Translations, rotations, reflections and symmetry

A transformation is a rule for positions, not a change of labels

Start with the point (2, −1). Moving it three units right and two units up gives (5, 1). Apply that same move to every point of a triangle: its shape and size stay the same while its position changes. A transformation describes such a move by one rule for every starting point.

Work in the Euclidean coordinate plane with perpendicular axes, a common length unit, x increasing to the right and y increasing upwards. A transformation assigns an image T(P) to every point P. If P is labelled A, its image may be written A′; the prime records correspondence, not a different rule. To transform a figure F, apply the same rule to every point of F. Its image T(F) is the set of all those images.

Translations, rotations and reflections are rigid transformations: they preserve the distance between every pair of points and have inverses. They move a figure without changing its shape or size. A reflection can reverse its orientation and still preserve congruence. A stretched drawing or a map that collapses the plane onto one line is a transformation, but it does not have these distance-preserving properties.

Specify the figure as well as the rule. A polygon boundary consists of its vertices and every point on the consecutive side segments, including the closing side. A circle is the complete set of points at a positive fixed distance from its centre; four marked points are only samples. If the figure also includes a filled region, those interior points must be transformed too. Decoration or labels count as part of the figure only when the problem includes them.

A translation adds the same directed change everywhere

For the move just described, (2, −1) becomes (2 + 3, −1 + 2) = (5, 1). A point at (0, 0) would become (3, 2), using the same changes. The ordered pair (3, 2) is the translation vector: it tells us how far and in which direction each coordinate changes.

A translation by vector v = (a, b) sends every point (x, y) to (x + a, y + b). The first component is the horizontal directed change and the second the vertical directed change, in the chosen length unit. Positive a moves right, negative a moves left, and zero a gives no horizontal change; b works similarly for upwards and downwards movement. Components may be negative, zero, fractional or any real values. They are not proposed positive side lengths.

For any two points, adding the same a and b cancels when subtracting their coordinates. Their horizontal and vertical differences are unchanged, so their squared distance remains Δx² + Δy². The movement distance of each point is √(a² + b²), a nonnegative length. For v = (0, 0), this distance is zero and the translation is the identity: every point stays where it was. No direction needs to be assigned to a zero vector.

The inverse translation uses −v = (−a, −b), because adding and then subtracting the same components restores every coordinate. A nonzero translation has no fixed point: x + a = x and y + b = y would force a = b = 0. This concerns a single point remaining in place; whether an entire figure returns to itself is a separate question.

Subtract the centre before a quarter-turn, then add it back

Place the centre at (1, 1) and start at (3, 1), two units to its right. A quarter-turn counterclockwise sends that radius upwards, so the point lands at (1, 3). The centre stays in place. Choosing the origin instead would turn about a different position and give a different image.

A rotation needs a centre C and a directed angle θ. We take positive angles counterclockwise and negative angles clockwise. The centre stays fixed. Every other point P moves to a point P′ with CP′ = CP, while ray CP turns through θ to ray CP′. A turn of 360° returns the same position as 0°, though the turn records differ. A rotation about a different centre generally gives a different transformation.

For C = (h, k), first express P relative to C: u = x − h and v = y − k. A counterclockwise quarter-turn sends the positive horizontal direction to the positive vertical direction and the positive vertical direction to the negative horizontal direction. The relative coordinate rule is therefore (u, v) ↦ (−v, u). A half-turn gives (−u, −v), and three counterclockwise quarter-turns give (v, −u). Finally add (h, k) to obtain the absolute image coordinates.

These rules use exact perpendicular directions rather than a decimal trigonometric calculation. For a 90° rotation about (1, 1), the point (3, 1) has relative coordinates (2, 0), which become (0, 2), so its image is (1, 3). Applying the origin rule directly would instead give (−1, 3), the image under a different rotation. The centre itself has relative coordinates (0, 0), so every listed rule returns it unchanged.

For quarter-turns, adding or subtracting full turns leaves the point mapping unchanged. In particular, −90° and 270° give the same images. Applying two rotations about the same centre adds their directed angles; this follows by following the turn of each ray while its radius stays unchanged. That statement does not allow the centres of two different rotations to be ignored.

The mirror line fixes its own points and bisects every off-line pair

Use the vertical mirror x = 1. The point (2, 1) lies one unit to its right, so its image is (0, 1), one unit to its left. Their joining segment is horizontal, perpendicular to the mirror, and its midpoint (1, 1) lies on the mirror. These are the features that determine the reflected point.

