ$ can take any value) AG

Note: The AG line (or equivalent) must be seen for the final A1 line to be awarded.

[4 marks]

Examiners report

In part (a), most candidates were unable to convert the Cartesian equation of a line into its vector form. In part (b), almost every candidate showed that the value of the determinant was zero. In part (c), the great majority of candidates failed to come up with any sort of strategy to solve the problem.

[N/A]

Syllabus sections

Topic 3—Geometry and trigonometry » AHL 3.9—Matrix transformations

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21N.2.AHL.TZ0.4a.iv:
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EXM.2.AHL.TZ0.15a:
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EXM.2.AHL.TZ0.16a:
Describe fully the geometrical transformation represented by B.
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EXM.2.AHL.TZ0.16d:
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EXN.3.AHL.TZ0.2c.i:
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