Composing transformations
A figure is slid and then flipped, or turned about one point and then about another. Where does it end up, and is there one motion that does the whole job?
Where you are 11 in Transformations
When you reach for it
A single translation, reflection or rotation is easy to describe. Many moves are not one of those on its own. A drawing program rotates a shape and then drags it somewhere else. A row of footprints in sand goes left foot, right foot, left foot, and each print is the one before it flipped and moved forward. A triangle on a grid sits where it does because it was turned and then slid, or flipped and then turned.
Composing two motions means doing one and then doing the other to the result. You reach for it in three situations.
- A sequence of motions is given, and you need to know where the figure ends up.
- A sequence is given, and you want one motion with the same effect: to name it, to undo it, or to repeat it.
- Two figures are given, and you need a sequence that carries one onto the other. That is exactly what it takes to show the figures are congruent.
The lesson runs over two days. Part 1 is doing motions in a row: following points, writing the combined coordinate rule, seeing when the order matters, and undoing a sequence. Part 2 is naming the single motion a sequence amounts to, which brings in the glide reflection, the one rigid motion that is not a slide, a turn or a flip.
The procedure
Part 1: one motion after another
Composition. If f and g are motions of the plane, their composition g \circ f sends each point P first through f, to P' = f(P), and then through g, to P'' = g(P'):
It is read “g after f”. The motion written nearest to P acts first, so a composition is read from right to left. This is the same convention as composing functions, and for the same reason: a motion of the plane is a function whose inputs and outputs are points.
Following a point. Apply the first motion, write down the image, and apply the second motion to that image. Each step is a single motion you already know how to do, so nothing new is needed. To move a polygon, follow each vertex and join the images in the same order. That is allowed because a rigid motion carries a segment onto the segment joining the images of its endpoints.
Composing coordinate rules. When f and g are rules in x and y, the rule for g \circ f comes from substitution. Write out f(x, y), then put its first coordinate wherever g’s rule has x and its second wherever g’s rule has y. The result is one rule that does both motions at once, and it can be checked on any single point.
Order matters. In general g \circ f and f \circ g are different motions, as Examples 1 and 2 show. They agree only in special cases. Two translations commute, since \langle a, b \rangle followed by \langle c, d \rangle sends (x, y) to (x + a + c,\ y + b + d), and addition of numbers does not care about order:
Two rotations about the same center commute too, since their angles add. A reflection and a translation along its own mirror commute, which is the glide reflection of Part 2. Outside cases like these, the order has to be read carefully.
A composition of rigid motions is rigid. Suppose f and g both keep every distance. Then for any two points P and Q, the distance from g(f(P)) to g(f(Q)) equals the distance from f(P) to f(Q), because g keeps distances, and that equals PQ, because f does. So g \circ f keeps distances as well, and the same argument works for angle measures. The final figure is always congruent to the first, whatever the sequence.
Undoing. Every rigid motion f has an inverse f^{-1} that puts every point back where it was. The translation \langle a, b \rangle is undone by \langle -a, -b \rangle, the rotation R_{C,\theta} by R_{C,-\theta}, and a reflection by itself, since flipping twice in one mirror returns every point. To undo g \circ f, undo the last motion first:
The reason is that the undoing has to peel the motions off from the outside in. Applying the right side to g(f(P)) gives
with each inverse canceling the motion right next to it. It is the rule for socks and shoes: they go on socks first, and come off shoes first.
Part 2: one motion that does the same job
Handedness. Label a triangle’s vertices so that going A \to B \to C runs counterclockwise. After a translation or a rotation, A' \to B' \to C' still runs counterclockwise. After a reflection it runs clockwise: a mirror image is a figure of the other hand. In a composition, every reflection reverses the direction and every translation or rotation leaves it alone. Two reversals cancel, so
- an even number of reflections in the sequence keeps the handedness, and
- an odd number reverses it.
That is a fact about the whole sequence that can be read off without moving a single point.
