Transformations — translation, reflection, rotation, enlargement (basic): Exercises

This page is a ladder. Start at L1 (just recognise what's happening) and climb to L5 (design your own transformation to hit a target). Every problem has a full solution hidden inside a collapsible box — try first, then reveal.

Everything here rests on the four rules from the parent note. We restate each rule the moment we use it, so you never need to flip back.

Prerequisites you may want open: Coordinate Geometry, Vectors, Distance Formula, Trigonometry, Similar Triangles.

Level 1 — Recognition

Here you only need to spot which transformation happened and read the rule off a picture.

Exercise 1.1 (L1)

A shape's point $(4, 1)$ became $(4, - 1)$ . Another point $(2, 3)$ became $(2, - 3)$ . Name the transformation.

Recall Solution

In both cases the first number ( $x$ ) stayed the same and the second number ( $y$ ) flipped sign. The rule is $(x, y) \to (x, - y)$ .

" $x$ unchanged" means no left/right movement. " $y$ flips" means each dot jumps to the mirror-opposite side of the horizontal line $y = 0$ — that line is the $x$ -axis.

Answer: reflection in the $x$ -axis.

Look at figure s01: the dots and their images sit like a shape and its puddle-reflection across the horizontal axis.

Figure — Transformations — translation, reflection, rotation, enlargement (basic)

Exercise 1.2 (L1)

Every point of a shape moved so that $(1, 1) \to (4, 3)$ and $(5, 2) \to (8, 4)$ . What single vector describes this?

Recall Solution

A translation vector $(a b)$ means "add $a$ to $x$ , add $b$ to $y$ " — $a$ is the rightward slide, $b$ the upward slide.

From the first point: $x$ went $1 \to 4$ , so $a = 4 - 1 = 3$ . $y$ went $1 \to 3$ , so $b = 3 - 1 = 2$ . Check with the second point: $5 + 3 = 8$ ✓ and $2 + 2 = 4$ ✓. The same slide worked for both, which is exactly what "translation" demands.

Answer: $(32)$ .

Level 2 — Application

Now you apply a stated rule to given points and do the arithmetic carefully.

Exercise 2.1 (L2)

Translate $A (- 2, 5)$ by $(6 - 8)$ .

Recall Solution

Rule: $(x^{'}, y^{'}) = (x + a, y + b)$ with $a = 6, b = - 8$ . $x^{'} = - 2 + 6 = 4, y^{'} = 5 + (- 8) = - 3.$ Answer: $A^{'} (4, - 3)$ .

Exercise 2.2 (L2)

Reflect $B (- 3, 4)$ in the $y$ -axis.

Recall Solution

The $y$ -axis is the vertical line $x = 0$ . Reflection across it keeps height ( $y$ ) but flips left/right: rule $(x, y) \to (- x, y)$ . $B (- 3, 4) \to B^{'} (3, 4) .$ $B$ was $3$ left of the axis; $B^{'}$ is $3$ right. Same height. Answer: $B^{'} (3, 4)$ .

Exercise 2.3 (L2)

Rotate $C (2, 5)$ by 90° anticlockwise about the origin.

Recall Solution

Why a rule and not "just spin it"? Rotation about the origin sends the point on angle $ϕ$ from the positive $x$ -axis to angle $ϕ + 90°$ , keeping the same distance $r = x^{2} + y^{2}$ from the origin. Plugging $θ = 90°$ ( $cos 90° = 0$ , $sin 90° = 1$ ) into the general rotation gives the shortcut $(x, y) \to (- y, x)$ . $C (2, 5) \to C^{'} (- 5, 2) .$ Sanity: $C$ was in quadrant I (right, up); after a quarter-turn anticlockwise it should land in quadrant II (left, up). $(- 5, 2)$ is left and up ✓. Distance check: $2^{2} + 5^{2} = 29$ before, $(- 5)^{2} + 2^{2} = 29$ after ✓. Answer: $C^{'} (- 5, 2)$ .

Exercise 2.4 (L2)

Enlarge $D (4, 6)$ with centre the origin and scale factor $k = \frac{1}{2}$ .

