Transformations

Study Sheet

Transformations

Slides, flips, turns, and resizes on the coordinate plane

Transformation Vocabulary

Concept

A transformation moves or changes a figure according to a rule.

  • Pre-image: the original figure (points AA, BB, CC).
  • Image: the new figure after the transformation (points A′A', B′B', C′C', read “AA prime”).
  • Mapping: the pairing of each pre-image point with its image point. We write A→A′A \to A'.
  • Rigid motion (also called an isometry): a transformation that keeps the same size and shape. Translations, reflections, and rotations are rigid.
  • Non-rigid transformation: changes the size. A dilation is non-rigid.

Translations, reflections and rotations are the rigid motions — they preserve distances and angles, so a figure and its image are congruent. Dilations scale distances by a factor and preserve angles, so the image is similar. Composing transformations gives new ones, and the rules in coordinates make every transformation a formula.

Example
Pre-image and image
xy−5−4−3−2−11234123ABCA′B′C′
Triangle ABC (solid) and its mirror image A'B'C' (dashed) across the y-axis.

A rigid motion moves triangle ABCABC to triangle A′B′C′A'B'C' without stretching it. In the figure the motion is a reflection across the yy-axis: A(1,1)→A′(−1,1)A(1,1) \to A'(-1,1), B(4,1)→B′(−4,1)B(4,1) \to B'(-4,1), C(1,3)→C′(−1,3)C(1,3) \to C'(-1,3). Every side keeps its length (AB=A′B′=3AB = A'B' = 3) and every angle keeps its measure, so the two triangles are congruent.

Side note
Where it goes wrong

Rotating 90∘90^\circ counterclockwise with the wrong rule. About the origin, (x,y)→(−y,x)(x, y) \to (-y, x); check with (1,0)→(0,1)(1, 0) \to (0, 1), which is a quarter turn up.

Tip

Tip: If you can slide, flip, or turn a figure onto its image, the two are congruent. If you must also grow or shrink it, they are only similar.

Try it
Quick check

A transformation sends P(2,3)P(2, 3) to P′(4,6)P'(4, 6) and Q(1,0)Q(1, 0) to Q′(2,0)Q'(2, 0). Could it be a rigid motion?

Show the answer

Compare lengths. PQ=12+32=10PQ = \sqrt{1^2 + 3^2} = \sqrt{10}, but P′Q′=22+62=40=210P'Q' = \sqrt{2^2 + 6^2} = \sqrt{40} = 2\sqrt{10}. A rigid motion keeps every length, so no — this looks like a dilation by 22 from the origin.

Translations (Slides)

Concept

A translation slides every point the same distance in the same direction. If you move aa units horizontally and bb units vertically:

(x,y)→(x+a,  y+b).(x,y) \to (x+a,\; y+b).

The pair ⟨a,b⟩\langle a,b\rangle is the translation vector. Positive aa is right, negative is left; positive bb is up, negative is down.

A translation is completely described by one arrow, the translation vector: every point moves along a copy of it. Because all the arrows are equal and parallel, each segment of the figure and its image form a parallelogram with two of the arrows, so the image segment is parallel to and as long as the original — the figure is congruent to its image and faces the same way.

Example
Translating a triangle
xy−11234561234ABCA′B′C′
Every point moves by the same arrow (2, 1).

Translate triangle A(1,1)A(1,1), B(4,1)B(4,1), C(1,3)C(1,3) by the rule (x,y)→(x+2, y+1)(x,y)\to(x+2,\,y+1).

A′(3,2),B′(6,2),C′(3,4).A'(3,2),\quad B'(6,2),\quad C'(3,4).
Side note
Where it goes wrong

Getting a direction's sign backwards. "33 left and 22 down" is (x,y)→(x−3,y−2)(x, y) \to (x - 3, y - 2); writing (x+3,y−2)(x + 3, y - 2) moves the figure right. Check one point on the grid before translating the rest.

Tip

Tip: “Right/left” changes only xx; “up/down” changes only yy. A slide never turns or flips the figure.

Try it
Quick check

A translation maps (3,−2)(3, -2) to (−1,4)(-1, 4). Where does it send (0,5)(0, 5)?