The glide reflection. A glide reflection is a reflection in a line \ell followed by a translation by a non-zero vector parallel to \ell. The line \ell is its axis. Footprints are the everyday picture: each print is the one before it flipped across the line down the middle of the trail and moved forward along it.
Place coordinates so that \ell is the x-axis and the slide is \langle a, 0 \rangle with a \ne 0. Every glide reflection can be set up this way, since any line can be chosen as the x-axis. Then the reflection sends (x, y) to (x, -y) and the slide adds a to the first coordinate, so the glide reflection is
and everything about it can be read from that rule.
- The order does not matter. Slide first and then reflect: (x + a, y) goes to (x + a, -y), the same point.
- It reverses handedness, since it contains one reflection.
- It has no fixed point. A fixed point would need x + a = x, which is impossible when a \ne 0.
- The midpoint of each point and its image lies on the axis. The midpoint of (x, y) and (x + a, -y) is \displaystyle \left(x + \frac{a}{2},\ 0\right).
- Doing it twice is a translation, by twice the slide: (x, y) \mapsto (x + 2a,\ y).
So a glide reflection is not a translation or a rotation, which keep handedness, and it is not a reflection, which fixes every point of its mirror. It is a fourth kind of rigid motion. In the rigid motions of a plane, the four kinds are all there is.
The four kinds. Every rigid motion of the plane is a translation, a rotation, a reflection or a glide reflection (counting the identity, which moves nothing, as a translation by \langle 0, 0 \rangle). Which one is decided by two questions, the handedness and the fixed points:
| Handedness | Points that do not move | The motion |
|---|---|---|
| kept | every point | the identity |
| kept | none | a translation |
| kept | exactly one, C | a rotation about C |
| reversed | every point of one line \ell | the reflection in \ell |
| reversed | none | a glide reflection |
That the list is complete is stated here, not proved. The proof writes every rigid motion as at most three reflections in a row and sorts the cases, and the case of two is two reflections in a row. What this lesson uses is the table itself: to name a composition, count its reflections and look for points that stay put.
Naming the motion from its rule. The coordinate rules in this lesson all have the same shape: a quarter turn, half turn or flip about the origin, followed by a shift. That is, f(P) = L(P) + v for one of the eight rules
as L, and a fixed vector v. Each coordinate of L is plus or minus x or y, so L subtracts term by term: L(P) - L(Q) = L(P - Q).
When L is a turn other than (x, y), solve f(C) = C for the fixed point C. Then for every point P,
because the shift v cancels. Read in words: the offset of the image from C is the offset of P from C, turned through L’s angle. That is precisely the rotation about C through that angle, the way rotating about any point does it. So f is L’s turn, moved to the new center C. When L is (x, y) itself, f is simply the translation by v.
When L is a flip, f reverses handedness, so by the table it is a reflection or a glide reflection. Work out the midpoint of (x, y) and f(x, y) with letters. If the midpoints all lie on one line, that line is the axis. Then compare f with the reflection in that line: if they agree, f is that reflection; if they differ by a slide along the line, f is a glide reflection with that slide.
Finding a sequence between two figures. Suppose two congruent figures are given, with the corresponding vertices named. Compare handedness first: if it is reversed, the sequence must contain an odd number of reflections, so at least one. Then compare the direction of one side with the direction of its image. The turn (or flip) that carries the first direction to the second is the piece L, and whatever is left over after doing it is a translation, read off from a single vertex. Finish by checking every vertex, since a sequence chosen from one side alone can still put the third vertex on the wrong side.
Worked examples
Examples 1 to 4 belong to Part 1, and Examples 5 to 9 to Part 2.
Two orders, two different triangles
Triangle ABC has vertices A(1, 1), B(5, 1) and C(5, 4). Let t be the translation \langle -6, 2 \rangle and r the reflection in the x-axis. Find the image of the triangle under r \circ t and under t \circ r.
