Transformations — translation, reflection, rotation, enlargement (basic): Question bank

A word you'll need throughout: a transformation is isometric (also called rigid) if it keeps every distance the same — the shape's size and side-lengths are untouched. If instead it changes distances, it is not an isometry. Keep that one idea in your pocket; half the traps below turn on it.

The pictures behind the traps

Before the questions, look at the four moves in one glance. In every panel, the faint shape is the original and the coloured shape is the image — read the arrows to see how each point travelled.

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

Notice the tell-tale signs you'll be quizzed on: the translation and rotation and reflection images are all the same size (isometries), while the enlargement image is bigger. And the reflection image is back-to-front — its corner-labels go the other way round — which is the visual meaning of "orientation reversed."

Read the figure: the horizontal run $x$ becomes the new vertical rise (so the new $y^{'} = x$ ), and the vertical rise $y$ becomes a new leftward run (so the new $x^{'} = - y$ ). That is precisely $(x, y) \to (- y, x)$ . Do the same swing the other way (clockwise) and you get $(x, y) \to (y, - x)$ — the horizontal run drops to become downward, the vertical rise swings right. No formula needed; the corner does the bookkeeping.

Read the reflection figure: trace the corners of the original in the order $A \to B \to C$ and note whether you turn clockwise or anticlockwise. Now trace $A^{'} \to B^{'} \to C^{'}$ on the image — the turning direction is reversed. That reversal is what "swaps handedness" means; a mirror can't be undone by any amount of sliding or spinning.

True or false — justify

A translation changes the shape's orientation the way a reflection does.

False. A translation slides every point by the same arrow (see figure s01/s03), so the corner-order stays the same way round — nothing is flipped. Only reflection reverses orientation.

Reflection and rotation are both isometries, so they are basically the same transformation.

False. Both keep distances (isometries), but reflection reverses orientation (mirror-image) while rotation preserves it. A right-handed shape stays right-handed under rotation but becomes left-handed under reflection.

Enlargement with scale factor

k = 1

does nothing at all.

True. Each point maps to

(a, b) + 1 \cdot [(x, y) - (a, b)] = (x, y)

, its own position, so every point is fixed regardless of where the centre is.

Enlargement is an isometry because the shape "looks the same."

False (unless

k = \pm 1

). It preserves shape/angles (the image is similar, not congruent), but distances scale by

∣ k ∣

— you can see the enlargement image is visibly bigger in figure s01 — so it is generally not an isometry.

Reflecting a point across the line

y = x

gives the same result as swapping its coordinates.

True. The map is

(x, y) \to (y, x)

; the line

y = x

treats the two axes symmetrically, so it exchanges the roles of

x

and

y

. Linked to Inverse Functions: reflecting a graph in

y = x

gives its inverse.

Rotating a point by

9 0^{\circ}

clockwise is the same as rotating it by

27 0^{\circ}

anticlockwise.

True. A full turn is

36 0^{\circ}

; going

9 0^{\circ}

one way lands you where

36 0^{\circ} - 9 0^{\circ} = 27 0^{\circ}

the other way does. Both send

(x, y) \to (y, - x)

The area of a shape after enlargement by factor

k

is multiplied by

k

False. Every length scales by

k

, and area is length

\times

length, so area scales by

k^{2}

. A factor-3 enlargement makes area 9 times bigger.

A negative scale factor is meaningless because you can't have a negative size.

False.

k < 0

is valid: the point goes to the opposite side of the centre. Size scales by

∣ k ∣

; the sign flips the figure through the centre (like a

18 0^{\circ}

rotation combined with scaling).

Spot the error

"To reflect

(4, 5)

in the

x

-axis I negate the

x

(- 4, 5)

Wrong axis logic. The

x

-axis is the line

y = 0

; reflection flips the vertical distance to the axis, so

(x, y) \to (x, - y)

, giving

(4, - 5)

. Negating

x

is reflection in the

y

-axis instead.

"Rotating

(3, 1)

9 0^{\circ}

anticlockwise: I just swap to get

(1, 3)

A swap alone is reflection in

y = x

, not rotation. As the swing in figure s02 shows, the run

x

becomes the new rise and the rise

y

becomes a leftward run, so the rule is

(x, y) \to (- y, x) = (- 1, 3)

— swap and negate the new first coordinate.

"Enlarging

(3, 2)

k = 2

just doubles both coordinates to

(6, 4)

Only true if the centre is the origin. From a general centre

(a, b)

you double the displacement from the centre:

(a, b) + 2 [(x, y) - (a, b)]

. With centre

(1, 1)

that gives

(5, 3)

, not

(6, 4)

"For a combined 'reflect then translate', order doesn't matter."