Show the answer

The vector is (−1−3, 4−(−2))=⟨−4,6⟩(-1 - 3,\ 4 - (-2)) = \langle -4, 6 \rangle. So (0,5)→(0−4, 5+6)=(−4,11)(0, 5) \to (0 - 4,\ 5 + 6) = \mathbf{(-4, 11)}.

Reflections (Flips)

Concept

A reflection flips a figure over a line of reflection, producing a mirror image. Coordinate rules:

over the x-axis:(x,y)→(x, −y)over the y-axis:(x,y)→(−x, y)over y=x:(x,y)→(y, x)over y=−x:(x,y)→(−y, −x)\begin{aligned} \text{over the } x\text{-axis:} &\quad (x,y)\to(x,\,-y)\\ \text{over the } y\text{-axis:} &\quad (x,y)\to(-x,\,y)\\ \text{over } y=x: &\quad (x,y)\to(y,\,x)\\ \text{over } y=-x: &\quad (x,y)\to(-y,\,-x) \end{aligned}

A reflection in a line mm sends each point to the point on the other side at the same distance, so mm is the perpendicular bisector of every segment joining a point to its image. The coordinate rules just express that: across the xx-axis the distance above becomes the same distance below, (x,y)→(x,−y)(x, y) \to (x, -y); across y=xy = x the roles of the two coordinates swap. Reflections keep lengths but reverse orientation — a clockwise A→B→CA \to B \to C becomes counterclockwise.

Example
Over the x-axis
xy−11234−4−3−2−1123ABCA′B′C′
Each point and its image are the same distance from the x-axis, on opposite sides.

Reflect triangle A(1,1)A(1,1), B(4,1)B(4,1), C(1,3)C(1,3) over the xx-axis using (x,y)→(x,−y)(x,y)\to(x,-y).

A′(1,−1),B′(4,−1),C′(1,−3).A'(1,-1),\quad B'(4,-1),\quad C'(1,-3).
Example
Over y=xy = x and y=−xy = -x
xyP(3, 5)P′(5, 3)P″(−5, −3)
The dashed mirrors are y = x and y = −-x; each faint segment crosses its mirror at a right angle.

Reflect P(3,5)P(3,5) over the line y=xy=x using (x,y)→(y,x)(x,y)\to(y,x): the coordinates swap, giving P′(5,3)P'(5,3). Over y=−xy=-x, the rule (x,y)→(−y,−x)(x,y)\to(-y,-x) gives P′′(−5,−3)P''(-5,-3).

Side note
Where it goes wrong

Swapping without the signs for y=−xy = -x. The rule is (x,y)→(−y,−x)(x, y) \to (-y, -x): for P(3,5)P(3, 5) that is (−5,−3)(-5, -3), not (5,3)(5, 3) (which is the reflection in y=xy = x). Check with a point on the mirror: (2,−2)(2, -2) must stay put, and (−(−2),−2)=(2,−2)(-(-2), -2) = (2, -2) does.

Tip

Tip: A point on the line of reflection does not move. Watch the signs: over the xx-axis only yy changes sign; over the yy-axis only xx changes sign.

Try it
Quick check

Reflect the triangle with vertices (2,1)(2, 1), (5,1)(5, 1), (5,3)(5, 3) over the line y=xy = x.

Show the answer

Swap the coordinates: (1,2)\mathbf{(1, 2)}, (1,5)\mathbf{(1, 5)}, (3,5)\mathbf{(3, 5)}.

Rotations (Turns) About the Origin

Concept

A rotation turns a figure about a fixed point (the center). Rotating counterclockwise about the origin:

90∘:(x,y)→(−y, x)180∘:(x,y)→(−x, −y)270∘:(x,y)→(y, −x)\begin{aligned} 90^\circ: &\quad (x,y)\to(-y,\,x)\\ 180^\circ: &\quad (x,y)\to(-x,\,-y)\\ 270^\circ: &\quad (x,y)\to(y,\,-x) \end{aligned}

A 90∘90^\circ clockwise turn equals a 270∘270^\circ counterclockwise turn (and vice versa). A 180∘180^\circ turn is the same either direction.

Where (x,y)→(−y,x)(x, y) \to (-y, x) comes from: the point (x,y)(x, y) is reached by going xx along and yy up. Turn that whole path a quarter turn counterclockwise — "along" becomes "up" and "up" becomes "left" — and you arrive at xx up and yy to the left, the point (−y,x)(-y, x). Doing it twice gives (−x,−y)(-x, -y), the half turn.