Recall Solution

Centre at origin gives the simplest enlargement rule $(x, y) \to (k x, k y)$ : multiply both coordinates by the same number $k$ . Here $k = \frac{1}{2}$ means "move halfway toward the centre". $x^{'} = \frac{1}{2} \cdot 4 = 2, y^{'} = \frac{1}{2} \cdot 6 = 3.$ Answer: $D^{'} (2, 3)$ . It sits exactly on the line joining the origin to $D$ , halfway along.

Level 3 — Analysis

Now you work backwards, combine ideas, or find the missing rule.

Exercise 3.1 (L3)

A point $P$ is reflected in the line $y = x$ and lands at $(7, - 2)$ . Where was $P$ ?

Recall Solution

Reflection in $y = x$ swaps coordinates: rule $(x, y) \to (y, x)$ . This transformation is its own inverse — doing it twice returns you home (see Inverse Functions). So to undo it, apply the same swap. Swap $(7, - 2)$ : $P = (- 2, 7)$ . Check: $(- 2, 7) \to (7, - 2)$ under $(x, y) \to (y, x)$ ✓.

Exercise 3.2 (L3)

Triangle vertices $E (1, 1), F (4, 1), G (1, 3)$ are enlarged from centre $(0, 0)$ to $E^{'} (3, 3), F^{'} (12, 3), G^{'} (3, 9)$ . Find the scale factor, and hence the ratio of the new area to the old area.

Recall Solution

Every coordinate got multiplied by the same $k$ : $1 \cdot k = 3 \Rightarrow k = 3$ . Confirm with $F$ : $4 \cdot 3 = 12$ ✓. Why area scales by $k^{2}$ : enlargement multiplies every length by $k$ . Area is (length $\times$ width)-shaped, so it multiplies by $k \times k = k^{2}$ (see Similar Triangles). Area ratio $= k^{2} = 3^{2} = 9$ . Direct check: old area of right triangle $= \frac{1}{2} \cdot 3 \cdot 2 = 3$ ; new legs are $9$ and $6$ , area $= \frac{1}{2} \cdot 9 \cdot 6 = 27$ ; ratio $27/3 = 9$ ✓. Answers: $k = 3$ , area ratio $= 9$ .

Exercise 3.3 (L3)

A single transformation sends $(x, y) \to (- x, - y)$ . Describe it two different ways.

Recall Solution

Both coordinates flip sign. Feeding $θ = 180°$ ( $cos 180° = - 1$ , $sin 180° = 0$ ) into rotation gives exactly $(x, y) \to (- x, - y)$ . So it is a rotation of 180° about the origin (clockwise or anticlockwise — a half-turn looks the same either way). It is also an enlargement from the origin with scale factor $k = - 1$ : negative $k$ scales by $∣ k ∣ = 1$ (same size) but sends every point through the centre to the opposite side. Answer: 180° rotation about the origin = enlargement with centre origin, $k = - 1$ .

Level 4 — Synthesis

Chain transformations and track a whole shape.

Exercise 4.1 (L4)

Point $H (3, 2)$ is rotated 90° anticlockwise about the origin, then translated by $(- 1 4)$ . Find the final image $H^{''}$ .

Recall Solution

Step 1 (rotate): rule $(x, y) \to (- y, x)$ . $H (3, 2) \to H^{'} (- 2, 3) .$ What this looks like: the dot swings a quarter-turn anticlockwise, staying $13$ from the origin. Step 2 (translate): rule $(x, y) \to (x - 1, y + 4)$ . $H^{'} (- 2, 3) \to H^{''} (- 3, 7) .$ Answer: $H^{''} (- 3, 7)$ .

Exercise 4.2 (L4)

Square with corners $J (2, 2), K (6, 2), L (6, 6), M (2, 6)$ is enlarged by $k = 2$ from centre $(2, 2) = J$ . Give the four images, and confirm $J$ is a fixed point.

Recall Solution

Rule with centre $(a, b) = (2, 2)$ : $x^{'} = a + k (x - a) = 2 + 2 (x - 2)$ and $y^{'} = 2 + 2 (y - 2)$ .

