Chapter 8: Transformations and Vectors

Slide, Flip, & Point Your Way to Victory 🏆

Introduction

1. Introduction

Ever wonder how your character moves so smoothly across the screen in a game, or how a Snapchat filter perfectly scales to your face? It's all about transformations and vectors! In this chapter, we're ditching the abstract and getting into the code behind the graphics. First, we'll master transformations - the art of sliding, flipping, rotating, and resizing shapes on a grid. Then, we'll jump into vectors, which are basically instructions for movement, like giving your mate directions with both distance and direction. You'll learn the rules of vector arithmetic (adding and subtracting them), and finally, we'll put it all together to solve some tricky vector geometry problems, proving you're a strategic thinker. Let's get this level cleared!

\begin{pmatrix} x \\ y \end{pmatrix}

Sometimes you'll do a combination, like a reflection followed by a translation. Just take it one step at a time, and you'll nail it!

Worked example

Let's Get This Triangle Transformed 🚀

T

Answer

T''

\mathbf{a} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}

|\mathbf{a}|

Worked example

Let's Get This Bread (Mathematically) 🍞

\mathbf{p} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}

1
$2\mathbf{q}$
$2\mathbf{q} = 2 \times \begin{pmatrix} -2 \\ 6 \end{pmatrix} = \begin{pmatrix} 2 \times -2 \\ 2 \times 6 \end{pmatrix} = \begin{pmatrix} -4 \\ 12 \end{pmatrix}$
2
$2\mathbf{q}$
$\mathbf{p} + 2\mathbf{q} = \begin{pmatrix} 3 \\ -4 \end{pmatrix} + \begin{pmatrix} -4 \\ 12 \end{pmatrix} = \begin{pmatrix} 3 + (-4) \\ -4 + 12 \end{pmatrix} = \begin{pmatrix} -1 \\ 8 \end{pmatrix}$
3
$\mathbf{p} - \mathbf{q}$
$\mathbf{p} - \mathbf{q} = \begin{pmatrix} 3 \\ -4 \end{pmatrix} - \begin{pmatrix} -2 \\ 6 \end{pmatrix} = \begin{pmatrix} 3 - (-2) \\ -4 - 6 \end{pmatrix} = \begin{pmatrix} 5 \\ -10 \end{pmatrix}$
4
$|\mathbf{v}| = \sqrt{x^2 + y^2}$
$|\mathbf{p} - \mathbf{q}| = \sqrt{(5)^2 + (-10)^2} = \sqrt{25 + 100} = \sqrt{125}$
5
$Don't forget to simplify the square root if you can, or give the decimal answer to 3 significant figures as is standard in IGCSE. Let's give both!$
$\sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5} \approx 11.2 \text{ (3 s.f.)}$

Answer

\sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5} \approx 11.2 \text{ (3 s.f.)}

\vec{OA}

\vec{PQ} = k \times \vec{RS}

Worked example

Let's Solve This Geometry Puzzle 🕵️‍♀️

\vec{OA} = \mathbf{a}

1
$\frac{2}{3}$
$\vec{AC} = \vec{OC} - \vec{OA} = \mathbf{c} - \mathbf{a} \\ \vec{OM} = \vec{OA} + \frac{2}{3}\vec{AC} = \mathbf{a} + \frac{2}{3}(\mathbf{c} - \mathbf{a}) \\ \vec{OM} = \mathbf{a} + \frac{2}{3}\mathbf{c} - \frac{2}{3}\mathbf{a} = \frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{c}$
2
$\mathbf{a}$
$\vec{CB} = \vec{OA} = \mathbf{a} \\ \vec{ON} = \vec{OC} + \vec{CN} = \vec{OC} + \frac{1}{2}\vec{CB} \\ \vec{ON} = \mathbf{c} + \frac{1}{2}\mathbf{a}$
3
$\vec{OM}$
$\vec{OM} = \frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{c} = \frac{2}{3} (\frac{1}{2}\mathbf{a} + \mathbf{c})$
4
$\vec{ON}$
$\text{Since } \vec{ON} = \frac{1}{2}\mathbf{a} + \mathbf{c}, \text{ we have:} \\ \vec{OM} = \frac{2}{3}\vec{ON}$
5
$\vec{OM} = k\vec{ON}$
$\text{Since } \vec{OM} \text{ is a scalar multiple of } \vec{ON} \text{ and both vectors share a common point O, the points O, M, and N are collinear.}$

Answer

\text{Since } \vec{OM} \text{ is a scalar multiple of } \vec{ON} \text{ and both vectors share a common point O, the points O, M, and N are collinear.}

Worked example

T

1
$x$
$A(2,1) \to A'(2,-1)$
2
$\binom{-3}{4}$
$A'(2,-1) \to A''(-1, 3)$
3
$T$
$T'': \ A''(-1,3), \ B''(2,3), \ C''(2,1)$
4
$Key exam takeaway: for any 'describe the combination' question, apply the transformations in order first, then compare original and final shapes to work out what single move matches — usually a reflection, rotation or glide reflection. Don't guess — check all three vertices.$
$\text{State vertices of } T'' \text{; describe the single transformation by testing candidates.}$

Answer

\text{State vertices of } T'' \text{; describe the single transformation by testing candidates.}

Worked example

T

Worked example

\overrightarrow{AB} = \binom{-8}{6}

Worked example

P

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