A reflection is specified by a complete line ℓ. A point P on ℓ remains fixed. For a point off ℓ, drop the perpendicular to ℓ at H and continue through H to the opposite side, placing P′ so that HP′ = HP. Thus H is the midpoint of PP′ and ℓ is its perpendicular bisector. Equal distance from the line alone would allow many points; the perpendicular direction and opposite-side condition determine the image.

For a vertical mirror x = c, only the horizontal coordinate changes: x′ = 2c − x and y′ = y. The midpoint's horizontal coordinate is c, and PP′ is horizontal when P is off the mirror. A horizontal mirror y = c similarly gives (x, 2c − y). These formulas also handle points on the line: they return the original point, and no positive segment or unique perpendicular bisector of a zero segment is claimed.

Reflection across y = x exchanges the two coordinates; across y = −x it exchanges them and reverses both signs. More generally, reflection across y = x + c sends (x, y) to (y − c, x + c). To check this last rule, the midpoint has coordinates ((x + y − c)/2, (x + y + c)/2), whose vertical coordinate exceeds its horizontal coordinate by c. For an off-line point, the joining segment has equal-magnitude, opposite-signed coordinate differences, perpendicular to the mirror's equal-coordinate direction. The midpoint and perpendicular conditions therefore give the specified reflection.

The complete mirror line is the fixed-point set. Reflecting twice in that same line returns every point: an off-line pair is exchanged back, while each on-line point was fixed already. A line drawn at the edge of a diagram is not automatically the mirror. Its equation or geometric specification must identify which complete line is intended.

Pairwise distance preservation explains congruence

The coordinate distance formula gives an exact check for the listed transformations. A translation leaves both coordinate differences unchanged. A quarter-turn swaps those differences and may change their signs, while a listed reflection swaps or changes signs in the corresponding way; any constant offsets cancel. In every case the sum Δx² + Δy² is unchanged. This proof covers every pair of real points, including coincident points and zero components, rather than just a particular triangle's three sides.

The geometric definitions give the same result beyond quarter-turns. A rotation keeps each distance to its centre and the ordinary included angle between any two noncollinear radius rays, so SAS preserves the distance between their endpoints. If the centre and endpoints are collinear, the endpoints stay on equal or opposite rays and their distance remains the difference or sum of the unchanged radii. If an endpoint is the centre, its distance is the other radius. For a reflection, choose perpendicular coordinate axes along the mirror and across it; the along-mirror coordinate stays the same and the across-mirror coordinate changes sign, preserving squared differences.

For a nondegenerate triangle, preserving all three sides gives SSS congruence, so corresponding interior angles are equal. Straight angles and right angles are preserved as well: lines and their incidence map to lines and the same incidence, and the adjacent nondegenerate angles agree. Corresponding segment lengths, polygon perimeters and triangular areas therefore agree. Decomposing a convex polygon into triangles gives equal polygon area. A rotation or translation preserves clockwise versus counterclockwise boundary order; one reflection reverses that order without changing the ordinary angle measures or area.

Congruence allows a different position or orientation. It does not require the image to coincide with the original set of points, and it does not require every point to stay fixed. Keep three claims separate: distances are preserved by the transformation, the original and image figures are congruent, and this transformation is a symmetry of the original figure. The first two hold for all the rigid movements here; the third needs a whole-figure check.

Undo the rule, and distinguish fixed points from repeated positions

An inverse undoes the transformation for every point. The inverse of translation by (a, b) is translation by (−a, −b). The inverse of rotation through θ about C is rotation through −θ about that same C. Reflection in a line is its own inverse. In symbols, T⁻¹(T(P)) = P and T(T⁻¹(P)) = P. These two identities ensure that distinct points have distinct images and that every target point has a preimage.

The identity fixes the whole plane. A nonzero translation fixes no point. For a 90°, 180° or 270° rotation, the only fixed point is its centre: solving the relative coordinate equations in the table gives u = v = 0. This conclusion must not be assigned to a 0° or full-turn rotation, which fixes every point. A reflection fixes every point of its mirror line and no point off it.

A point need not be fixed by one application to return after several. Four 90° rotations about one centre give the identity, so all points return after four applications. Two reflections in one line also give the identity. A square may already coincide with itself after one quarter-turn even though all its vertices move to different vertices. Returning the point set and fixing each labelled point are different requirements.

The second rule acts on the first image

Start at (2, 0). Translate one unit right to (3, 0), then turn a quarter-turn counterclockwise about the origin to reach (0, 3). Reverse the order: turn first to (0, 2), then move one unit right to reach (1, 2). The same two instructions can give different final positions.

