Transformations

Take each corner of the triangle in turn.

Count how many squares it is from the mirror line, then plot the new corner the same distance on the other side.

Join the new corners to make the image.

Take each corner of the shape in turn and count its distance to the mirror line.

Plot the image corner the same distance on the other side, then join the corners up.

(a) The horizontal edge of A becomes the vertical edge of B, so the mirror line is diagonal. Draw the line that is the same distance from each corner of A as from the matching corner of B.

(b) Reflecting in the $x$-axis flips the shape down, so the image should be below the $x$-axis (between $y = -2$ and $y = -4$). Alex flipped it across the $y$-axis instead.

Take each corner of A and count its distance to the mirror line.

Plot the image point the same distance on the other side, then join the points.

Take each corner in turn. Count how many squares it is from the mirror line, then plot the image the same distance on the other side.

The corner nearest the line (2 squares away) stays 2 squares away; the far edge (4 squares away) goes 4 squares to the left. Join the new corners.

(a) Each vertex keeps its $x$ value and moves to the same distance on the other side of $y = 1$:

$(1,2)\to(1,0)$, $(1,3)\to(1,-1)$, $(3,4)\to(3,-2)$, $(4,2)\to(4,0)$

(b) Every point on the line has $y$-coordinate 1, so it is $y = 1$.

(a) A 180° rotation about $(0,0)$ sends $(x, y)$ to $(-x, -y)$.

$(1,4)\to(-1,-4)$, $(5,4)\to(-5,-4)$, $(4,2)\to(-4,-2)$, $(2,2)\to(-2,-2)$. Join these points.

(b) $x = 3$ is a vertical line through 3 on the $x$-axis. Mike's image is below A, so he reflected in the horizontal line $y = 3$.

The triangle has base 2 squares and height 3 squares, with its top above the middle of the base.

Multiply every length by 3: base $2 \times 3 = 6$, height $3 \times 3 = 9$.

Draw a base of 6 squares and put the top 9 squares above its midpoint.

(a)(i) The top side of A is 3 squares; in B it is 6 squares. $6 \div 3 = 2$.

(a)(ii) Draw lines through matching corners of B and A and extend them; they meet at the centre.

(b) Corner $(1, 4)$ of C maps to $(4, 1)$ of D: the coordinates swap. That is a reflection in $y = x$.

Count the length of each side of the shape in squares, including how far across and up each sloping side goes.

Multiply every horizontal and vertical step by 3.

For example, a side 2 squares long becomes 6 squares long, and a slope of 3 across, 1 up becomes 9 across, 3 up.

Each point moves to the same distance on the other side of $y = 3$.

$(1, 4)$ is 1 above, so it goes to $(1, 2)$. $(2, 4)$ goes to $(2, 2)$. $(1, 7)$ is 4 above, so it goes to $(1, -1)$.

Each point of A has the same $x$-coordinate in B, with the $y$-coordinate's sign changed, e.g. $(1, 1) \to (1, -1)$.

So it is a **reflection in the $x$-axis**.

Shape B is a mirror image of A, the same distance from the $y$-axis on the other side.

So it is a reflection in the $y$-axis ($x = 0$).

Rotating 90° clockwise about $(0,0)$ sends $(x, y)$ to $(y, -x)$.

$(1,2) \to (2,-1)$

$(1,4) \to (4,-1)$

$(2,2) \to (2,-2)$

Plot and join these three points.

Corner $(-3, 1)$ of A moves to $(2, -3)$ on B.

That is 5 right and 4 down, with no turning or flipping.

Translation by $\begin{pmatrix}5\\-4\end{pmatrix}$

The triangle is bigger, so it is an enlargement.

Sides of A double in B (e.g. height 2 becomes 4), so the scale factor is 2.

Lines through matching corners, e.g. $(3, 1)$ to $(6, 2)$, meet at the origin, so the centre is $(0, 0)$.

The shape has turned, so it is a rotation.

The vertex $(2, -2)$ goes to $(2, 2)$, $(5, -2)$ to $(2, 5)$ and $(2, -4)$ to $(4, 2)$: each point $(x, y)$ becomes $(-y, x)$.

That is a rotation of $90°$ anticlockwise about the origin $(0, 0)$.

The shape is the same size and way round, so it is a translation.

Pick a corner: $(1, -5)$ on S moves to $(-4, 1)$ on T.

That is 5 left and 6 up: translation by $\begin{pmatrix}-5\\6\end{pmatrix}$.