Order usually matters. Reflecting then translating generally gives a different image from translating then reflecting, because the mirror sits in a fixed place and the reflection acts on wherever the point currently is.

18 0^{\circ}

rotation and a reflection give the same picture."

Not generally. A

18 0^{\circ}

rotation

(x, y) \to (- x, - y)

preserves orientation (corner-order unchanged); a reflection reverses it (see figure s03). They coincide only for special symmetric figures, never as transformations in general.

"To rotate about a point that isn't the origin, I can use

(x, y) \to (- y, x)

directly."

That formula rotates about the origin only. For centre

(a, b)

you first subtract the centre, rotate, then add it back — otherwise the whole figure drifts.

"Translating by

(0 0)

is not a real transformation."

It is — it's the identity translation, mapping every point to itself. "Does nothing" still counts as a valid (and useful) transformation.

Why questions

Why does reflection reverse orientation but rotation does not?

A mirror swaps left and right, so a clockwise-labelled corner-loop becomes anticlockwise — exactly what figure s03 shows. Rotation just turns the whole shape as one rigid piece, so the corner-order (its "handedness") is preserved.

Why does the enlargement formula subtract the centre before scaling?

Because scaling must happen relative to the centre, not relative to the origin. Subtracting the centre measures how far the point sticks out from it; we stretch that stick-out by

k

, then re-anchor by adding the centre back.

Why is the distance from the origin unchanged by rotation?

A rotation turns the whole radius-arrow from the origin as a rigid rod (see the equal-length arrows in figure s02). It only changes which direction the rod points, never its length

x^{2} + y^{2}

— and that length is exactly the distance, so the distance is preserved. This is why rotation is an isometry.

Why can every reflection and rotation be written as a matrix times

(x, y)

, but a translation cannot (in

2 \times 2

form)?

Reflection and rotation fix the origin, so

(0, 0) \to (0, 0)

, which a

2 \times 2

matrix always does. Translation moves the origin, and no plain

2 \times 2

matrix can send

(0, 0)

anywhere but itself — you need an added vector.

Why does a

27 0^{\circ}

anticlockwise rotation equal a

9 0^{\circ}

clockwise one?

Direction and angle trade off around a full

36 0^{\circ}

circle. Turning

27 0^{\circ}

one way leaves the same

9 0^{\circ}

gap you'd cover turning the other way, so both end at the same spot:

(x, y) \to (y, - x)

Why does enlargement produce similar (not congruent) figures?

All angles stay equal (shape preserved) but all lengths scale by

k

. Equal angles + proportional sides is exactly the definition of similar; congruent would additionally require the sides to be identical, i.e.

∣ k ∣ = 1

Why does reflecting a function's graph in

y = x

give its inverse?

Reflecting in

y = x

swaps

x

and

y

on every point. Swapping input and output is precisely what taking an inverse does, so the reflected graph is the graph of the inverse relation.

Edge cases

What is the image of a point that lies on the mirror line during a reflection?

It maps to itself. Its perpendicular distance to the mirror is zero, so the "same distance on the other side" is the same point — every point of the mirror line is fixed.

What happens to the centre of enlargement itself under that enlargement?

It stays fixed for any

k

. Its displacement from the centre is

(0, 0)

, and

k \cdot (0, 0) = (0, 0)

, so it never moves — the centre is the one guaranteed fixed point.

What is the image of the origin under a rotation about the origin, for any angle?

Always the origin. The rotation centre is fixed, and here the centre is the origin, so it can't move no matter the angle.

If you enlarge with

k = 0

, where does every point go?

They all collapse onto the centre

(a, b)

, since

(a, b) + 0 \cdot [(x, y) - (a, b)] = (a, b)

. The shape shrinks to a single point — a degenerate, non-invertible "enlargement."

If the mirror line passes through a shape, is the whole shape moved?

Not entirely — points on the line stay put while points off the line swap sides. If the shape is symmetric about that line, the image looks identical to the original even though individual off-line points have moved.

What does a translation do to a point already sitting at the destination of another point?

Nothing special — translation ignores where points sit relative to each other; every point, occupied "spot" or not, shifts by the same vector. Two points can never collide, since both move together.

What is the image of a shape under enlargement with

k = - 1

A point-reflection through the centre: each point lands the same distance away but on the opposite side. This equals a

18 0^{\circ}

rotation about the centre — same size, orientation preserved, flipped through the centre.

Recall Quick self-test

Which two of the four basic transformations are not always isometries? ::: Only enlargement can fail to be an isometry (when $∣ k ∣ \neq = 1$ ); translation, reflection and rotation are always isometries. A transformation with a matrix that fixes the origin but reverses orientation — which is it? ::: A reflection (through a line via the origin).