Example
A quarter turn
xy−4−3−2−11234−11234ABCA′B′C′
B(4, 1) turns a quarter circle about O to B'(−-1, 4); OB and OB' are perpendicular.

Rotate triangle A(1,1)A(1,1), B(4,1)B(4,1), C(1,3)C(1,3) by 90∘90^\circ counterclockwise using (x,y)→(−y,x)(x,y)\to(-y,x).

A′(−1,1),B′(−1,4),C′(−3,1).A'(-1,1),\quad B'(-1,4),\quad C'(-3,1).
Side note
Where it goes wrong

Using (x,y)→(y,−x)(x, y) \to (y, -x) for a counterclockwise turn. That is the clockwise quarter turn. Test with (1,0)(1, 0): counterclockwise should carry it up to (0,1)(0, 1), and (−y,x)=(0,1)(-y, x) = (0, 1) does.

Tip

Tip: For 90∘90^\circ turns, swap the coordinates first, then fix signs. Counterclockwise is the positive direction in math.

Try it
Quick check

Rotate the point (4,−3)(4, -3) by 90∘90^\circ clockwise about the origin.

Show the answer

Clockwise 90∘90^\circ is counterclockwise 270∘270^\circ: (x,y)→(y,−x)(x, y) \to (y, -x), so (4,−3)→(−3,−4)(4, -3) \to \mathbf{(-3, -4)}.

Compositions of Transformations

Concept

A composition applies one transformation and then another, in order. Do the first transformation, write the image, then apply the second transformation to that image.

A composition is itself a transformation, and knowing which one saves work: two reflections in perpendicular lines make a half turn, two reflections in parallel lines make a translation, and any two rigid motions make a rigid motion. Order matters because the second step acts on the image of the first: reflecting (1,2)(1, 2) over the xx-axis and then sliding right gives (4,−2)(4, -2), and here the reverse order happens to agree — but reflect-then-rotate and rotate-then-reflect usually do not.

Example
Reflect, then translate
xy−11234−3−2−112(1, 2)(1, −2)(4, −2)
Step 1 flips across the x-axis; step 2 slides 3 right.

Reflect (1,2)(1,2) over the xx-axis, then translate by (x,y)→(x+3,y)(x,y)\to(x+3,y).

Step 1 (reflect):(1,2)→(1,−2)Step 2 (translate):(1,−2)→(4,−2)\begin{aligned} \text{Step 1 (reflect):} &\quad (1,2)\to(1,-2)\\ \text{Step 2 (translate):} &\quad (1,-2)\to(4,-2) \end{aligned}

The final image is (4,−2)(4,-2).

Example
Two reflections make a half turn
xy
Solid →\to dashed (across the x-axis) →\to shaded (across the y-axis): the same as a 180∘^\circ turn.

Two reflections over the xx-axis and then the yy-axis have the same effect as one 180∘180^\circ rotation about the origin, because (x,y)→(x,−y)→(−x,−y)(x,y)\to(x,-y)\to(-x,-y).

Side note
Where it goes wrong

Applying the second rule to the original point. For "reflect over the xx-axis, then over y=xy = x" starting at (3,1)(3, 1): the first step gives (3,−1)(3, -1), and the second must swap that, giving (−1,3)(-1, 3) — not the swap of (3,1)(3, 1), which is (1,3)(1, 3).

Tip

Tip: Order matters! Reflecting then rotating usually gives a different image than rotating then reflecting. Always finish step 1 completely before starting step 2.

Try it
Quick check

Rotate (2,5)(2, 5) by 180∘180^\circ about the origin, then translate by (x,y)→(x−3,y+1)(x, y) \to (x - 3, y + 1).

Show the answer

Half turn: (2,5)→(−2,−5)(2, 5) \to (-2, -5). Translate: (−2−3,−5+1)=(−5,−4)(-2 - 3, -5 + 1) = \mathbf{(-5, -4)}.

Dilations (Resizing) About the Origin

Concept

A dilation centered at the origin multiplies every coordinate by the same scale factor kk:

(x,y)→(kx, ky).(x,y)\to(kx,\,ky).
  • k>1k>1: an enlargement (the figure grows).
  • 0<k<10<k<1: a reduction (the figure shrinks).
  • k=1k=1: the figure is unchanged.