The shape has vertices $(-4,-2)$, $(-1,-2)$, $(-1,-4)$, $(-2,-4)$.

A 90° anticlockwise turn about O sends $(x, y)$ to $(-y, x)$.

New vertices: $(2,-4)$, $(2,-1)$, $(4,-1)$, $(4,-2)$.

The shape keeps the same angles but gets bigger, so it is an enlargement.

AB is 1 unit and AD is 4 units, so the scale factor is 4.

Point A stays still at $(1, 1)$, so that is the centre.

Enlargement, scale factor 4, centre $(1, 1)$.

Corners of A: $(1,5), (3,5), (1,2)$. Corners of B: $(-5,-1), (-5,-3), (-2,-1)$.

Each point $(a, b)$ goes to $(-b, -a)$, which is a reflection.

The mirror line is $y = -x$, so the answer is **reflection in the line $y = -x$**.

Triangle B is triangle A turned upside down, so it is a half-turn.

A vertex at $(-5, 5)$ maps to $(3, -5)$. The centre is the midpoint: $\left(\frac{-5+3}{2}, \frac{5-5}{2}\right) = (-1, 0)$.

Rotation $180^\circ$ about $(-1, 0)$.

Reflect A in the $x$-axis: $(3, 2) \to (3, -2)$, $(5, 2) \to (5, -2)$, $(3, 5) \to (3, -5)$

B has matching corners at $(-3, -3)$, $(-1, -3)$, $(-3, -6)$.

Move from $(3, -2)$ to $(-3, -3)$: 6 left and 1 down.

So $c = -6$ and $d = -1$

A 180° turn about $(1, 0)$ sends $(x, y)$ to $(2 - x, -y)$.

$(-3, 5) \to (5, -5)$, $(-2, 5) \to (4, -5)$, $(-2, 3) \to (4, -3)$, $(-3, 4) \to (5, -4)$

Plot and join these points.

B's sides are $\frac{2}{3}$ of A's (height 6 → 4, width 3 → 2), so it is an enlargement with scale factor $\frac{2}{3}$.

Join matching corners, e.g. $(5, 7)$ to $(2, 4)$ and $(5, 1)$ to $(2, 0)$, and extend the lines: they meet at $(-4, -2)$.

Q's top edge is 3 units, P's is 9 units, so the scale factor is $\frac{3}{9} = \frac{1}{3}$.

Draw lines through matching corners, e.g. $(6, 14)$ to $(2, 6)$ and $(15, 14)$ to $(5, 6)$: they meet at $(0, 2)$.

Enlargement, scale factor $\frac{1}{3}$, centre $(0, 2)$.

Reflecting in two parallel lines gives a translation.

The distance moved is twice the gap between the lines: $2 \times (6 - 2) = 8$ units to the right.

Translation by $\begin{pmatrix} 8 \\ 0 \end{pmatrix}$

(a) P has vertices $(1,2), (1,5), (2,5)$.

Rotating $180^\circ$ about O gives Q: $(-1,-2), (-1,-5), (-2,-5)$.

Translating by $\binom{5}{-2}$ gives R: $(4,-4), (4,-7), (3,-7)$.

R is upside down relative to P, so it is a rotation of $180^\circ$. The centre is the midpoint of $(1,2)$ and its image $(4,-4)$: $(2.5, -1)$.

(b) The centre of rotation does not move, so $A = (2.5, -1)$.

Centre $(0, 1)$. Triangle A has vertices $(1, 1)$, $(1, 3)$ and $(2, 3)$.

From the centre, these are $(1, 0)$, $(1, 2)$ and $(2, 2)$ away. Multiply by 2.5: $(2.5, 0)$, $(2.5, 5)$, $(5, 5)$.

Add back to the centre: $(2.5, 1)$, $(2.5, 6)$, $(5, 6)$.

Reflect T in $x = -1$ to get R, then reflect R in $y = -2$ to get S.

Two reflections in perpendicular lines make a half turn about the point where the lines cross.

The lines meet at $(-1, -2)$, so T to S is a rotation of $180^\circ$ about $(-1, -2)$.

Measure each vertex from the centre $(0, 1)$ and multiply by $\frac{1}{3}$.

$(3, 4)$ is $(3, 3)$ from the centre → $(1, 1)$ → image $(1, 2)$

$(3, 10)$ → $(1, 3)$ → $(1, 4)$; $(6, 10)$ → $(2, 3)$ → $(2, 4)$; $(6, 7)$ → $(2, 2)$ → $(2, 3)$

P has vertices $(-1, 3), (3, 3), (3, 5)$.