A composition applies transformations successively. We write U ∘ T for first T, then U, so (U ∘ T)(P) = U(T(P)). Substitute the first image into the second rule. The rightmost named rule acts first under this notation. If the problem states the order in words, follow those words and write the convention down before calculating.

Order can change the result. Let T translate by (1, 0), and let R rotate 90° about the origin. Starting at (2, 0), first T then R gives (3, 0) then (0, 3). First R then T gives (0, 2) then (1, 2). The different images show that these two plane transformations do not commute. A drawing of only the final shape can hide this difference if that shape has a symmetry; a labelled point still records it.

Some transformations do commute. Translations by v and w, in either order, add the same total vector v + w. Rotations about one shared centre add directed angles in either order. Neither fact gives permission to exchange arbitrary rotations, reflections or translations. For example, two reflections in the same line give the identity, while different mirror lines can produce a different movement.

Undo a sequence in reverse order: the inverse of U ∘ T is T⁻¹ ∘ U⁻¹. Starting from U(T(P)), first U⁻¹ returns T(P), then T⁻¹ returns P. Keeping the same written order while replacing the rules by their inverses would generally undo the wrong sequence. Each rule must retain its original centre, mirror line or vector when forming the inverse.

Symmetry requires the image to be the same entire figure

Turn a centred square a quarter-turn: its vertices move, but its complete boundary occupies the same position. Try the same turn with a centred nonsquare rectangle. The image still has the same size and shape, yet its boundary occupies a different position. Symmetry asks whether the whole original figure returns, rather than merely whether its shape is preserved.

A rigid transformation is a symmetry of F when T(F) = F as sets of points. It may exchange vertices or edges rather than fixing each individually. The identity always satisfies this equality; when discussing a nontrivial symmetry, we exclude the identity transformation of the plane. A line of symmetry is a line whose reflection maps the entire figure onto itself. A rotational symmetry specifies a centre and a rotation that has the same whole-figure property.

For the centred square with vertices (−1, −1), (1, −1), (1, 1) and (−1, 1) in boundary order, a 90° rotation about the origin cycles these four vertices and their adjacent edges. For the nonsquare rectangle with vertices (−2, −1), (2, −1), (2, 1) and (−2, 1), a 180° rotation exchanges opposite vertices and edges. A 90° rotation fails: (2, 1) would become (−1, 2), outside its boundary. The rectangle's image is still congruent to it, but this quarter-turn is not a symmetry of the specified rectangle.

An isosceles triangle with vertices (−1, 0), (1, 0) and (0, 2) has reflection symmetry in x = 0: its base endpoints are exchanged, its apex is fixed and its sides are paired correctly. A scalene triangle cannot have a reflection symmetry: the reflection's vertex permutation must exchange two vertices and fix the third, since repeating it is the identity and three noncollinear vertices cannot all lie on its fixed line. Distance preservation would then make the two sides from the fixed vertex equal, contradicting scalene side lengths. The mirror's position also matters. Reflecting the centred square in x = 1 sends (−1, −1) to (3, −1), so it fails symmetry even though the image remains congruent.

For a figure with a finite rotational symmetry order, count distinct rotation mappings about the specified centre with angles from 0° up to but not including 360°, including the identity. The square has order 4: 0°, 90°, 180° and 270°. No other angle works because the image of any one vertex must be one of its four equal-radius vertices, whose rays are separated by those quarter-turns. The nonsquare centred rectangle has order 2. Its long horizontal sides must map to long horizontal sides, so the rotation must preserve the horizontal line direction, forcing 0° or 180° modulo a full turn. Both work. A circle requires a different, continuous argument.

An exact finite polygon certificate includes the joining segments

For a simple polygon boundary, list every vertex in boundary order and every consecutive edge, including the closing edge. To prove a proposed rigid symmetry, first show that its image vertices give a bijection of the original vertices. Then check adjacency: the image of each original edge must be an actual boundary edge joining its two image endpoints. Orientation may be reversed by a reflection, so an edge is the same segment in either endpoint order. Permuting some chosen markers is insufficient when the joins are omitted.

Why does this finite certificate cover the continuous sides? For the coordinate rules here, a point on segment AB can be written A + t(B − A), where 0 ≤ t ≤ 1 and coordinate pairs are added and multiplied componentwise. Direct substitution into a translation, quarter-turn or listed reflection rule gives T(A + t(B − A)) = T(A) + t(T(B) − T(A)). Thus every point of AB maps to the corresponding image segment, and every point of that segment comes from the same parameter t. This identity is a statement for all t in the interval, not a numerical sample.