A dilation is not a rigid motion, but it keeps the same shape, so the image is similar to the pre-image.

A dilation pushes every point along its ray from the center, multiplying its distance by kk. Every segment then becomes kk times as long (the triangle made by the center, a point and another point is scaled by SAS∼\sim) and parallel to the original, and every angle is unchanged — which is exactly what "similar" means.

Example
Scale factor 2
xy12341234ABCA′B′C′
Each image point is twice as far from the origin along the same ray.

Dilate triangle A(1,1)A(1,1), B(2,1)B(2,1), C(1,2)C(1,2) by scale factor k=2k=2 using (x,y)→(2x,2y)(x,y)\to(2x,2y).

A′(2,2),B′(4,2),C′(2,4).A'(2,2),\quad B'(4,2),\quad C'(2,4).
Side note
Where it goes wrong

Adding kk instead of multiplying: (1,1)→(3,3)(1, 1) \to (3, 3) is a translation by ⟨2,2⟩\langle 2, 2 \rangle, not a dilation by 22. It keeps the size; the dilation must double every length, giving (2,2)(2, 2).

Tip

Tip: The scale factor of the lengths is kk, but the area scales by k2k^2. Doubling the sides (k=2k=2) makes the area 44 times as large.

Try it
Quick check

Dilate (6,−9)(6, -9) by scale factor 13\tfrac13 about the origin. By what factor does the area of a figure change under this dilation?

Show the answer

(6,−9)→(13⋅6,13⋅(−9))=(2,−3)(6, -9) \to \left(\tfrac13 \cdot 6, \tfrac13 \cdot (-9)\right) = \mathbf{(2, -3)}. Areas scale by k2=19k^2 = \mathbf{\tfrac19}.

Symmetry

Concept

Line (reflectional) symmetry: a figure has line symmetry if a reflection over some line maps the figure exactly onto itself. That line is a line of symmetry.

Rotational symmetry: a figure has rotational symmetry if a rotation of less than 360∘360^\circ about its center maps it onto itself. The order is how many times it matches in one full turn; the angle of rotation is 360∘÷order360^\circ \div \text{order}.

The angle of rotational symmetry must divide 360∘360^\circ evenly: if a turn of θ\theta maps the figure onto itself, so does every multiple of θ\theta, and the turns must come back to the start after one full revolution. That is why the smallest angle is 360∘÷order360^\circ \div \text{order}. A rectangle's diagonal is not a mirror line: reflecting across it sends a long side onto a short one.

Example
Regular hexagon
6 lines: 3 through opposite vertices, 3 through opposite side midpoints.

A regular hexagon has 66 lines of symmetry. Its rotational symmetry has order 66, with smallest angle 360∘÷6=60∘360^\circ\div 6 = 60^\circ.

Example
Square and rectangle
The square has 4 lines of symmetry; the rectangle only 2 (its diagonals are not mirrors).

A square has 44 lines of symmetry (two diagonals and two through opposite side midpoints) and rotational symmetry of order 44 (90∘90^\circ). A non-square rectangle has only 22 lines of symmetry and rotational symmetry of order 22 (180∘180^\circ).

Side note
Where it goes wrong

Counting a rectangle's diagonals as lines of symmetry. Fold a non-square rectangle along a diagonal and the corners do not meet. Only for a square does the diagonal fold work.

Tip

Tip: Every regular polygon with nn sides has exactly nn lines of symmetry and rotational symmetry of order nn.

Try it
Quick check

How many lines of symmetry and what order of rotational symmetry does (a) a regular pentagon have? (b) the letter Z?

Show the answer

(a) 5 lines, rotational order 5 (smallest angle 72∘72^\circ). (b) Z has no line of symmetry, but a half turn maps it onto itself: order 2.

Congruence vs. Similarity

Concept
Rigid Motions vs. Dilations
  • Rigid motions (translations, reflections, rotations, and their compositions) preserve size and shape. The image is congruent to the pre-image.
  • Dilations (scale factor k≠1k\ne 1) preserve shape but change size. The image is similar, but not congruent, to the pre-image.

Two figures are congruent if a sequence of rigid motions maps one onto the other. They are similar if a sequence of rigid motions and dilations does so.

