21M.1.AHL.TZ2.14

pestleMathematicsAIHLPaper 121M· ahl-3-9-matrix-transformationssource ↗

A geometric transformation $T : (\begin{matrix} x \\ y \end{matrix}) \mapsto (\begin{matrix} x' \\ y' \end{matrix})$ is defined by

$T : (\begin{matrix} x' \\ y' \end{matrix}) = (\begin{matrix} 7 & - 10 \\ 2 & - 3 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) + (\begin{matrix} - 5 \\ 4 \end{matrix})$ .

Find the coordinates of the image of the point $(6, - 2)$ .

[2]

Given that $T : (\begin{matrix} p \\ q \end{matrix}) \mapsto 2 (\begin{matrix} p \\ q \end{matrix})$ , find the value of $p$ and the value of $q$ .

[3]

A triangle $L$ with vertices lying on the $x y$ plane is transformed by $T$ .

Explain why both $L$ and its image will have exactly the same area.

[2]

Markscheme / solution

$(\begin{matrix} 7 & - 10 \\ 2 & - 3 \end{matrix}) (\begin{matrix} 6 \\ - 2 \end{matrix}) + (\begin{matrix} - 5 \\ 4 \end{matrix})$ (M1)

$= (\begin{matrix} 57 \\ 22 \end{matrix})$ OR $(57, 22)$ A1

[2 marks]

$(\begin{matrix} 2 p \\ 2 q \end{matrix}) = (\begin{matrix} 7 & - 10 \\ 2 & - 3 \end{matrix}) (\begin{matrix} p \\ q \end{matrix}) + (\begin{matrix} - 5 \\ 4 \end{matrix})$ (M1)

$7 p - 10 q - 5 = 2 p$

$2 p - 3 q + 4 = 2 q$ (A1)

solve simultaneously:

$p = 13, q = 6$ A1

Note: Award A0 if $13$ and $6$ are not labelled or are labelled the other way around.

[3 marks]

$det (\begin{matrix} 7 & - 10 \\ 2 & - 3 \end{matrix}) = - 1 (OR |det (\begin{matrix} 7 & - 10 \\ 2 & - 3 \end{matrix})| = 1)$ A1

scale factor of image area is therefore $(|- 1| =) 1$ (and the translation does not affect the area) A1

[2 marks]

Examiners’ report

[N/A]