$J (2, 2)$ : $x^{'} = 2 + 2 (0) = 2$ , $y^{'} = 2 + 2 (0) = 2 \Rightarrow J^{'} (2, 2)$ . The centre never moves — it's the fixed point, because its displacement from itself is zero and $2 \times 0 = 0$ .
$K (6, 2)$ : $x^{'} = 2 + 2 (4) = 10$ , $y^{'} = 2 \Rightarrow K^{'} (10, 2)$ .
$L (6, 6)$ : $x^{'} = 10$ , $y^{'} = 2 + 2 (4) = 10 \Rightarrow L^{'} (10, 10)$ .
$M (2, 6)$ : $x^{'} = 2$ , $y^{'} = 10 \Rightarrow M^{'} (2, 10)$ . Side length went from $4$ to $8$ — doubled, as $k = 2$ demands. Answers: $J^{'} (2, 2), K^{'} (10, 2), L^{'} (10, 10), M^{'} (2, 10)$ ; $J$ is fixed.

Level 5 — Mastery

Design a transformation to meet a target, or reason about invariants.

Exercise 5.1 (L5)

Find the single translation vector that has the same effect on every point as: "translate by $(3 - 1)$ , then by $(- 5 4)$ ".

Recall Solution

Two translations in a row just add: the total slide is the sum of the vectors (this is vector addition, see Vectors). $(3 - 1) + (- 5 4) = (3 + (- 5) - 1 + 4) = (- 2 3) .$ Answer: $(- 2 3)$ . (Test on $(0, 0)$ : first rule $\to (3, - 1)$ , second $\to (- 2, 3)$ ✓.)

Exercise 5.2 (L5)

You want a single rotation about the origin that sends $(1, 0)$ to $(0, - 1)$ . What angle and direction? Then apply that same rotation to $(4, 3)$ .

Recall Solution

The point $(1, 0)$ lies on the positive $x$ -axis (angle $0°$ ). Its image $(0, - 1)$ is on the negative $y$ -axis (angle $- 90°$ , i.e. pointing straight down). So the turn is 90° clockwise (equivalently 270° anticlockwise). The rule for 90° clockwise: $(x, y) \to (y, - x)$ . Check: $(1, 0) \to (0, - 1)$ ✓. Apply to $(4, 3)$ : $(4, 3) \to (3, - 4)$ . Distance preserved: $4^{2} + 3^{2} = 5 = 3^{2} + (- 4)^{2}$ ✓. Answer: 90° clockwise; $(4, 3) \to (3, - 4)$ .

Exercise 5.3 (L5)

A shape is enlarged by $k = 3$ from centre $(1, 0)$ . The point $(1, 0)$ obviously stays fixed. Show that the image of $(2, 0)$ and the image of $(0, 0)$ still lie on the same straight line through the centre — confirming enlargement preserves alignment.

Recall Solution

Rule: $x^{'} = 1 + 3 (x - 1)$ , $y^{'} = 3 y$ .

$(2, 0) \to (1 + 3 (1), 0) = (4, 0)$ .
$(0, 0) \to (1 + 3 (- 1), 0) = (- 2, 0)$ . The centre $(1, 0)$ , plus images $(4, 0)$ and $(- 2, 0)$ , all have $y = 0$ — they lie on the $x$ -axis, one straight line through the centre. Distances from the centre: $∣4 - 1∣ = 3$ and $∣ - 2 - 1∣ = 3$ , each is $3 \times$ the original distances ( $1$ and $1$ ). Alignment is kept and every length scaled by $k = 3$ . Answer: all three points are collinear on $y = 0$ ; distances scaled by $3$ ✓.

Recall Quick self-quiz

Rule for reflection in the $x$ -axis? ::: $(x, y) \to (x, - y)$ Rule for 90° anticlockwise about origin? ::: $(x, y) \to (- y, x)$ How does area scale under enlargement factor $k$ ? ::: by $k^{2}$ Reverse of "translate by $v$ "? ::: translate by $- v$ Which reflection is a pure coordinate swap $(x, y) \to (y, x)$ ? ::: reflection in the line $y = x$