Defining congruence and similarity by motions makes the triangle shortcuts theorems rather than extra rules: SSS, SAS and ASA say that three matching parts are enough to build the sequence of rigid motions, and AA says a dilation followed by rigid motions will line up two triangles with equal angles. It also explains why congruent figures are automatically similar — a dilation by 11 is allowed.

Example
A similarity by three moves
xy−3−2−1123−112345ABCA″B″C″
Solid: the original. Dashed: its image after the three moves.

Show that the triangle A(0,0)A(0,0), B(2,0)B(2,0), C(0,1)C(0,1) is similar to the triangle A′′(1,1)A''(1,1), B′′(1,5)B''(1,5), C′′(−1,1)C''(-1,1), and decide whether they are congruent.

Find a sequence of moves:

rotate 90∘ (x,y)→(−y,x):(0,0), (0,2), (−1,0)dilate by 2:(0,0), (0,4), (−2,0)translate ⟨1,1⟩:(1,1), (1,5), (−1,1)\begin{aligned} \text{rotate } 90^\circ\ (x,y)\to(-y,x): &\quad (0,0),\ (0,2),\ (-1,0)\\ \text{dilate by } 2: &\quad (0,0),\ (0,4),\ (-2,0)\\ \text{translate } \langle 1, 1\rangle: &\quad (1,1),\ (1,5),\ (-1,1) \end{aligned}

Rigid motions plus one dilation carry one triangle onto the other, so they are similar with scale factor 22. They are not congruent: AB=2AB = 2 but A′′B′′=4A''B'' = 4.

Side note
Where it goes wrong

Deciding congruence from shape alone. The two triangles above have the same angles, but one is twice the size; "same shape" is similarity, and congruence needs the lengths to match too.

Tip

Tip: Congruent ⇒\Rightarrow similar (with ratio 11), but similar does not always mean congruent. Ask: “Did the size change?” If yes, it is similar only.

Try it
Quick check

Is the triangle P(0,0)P(0,0), Q(3,0)Q(3,0), R(0,4)R(0,4) congruent to P′(5,1)P'(5,1), Q′(5,4)Q'(5,4), R′(1,1)R'(1,1)? If so, give a sequence of rigid motions.

Show the answer

Rotate 90∘90^\circ counterclockwise about the origin: P→(0,0)P \to (0,0), Q→(0,3)Q \to (0,3), R→(−4,0)R \to (-4,0). Then translate by ⟨5,1⟩\langle 5, 1\rangle: (5,1)(5,1), (5,4)(5,4), (1,1)(1,1) — exactly P′Q′R′P'Q'R'. Rigid motions only, so the triangles are congruent (both are 33-44-55 right triangles).

Going Deeper: Advanced Transformations

Concept
Composing Two Reflections

Reflecting twice in a row is always a single rigid motion. Which one depends on the two mirror lines:

  • Parallel mirrors →\to a translation. The shift is perpendicular to the mirrors, in the direction from the first mirror to the second, and its distance is 2d2d, where dd is the gap between the mirrors.
  • Intersecting mirrors →\to a rotation about the point where they cross. If the angle from the first mirror to the second is θ\theta, the rotation angle is 2θ2\theta (same turning direction).

Either way, two flips restore the original orientation, so the result is an “even” (direct) isometry.

Example
Parallel mirrors
xy−11234123PP′P″
Mirrors x = 1 and x = 3 (dashed), 2 apart: the net move is 4 to the right.

Parallel mirrors give a translation. Reflect P(0,2)P(0,2) over the line x=1x=1, then over x=3x=3. The mirrors are d=2d=2 apart, so the net move is a translation of 2d=42d=4 units to the right.

Reflect over x=1:(0,2)→(2,2)Reflect over x=3:(2,2)→(4,2)\begin{aligned} \text{Reflect over } x=1: &\quad (0,2)\to(2,2)\\ \text{Reflect over } x=3: &\quad (2,2)\to(4,2) \end{aligned}

Net effect: (0,2)→(4,2)(0,2)\to(4,2), i.e. (x,y)→(x+4,y)(x,y)\to(x+4,y).