A vertex bijection preserving all consecutive edges therefore maps the whole boundary onto itself. For the convex filled polygons used here, the interior is covered too: every point of the closed region is a convex combination of its vertices, meaning a weighted coordinate sum with nonnegative weights adding to one. The same coordinate substitution maps it to the corresponding weighted sum of image vertices. The inverse supplies the reverse inclusion, so the complete filled region agrees as well.

To disprove a proposed symmetry, one exact counterexample is enough: find a point in the figure whose image is outside it. To prove symmetry, checking only one point or a few midpoints is not enough. A displayed list becomes a proof only when it exhausts the required finite vertices and edges and includes the argument covering every point on them. For a curve such as a circle, a finite list of boundary markers does not exhaust the boundary.

A circle's definition supplies the continuous symmetry proof

Let F be the circle of centre O and radius r > 0: F consists exactly of all P with OP = r. For any rotation about O, the centre is fixed and distances are preserved. Therefore OP′ = OP = r for every P on F, so every image point lies on the original circle. The inverse rotation has the same property, proving that every point of the original circle is also an image. Both inclusions establish T(F) = F for the entire circumference.

Reflection in any line through O also fixes O and preserves distances; applying the same argument, with the reflection itself as inverse, proves a symmetry of the whole circle. There are infinitely many permitted rotation angles and infinitely many such mirror lines. Four selected boundary points might have only fourfold rotational symmetry as a finite point set, but that is a different figure from their complete circle. A particular 90° rotation still returns every point after four applications.

A translation changes the circle's centre while preserving its radius, producing a congruent circle. For the unit circle centred at the origin, translation by (1, 0) takes (1, 0) to (2, 0); its distance from the original centre becomes 2 instead of 1. Thus this translation is not a symmetry. It does not contradict the translation's distance preservation: distance to the translated centre (1, 0) remains 1. Always compare to the centre belonging to the figure whose symmetry is claimed.

The defining equation, distance preservation and inverse prove more than an apparently overlapping plotted arc. Use exact coordinates and radii for the certificate. Screen positions and line thickness approximate those quantities, so neither a small apparent gap nor a near match decides a whole-circle symmetry.

Compare exact image rules with complete symmetry certificates

The explorer has four scenes with eight choices each, giving 32 exact cases. Translation, rotation and reflection scenes use the triangle P = (0, 0), Q = (2, 0), R = (0, 1). Its squared pair distances are PQ² = 4, PR² = 1 and QR² = 5. The controls include positive, negative, zero and fractional translation components, quarter-turns about several centres, and vertical, horizontal and diagonal mirrors with offsets. A coordinate certificate records the specified rule and corresponding image points.

Follow the centre or mirror actually selected. The zero translation and zero rotation leave every point fixed; a nonzero quarter-turn fixes its centre, and a reflection fixes its complete mirror line. The explorer labels its rotation centre C₀ to distinguish it from any polygon vertex labelled C. A centre or mirror marker on the picture is part of the transformation specification rather than an additional vertex of the triangle. Preserved pair distances illustrate the general algebraic proof, while the inverse verifies how the image returns to the source.

The symmetry scene tests a square, a nonsquare rectangle, isosceles and scalene triangles, and a circle. A positive polygon verdict must include the complete vertex matching and every boundary-edge matching; a negative verdict can identify a vertex outside the original figure. The circle verdict uses the full radius-distance argument rather than treating its displayed markers as a polygon. All images remain congruent under these rigid rules, including the choices that fail symmetry.

Worked examples

Try it yourself

1. A translation takes A = (−2, 3) to A′ = (1, −1). Find its vector and the image of B = (1/2, 2), then give the inverse. Are negative components invalid lengths?

   Show the answer and reasoning

   The vector is (3, −4), B′ = (7/2, −2), and the inverse vector is (−3, 4). Negative components are permitted directed changes.

   Subtract original coordinates from image coordinates: (1 − (−2), −1 − 3) = (3, −4). Apply the same change to B. The movement length is √(3² + (−4)²) = 5 units, which is positive; a negative component records direction rather than a negative physical length.

2. Rotate P = (3/2, 3/2) through 270° counterclockwise about C = (1/2, −1/2). Find P′, compare this with a −90° rotation, and check the radius distance and inverse.

   Show the answer and reasoning

   P′ = (5/2, −3/2). A −90° rotation about the same centre gives the same point. Both radius-distance squares are 5; the inverse is 90° about C.

   The relative pair is (1, 2). The 270° rule gives (2, −1), and adding C gives (5/2, −3/2). The original and image relative squares are 1² + 2² = 5 and 2² + (−1)² = 5. Applying the inverse relative rule (u, v) ↦ (−v, u) returns (1, 2).