Reflect in $y = -x$: Q is $(-3, 1), (-3, -3), (-5, -3)$.

Reflect in $x = -1$: R is $(1, 1), (1, -3), (3, -3)$.

Turning R through $90°$ anticlockwise about $(-1, 1)$ lands it exactly on P.

A has vertices $(-3, 2), (-2, 2), (-2, 4)$.

Translate by $\binom{6}{-4}$: B is $(3, -2), (4, -2), (4, 0)$.

Rotate B $90^\circ$ clockwise about $(1, 2)$: C is $(-3, 0), (-3, -1), (-1, -1)$.

Comparing A and C, C is A turned $90^\circ$ clockwise; tracing paper or perpendicular bisectors give the centre $(-4, 1)$.

Single transformation: **rotation $90^\circ$ clockwise about $(-4, 1)$**

T has vertices $(1, 2)$, $(1, 3)$ and $(3, 2)$.

Multiply each coordinate by $-2$: $(-2, -4)$, $(-2, -6)$, $(-6, -4)$.

Plot and join these points.

The shape's vertices are $(-6, -4)$, $(-5, -2)$, $(-3, -2)$, $(-2, -4)$.

Multiply each coordinate by $-2$: $(12, 8)$, $(10, 4)$, $(6, 4)$, $(4, 8)$.

The image is twice as big, upside down, on the opposite side of the origin.

B is upside down and on the opposite side, and its sides are longer: A's short sides are 2, B's are 3. So it is an enlargement with scale factor $-\frac{3}{2} = -1.5$.

Join matching corners, e.g. $(5, 5)$ to $(-5, -5)$ and $(3, 3)$ to $(-2, -2)$. The lines cross at $(1, 1)$.

Enlargement, scale factor $-1.5$, centre $(1, 1)$.

B is smaller and upside down compared with A, so it is an enlargement with a negative scale factor.

A's side of length 6 becomes 2, so the scale factor is $-\frac{1}{3}$.

Join matching corners, e.g. $(-4,-4)$ to $(4,4)$ and $(2,-4)$ to $(2,4)$: the lines cross at $(2, 2)$.

Enlargement, scale factor $-\frac{1}{3}$, centre $(2, 2)$.

Triangle T has vertices $(-6, -2)$, $(-7, -5)$, $(-5, -5)$.

From the centre $(-2, -2)$ each vertex is $(-4, 0)$, $(-5, -3)$, $(-3, -3)$ away.

Multiply by $-2$: $(8, 0)$, $(10, 6)$, $(6, 6)$.

Add back to the centre: $(6, -2)$, $(8, 4)$, $(4, 4)$. The image is upside down and twice the size.

(a) $P$ has vertices $(-8, 1)$, $(-8, 5)$, $(-6, 1)$. From the centre $(-2, -1)$ these are $(-6, 2)$, $(-6, 6)$, $(-4, 2)$ away.

Multiply by $-1.5$: $(9, -3)$, $(9, -9)$, $(6, -3)$. Add back to $(-2, -1)$: $Q$ is $(7, -4)$, $(7, -10)$, $(4, -4)$.

(b) Rotating $90°$ anticlockwise about $O$ sends $(x, y)$ to $(-y, x)$, so $(-8, 1) \to (-1, -8)$.

To send it back to $(-8, 1)$, translate by $\begin{pmatrix} -7 \\ 9 \end{pmatrix}$.

Triangle A has corners $(-4, 1)$, $(-3, 1)$, $(-3.5, 4)$.

Rotating 180° about $(-2, 0)$ sends $(x, y)$ to $(-4 - x, -y)$. Then translating by $\binom{-3}{2}$ gives $(-7 - x, 2 - y)$.

The corners go to $(-3, 1)$, $(-4, 1)$, $(-3.5, -2)$. The base maps onto itself reversed, so its midpoint stays put.

Check: $-7 - (-3.5) = -3.5$ and $2 - 1 = 1$, so the point is $(-3.5, 1)$.

Reflecting in $x = -1$ sends $A(1, 4) \to (-3, 4)$, $B(3, 4) \to (-5, 4)$, $C(3, 2) \to (-5, 2)$, $D(1, 2) \to (-3, 2)$.

To bring $B$ and $D$ back to where they started, rotate 90° clockwise about $(-1, 0)$: $(-5, 4) \to (3, 4)$ and $(-3, 2) \to (1, 2)$.

So B and D are invariant.