Side note
Where it goes wrong

Translating by the gap between the mirrors instead of twice the gap. Mirrors x=1x = 1 and x=3x = 3 are 22 apart, but the point moves 44: from (0,2)(0,2) to (4,2)(4,2). Each reflection doubles the distance to its mirror.

Example
Mirrors crossing at 45∘45^\circ
xy(3, 1)(3, −1)(−1, 3)
Across the x-axis, then across y = x (dashed): a 90∘^\circ turn about the origin.

Intersecting mirrors give a rotation. Reflect P(3,1)P(3,1) over the xx-axis, then over the line y=xy=x. These mirrors cross at the origin, and the angle from the xx-axis to y=xy=x is 45∘45^\circ, so the result is a rotation of 2(45∘)=90∘2(45^\circ)=90^\circ about the origin.

Reflect over x-axis:(3,1)→(3,−1)Reflect over y=x:(3,−1)→(−1,3)\begin{aligned} \text{Reflect over } x\text{-axis}: &\quad (3,1)\to(3,-1)\\ \text{Reflect over } y=x: &\quad (3,-1)\to(-1,3) \end{aligned}

Indeed (3,1)→(−1,3)(3,1)\to(-1,3) is exactly the 90∘90^\circ counterclockwise rule (x,y)→(−y,x)(x,y)\to(-y,x).

Concept
Rotations About Any Center & General Lines

Rotation about a center (a,b)(a,b): shift the center to the origin, rotate, then shift back.

(x,y)  →subtract  (x−a, y−b)  →rotate  ⋯  →add  (result).(x,y)\;\xrightarrow{\text{subtract}}\;(x-a,\,y-b)\;\xrightarrow{\text{rotate}}\;\cdots\;\xrightarrow{\text{add}}\;(\text{result}).

For a 90∘90^\circ counterclockwise turn about (a,b)(a,b) this collapses to

(x,y)→(a−(y−b),    b+(x−a)).(x,y)\to\bigl(a-(y-b),\;\; b+(x-a)\bigr).

Reflection over a general line y=mx+cy=mx+c: it still sends each point to the opposite side at equal perpendicular distance. A clean way is to compose known moves — translate so the line passes through the origin, rotate the line onto the xx-axis, reflect over the xx-axis, then undo the rotation and translation.

Example
About a point other than the origin
xy−11234−1123O(2, 1)P(4, 1)P′(2, 3)
P swings a quarter turn about O(2, 1), keeping its distance 2.

Rotate about a non-origin center. Rotate P(4,1)P(4,1) by 90∘90^\circ counterclockwise about the center O(2,1)O(2,1).

Subtract center:(4,1)−(2,1)=(2,0)Rotate 90∘ (x,y)→(−y,x):(2,0)→(0,2)Add center back:(0,2)+(2,1)=(2,3)\begin{aligned} \text{Subtract center:} &\quad (4,1)-(2,1)=(2,0)\\ \text{Rotate } 90^\circ\ (x,y)\to(-y,x): &\quad (2,0)\to(0,2)\\ \text{Add center back:} &\quad (0,2)+(2,1)=(2,3) \end{aligned}

So P′(2,3)P'(2,3): the point swings a quarter turn around OO, not around the origin.

Concept
Transformations as Matrices (Preview)

Every rotation, reflection, and dilation about the origin can be written as multiplication by a 2×22\times 2 matrix acting on the column vector [xy]\begin{bmatrix}x\\y\end{bmatrix}:

Rotate 90∘ CCW:[0−110]Reflect over x-axis:[100−1]Reflect over y=x:[0110]Dilation, factor k:[k00k]\begin{aligned} \text{Rotate } 90^\circ\ \text{CCW}: &\quad \begin{bmatrix}0 & -1\\ 1 & 0\end{bmatrix} & \text{Reflect over } x\text{-axis}: &\quad \begin{bmatrix}1 & 0\\ 0 & -1\end{bmatrix}\\[2pt] \text{Reflect over } y=x: &\quad \begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix} & \text{Dilation, factor } k: &\quad \begin{bmatrix}k & 0\\ 0 & k\end{bmatrix} \end{aligned}

For example, the 90∘90^\circ rotation of (3,1)(3,1) is

[0−110][31]=[−13],\begin{bmatrix}0 & -1\\ 1 & 0\end{bmatrix}\begin{bmatrix}3\\ 1\end{bmatrix}=\begin{bmatrix}-1\\ 3\end{bmatrix},

matching (x,y)→(−y,x)(x,y)\to(-y,x). A translation is not linear, so it needs a separate added vector [ab]\begin{bmatrix}a\\ b\end{bmatrix} (or a 3×33\times 3 “homogeneous” matrix you will meet later). Composing transformations then corresponds to multiplying their matrices.