3. Reflect P = (2, −1) in y = x + 1/2. Find P′ and the midpoint of PP′, and explain why the mirror is the perpendicular bisector. Check a second reflection.

   Show the answer and reasoning

   P′ = (−3/2, 5/2); the midpoint is (1/4, 3/4). The mirror contains this midpoint and is perpendicular to PP′. Reflecting again returns (2, −1).

   Use (x, y) ↦ (y − 1/2, x + 1/2). The midpoint satisfies 3/4 = 1/4 + 1/2. The off-line joining segment has changes (−7/2, 7/2), so its equal-magnitude opposite-coordinate direction is perpendicular to the mirror's equal-coordinate direction. Reapplying the rule gives (5/2 − 1/2, −3/2 + 1/2) = (2, −1).

4. Give the full fixed-point set for translation by (−2, 1), translation by (0, 0), rotation through 180° about (−2, 1), and reflection in x = 1/2. What is fixed after two applications of that same reflection?

   Show the answer and reasoning

   Respectively: no points; the whole plane; only (−2, 1); every point (1/2, y) with y real. After two identical reflections, the whole plane is fixed.

   Translation equality forces its vector to be zero. The half-turn relative equations (−u, −v) = (u, v) force u = v = 0. The reflection equation 1 − x = x gives x = 1/2 without restricting y. A reflection exchanges each off-line pair twice and fixes each on-line point throughout, so its square is the identity.

5. Let T translate by (1, 0) and R rotate 90° counterclockwise about the origin. Calculate (R ∘ T)(2, 0) and (T ∘ R)(2, 0). Give the general inverse of R ∘ T and use it to restore the first result.

   Show the answer and reasoning

   The images are (0, 3) and (1, 2). For an image (u, v), the inverse of R ∘ T is (v − 1, −u), obtained by rotating clockwise and then translating by (−1, 0). It takes (0, 3) back to (2, 0).

   R(T(x, y)) = (−y, x + 1), whereas T(R(x, y)) = (1 − y, x). To undo the first sequence, first R⁻¹ sends (u, v) to (v, −u), then T⁻¹ gives (v − 1, −u). Substitution restores both coordinates for every original point, not only the sample.

6. The rectangle has boundary-order vertices (−2, −1), (2, −1), (2, 1), (−2, 1). Test 90° and 180° rotations about the origin, including its whole side segments. Would reflection in x = 1 be a symmetry of the centred square with vertices (±1, ±1)?

   Show the answer and reasoning

   The rectangle's 90° rotation fails symmetry, while its 180° rotation is a symmetry of the whole boundary. The proposed square reflection also fails.

   At 90°, (2, 1) moves to (−1, 2), outside the rectangle boundary. At 180°, (x, y) ↦ (−x, −y) exchanges opposite vertices and sends every consecutive side onto its opposite side; the segment parameter identity covers all side points. For the square, (−1, −1) moves to (3, −1), outside it. All these image figures remain congruent.

7. A circle has centre C = (1, −1) and radius 2. Prove that rotation through 90° about C is a symmetry of its whole circumference. Test translation by (1, 0), and explain why congruence and a few marked points do not decide symmetry.

   Show the answer and reasoning

   The rotation is a whole-circle symmetry; translation by (1, 0) is not. The translated circle is congruent but has centre (2, −1).

   For every point P with CP = 2, rotation fixes C and gives CP′ = 2. Its inverse rotation supplies every original circle point as an image, proving equality of the full sets. Translation takes the original circle point (3, −1) to (4, −1), whose distance from C is 3, so it fails symmetry. A finite marker list neither defines nor exhausts the circumference.

What to carry forward

• Specify a transformation as one rule for every point, and specify whether the figure means a boundary, a filled region or a decorated object.
• A translation adds the same directed vector everywhere; negative, zero and fractional components are permitted, and the zero vector gives the identity.
• A rotation needs both its centre and directed angle. For exact quarter-turns, subtract the centre, rotate the relative pair and add the centre back.
• A reflection fixes its complete mirror line; for each off-line point, that line is the perpendicular bisector of the original-image segment.
• Rigid transformations preserve distances and congruence. Use their exact inverses, and undo a composition in reverse order because arbitrary movements need not commute.
• Whole-figure symmetry means T(F) = F, allowing points to be exchanged. Identity, a movement's fixed points and nontrivial symmetry are different claims.
• Prove polygon symmetry with all vertices and adjacent edges plus the segment argument; prove circle symmetry for every radius point and use the inverse. Approximate pictures and selected markers do not replace those certificates.