Rotations, reflections and dilations | Year 11

01 / LEARN

Understand the idea.

Test a standard transformation

Use coordinate rules to recognise standard maps about the origin. Reflection in the $x$ -axis sends $(x, y)$ to $(x, - y)$ ; reflection in $y = x$ sends it to $(y, x)$ . A point on the mirror line must remain fixed.

A dilation of factor $k$ about the origin has matrix $k I$ : every position vector is multiplied by $k$ . For a rotation about another centre, first translate that centre to the origin, rotate, and translate back.

A counterclockwise rotation

For a counterclockwise angle $θ$ , the unit-vector images are ${(cos θ, sin θ)}^{T}$ and ${(- sin θ, cos θ)}^{T}$ . Placing these in columns gives the rotation matrix below. A negative angle describes a clockwise rotation.

Check the direction using the positive $x$ -axis: a $90^{\circ}$ counterclockwise rotation sends $(1, 0)$ to $(0, 1)$ . Thus a quarter-turn has rule $(x, y) \mapsto (- y, x)$ ; a sketch should agree with the computed coordinates.

R_{θ} = (\begin{array}{cccccccccc} cos θ & - sin θ \\ sin θ & cos θ \end{array})

One-unit square grid from negative five to five on both axes. The unfilled gold triangle has vertices zero, zero; two, zero; one, two. Its unfilled sage image under a counterclockwise quarter-turn has vertices zero, zero; zero, two; negative two, one. Both boundaries are closed with vertex markers.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A quarter-turn

Rotate $(2, - 3)$ through $90^{\circ}$ counterclockwise.

Worked solution

1
Substitute the quarter-turn angle.
$cos 90^{\circ} = 0, sin 90^{\circ} = 1$
The positive $x$ -axis turns to the positive $y$ -axis.
2
Write the counterclockwise rotation matrix.
$R = (\begin{array}{cccccccccc} 0 & - 1 \\ 1 & 0 \end{array})$
The columns are the rotated images of the two unit vectors.
3
Apply it to the point.
$\begin{aligned} R (\begin{array}{cccccccccc} 2 \\ - 3 \end{array}) & = (\begin{array}{cccccccccc} 0 (2) - (- 3) \\ 1 (2) + 0 (- 3) \end{array}) \\ = (\begin{array}{cccccccccc} 3 \\ 2 \end{array}) \end{aligned}$
The coordinate rule is $(- y, x)$ , so the negative original height becomes positive horizontal position.

Answer

$(3, 2)$ .

EXAMPLE 02

A reflection and a dilation

Reflect $(1, 4)$ in $y = x$ , then dilate by factor 2.

Worked solution

1
Reflect by swapping coordinates.
$F = (\begin{array}{cccccccccc} 0 & 1 \\ 1 & 0 \end{array}), F (\begin{array}{cccccccccc} 1 \\ 4 \end{array}) = (\begin{array}{cccccccccc} 4 \\ 1 \end{array})$
Reflection in $y = x$ exchanges horizontal and vertical coordinates.
2
Dilate the reflected point by 2.
$D = (\begin{array}{cccccccccc} 2 & 0 \\ 0 & 2 \end{array}), D (\begin{array}{cccccccccc} 4 \\ 1 \end{array}) = (\begin{array}{cccccccccc} 8 \\ 2 \end{array})$
Dilation about the origin multiplies both components by the same factor.
3
Combine the matrices in action order.
$\begin{aligned} D F & = (\begin{array}{cccccccccc} 2 & 0 \\ 0 & 2 \end{array}) (\begin{array}{cccccccccc} 0 & 1 \\ 1 & 0 \end{array}) \\ = (\begin{array}{cccccccccc} 0 & 2 \\ 2 & 0 \end{array}) \end{aligned}$
The reflection acts first, so its factor is rightmost.
4
Check the combined map.
$(\begin{array}{cccccccccc} 0 & 2 \\ 2 & 0 \end{array}) (\begin{array}{cccccccccc} 1 \\ 4 \end{array}) = (\begin{array}{cccccccccc} 8 \\ 2 \end{array})$
The single product agrees with the two successive transformations.

Answer

$(8, 2)$ ; combined matrix $(\begin{array}{cccccccccc} 0 & 2 \\ 2 & 0 \end{array})$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Give the matrix for a $180^{\circ}$ rotation.

Show answer & reasoning

$(\begin{array}{cccccccccc} - 1 & 0 \\ 0 & - 1 \end{array})$ .

A $180^{\circ}$ turn sends both unit vectors to their negatives. Its columns are ${(- 1, 0)}^{T}$ and ${(0, - 1)}^{T}$ , giving $- I$ and the rule $(x, y) \mapsto (- x, - y)$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Give the matrix for reflection in the $y$ -axis.
2 marks
Rotate $(1, 2)$ through $90^{\circ}$ clockwise.
2 marks
Give the matrix for dilation of factor $- 2$ .
2 marks
Reflect $(3, - 1)$ in $y = x$ .
2 marks
Rotate $(2, 0)$ through $60^{\circ}$ counterclockwise using exact values.
2 marks
Describe the transformation $(\begin{array}{cccccccccc} 0 & - 1 \\ - 1 & 0 \end{array})$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Give the matrix for reflection in the $y$ -axis.
2 marks
Change the $x$ sign but keep $y$ .
$(\begin{array}{cccccccccc} - 1 & 0 \\ 0 & 1 \end{array})$ .
Rotate $(1, 2)$ through $90^{\circ}$ clockwise.
2 marks
Use $(x, y)$ to $(y, - x)$ .
$(2, - 1)$ .
Give the matrix for dilation of factor $- 2$ .
2 marks
Multiply both coordinates by $- 2$ .
$(\begin{array}{cccccccccc} - 2 & 0 \\ 0 & - 2 \end{array})$ .
Reflect $(3, - 1)$ in $y = x$ .
2 marks
Swap the coordinates.
$(- 1, 3)$ .
Rotate $(2, 0)$ through $60^{\circ}$ counterclockwise using exact values.
2 marks
Use $2 cos 60^{\circ} = 1$ and $2 sin 60^{\circ} = \sqrt{3}$ .
$(1, \sqrt{3})$ .
Describe the transformation $(\begin{array}{cccccccccc} 0 & - 1 \\ - 1 & 0 \end{array})$ .
2 marks
It sends $(x, y)$ to $(- y, - x)$ .
Every point $(t, - t)$ on that mirror line is fixed.
Reflection in $y = - x$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: matrices and transformations; original lessons follow the approved school-sized chapter structure.

Official curriculum source ↗

In this lesson

Understand the idea.

Test a standard transformation

A counterclockwise rotation

See the method in action.

A quarter-turn

Substitute the quarter-turn angle.

Write the counterclockwise rotation matrix.

Apply it to the point.

A reflection and a dilation

Reflect by swapping coordinates.

Dilate the reflected point by 2.

Combine the matrices in action order.

Check the combined map.

Try it yourself.

Check your understanding

Keep practising