Concept
Glide Reflections
xyTT′
T (solid) reflected across the x-axis and slid 2 right lands on T' (dashed); repeating gives the faint copy, like footprints.

A glide reflection is a reflection over a line followed by a translation parallel to that line (the order does not matter). It is the fourth basic rigid motion of the plane, alongside translations, rotations, and reflections.

  • It reverses orientation (like a single reflection), so it is an “odd” isometry.
  • It has no fixed points and no fixed line pointwise — the classic footprint pattern of a walking person is a glide reflection.

Here TT is reflected over the xx-axis and slid 22 units right to land on T′T'.

Concept
Symmetry Groups & Order of Rotational Symmetry

The set of all rigid motions that map a figure onto itself forms its symmetry group.

  • Cyclic group CnC_n: only rotational symmetry — nn rotations (including the 360∘360^\circ “do nothing”), and no lines of symmetry. A pinwheel or “S” shape is like this.
  • Dihedral group DnD_n: rotational and reflective symmetry — nn rotations plus nn reflections, giving 2n2n symmetries in all.

A regular nn-gon has symmetry group DnD_n: order-nn rotational symmetry (smallest angle 360∘/n360^\circ/n) and nn lines of symmetry, for 2n2n total symmetries. An equilateral triangle is D3D_3 (66 symmetries); a square is D4D_4 (88 symmetries).

Tip

Dilation — area & orientation: Under a dilation with scale factor kk, lengths scale by ∣k∣|k|, areas scale by k2k^2, and angles are unchanged. When k>0k>0 orientation is preserved; a negative kk acts like a positive dilation of factor ∣k∣|k| combined with a 180∘180^\circ rotation about the center, which also preserves orientation. So a dilation never produces a mirror image — only a resized, similar copy.

Try it
Quick check

Reflect P(5,2)P(5, 2) over the line x=1x = 1 and then over the line x=4x = 4. What single transformation has the same effect?

Show the answer

Over x=1x = 1: 55 is 44 right of the mirror, so it lands 44 left of it, (−3,2)(-3, 2). Over x=4x = 4: −3-3 is 77 left of that mirror, so it lands at (11,2)(11, 2). The net move is ⟨6,0⟩\langle 6, 0 \rangle — a translation of 66 to the right, twice the 33-unit gap between the parallel mirrors.

Why It Works

Theorem
The reason behind the rule
xy(3, 1)(1, 3)(−1, 3)
Across y = x, then across the y-axis: (3, 1) →\to (1, 3) →\to (−-1, 3), a quarter turn.

A reflection across the yy-axis sends (x,y)(x, y) to (−x,y)(-x, y): the mirror keeps the height and negates the horizontal position. Rotation by 90∘90^\circ counterclockwise is two reflections in a row (across y=xy = x, then across the yy-axis): (x,y)→(y,x)→(−y,x)(x, y) \to (y, x) \to (-y, x). Every rigid motion of the plane is a composition of at most three reflections — a fact that explains why they all preserve distance: each reflection does, so any composition of them does too.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
Rigid motions in the plane
(x,y)→(x+a, y+b),(x,y)→(x,−y),(x,y)→(−y,x) (90∘ CCW)(x,y)\to(x+a,\,y+b),\quad (x,y)\to(x,-y),\quad (x,y)\to(-y,x)\ (90^\circ\text{ CCW})

What it means. Translation, reflection in the xx-axis, and rotation about the origin.

Example. Reflecting (3,2)(3,2) over the xx-axis gives (3,−2)(3,-2).

Why it works. Rotating 90∘90^\circ counter-clockwise sends the direction "right" to "up" and "up" to "left", i.e. (1,0)→(0,1)(1,0)\to(0,1) and (0,1)→(−1,0)(0,1)\to(-1,0); applying that to (x,y)(x,y) gives (−y,x)(-y,x).

Tip. Rigid motions preserve lengths and angles, so the image is congruent. Only dilations change size.