Solutionr \circ t means t first. The translation adds -6 to x and 2 to y, and the reflection in the x-axis then changes the sign of y and keeps x:
t \circ r means r first, so the signs change before the slide:
The two results are different triangles, 4 units apart vertically. The reason is the upward part of the slide: under r \circ t the upward 2 is applied before the flip, so the flip turns it into a downward 2, while under t \circ r it is applied after the flip and stays upward. Both images are congruent to \triangle ABC, as every composition of rigid motions must be, and both are mirror images of it, since each sequence contains one reflection.
Composing two rules by substitution
Let f be the translation (x, y) \mapsto (x + 3,\ y - 1) and g = R_{O,90^\circ}, the quarter turn counterclockwise about the origin, (x, y) \mapsto (-y, x). Find rules for g \circ f and for f \circ g, and check each on the origin.
SolutionFor g \circ f, first f gives (x + 3, y - 1). The rule for g says “new first coordinate is minus the old second, new second is the old first”, so feed it the pair (x + 3, y - 1):
For f \circ g, first g gives (-y, x), and then f adds 3 to the first coordinate and subtracts 1 from the second:
The rules are different, so the motions are. Check on the origin. Doing f \circ g by hand: the quarter turn leaves O where it is, and then the slide takes it to (3, -1); the rule gives (3 - 0, 0 - 1) = (3, -1). Doing g \circ f by hand: the slide takes O to (3, -1), and the quarter turn takes that to (1, 3); the rule gives (1 - 0, 0 + 3) = (1, 3).
When the order does not matter, and when it does
Let r be the reflection in the x-axis, (x, y) \mapsto (x, -y). For each pair, find both compositions and say whether the order matters.
(a) the translations \langle 2, 5 \rangle and \langle -6, 1 \rangle \qquad (b) r and the translation \langle -3, 0 \rangle \qquad (c) r and the translation \langle 0, 4 \rangle
Solution(a) Either way round, the vectors add, and 2 + (-6) = -4 and 5 + 1 = 6. Both orders give the translation \langle -4, 6 \rangle. The order does not matter.
(b) Reflect first: (x, -y), then slide left: (x - 3,\ -y). Slide first: (x - 3, y), then reflect: (x - 3,\ -y). The same rule, so the order does not matter. The slide runs straight left, along the mirror, and the reflection does not touch the first coordinate, so neither motion interferes with the other.
(c) Reflect first and then slide: (x, -y) \mapsto (x,\ -y + 4). Slide first and then reflect: (x, y + 4) \mapsto (x,\ -(y + 4)) = (x,\ -y - 4). These differ, so the order matters.
Each of the two in (c) is itself a single reflection. The rule (x, 4 - y) keeps x and sends y to 4 - y, and the midpoint of y and 4 - y is always 2: it is the reflection in the line y = 2. In the same way (x, -y - 4) is the reflection in y = -2. A slide across a mirror, added to the reflection, moves the mirror by half the slide, in the direction of the slide when the slide comes second and against it when it comes first.
Undoing a composition
A point P is turned by f = R_{O,90^\circ} and then slid by g = \langle 2, -3 \rangle. It ends at (5, 1). Where did it start? Write a rule for (g \circ f)^{-1}.
SolutionUndo the last motion first. Taking off the slide means sliding by \langle -2, 3 \rangle:
Then take off the quarter turn by turning a quarter turn clockwise, R_{O,-90^\circ}, whose rule is (x, y) \mapsto (y, -x):
So P = (4, -3). Check forwards: R_{O,90^\circ} sends (4, -3) to (3, 4), and the slide sends that to (5, 1).
For the rule, (g \circ f)^{-1} = f^{-1} \circ g^{-1}: subtract the slide, then turn clockwise.
On (5, 1) this gives (4, -3), as before. Undoing in the other order, turning back first and then removing the slide, would send (5, 1) to (1, -5) and then to (-1, -2), which is not P. The turn was done before the slide, so it can only be taken off after the slide has been.
A reflection and a turn make a reflection
Let f be the reflection in the line y = x, (x, y) \mapsto (y, x), and g = R_{O,90^\circ}. Name the single motion g \circ f, and the single motion f \circ g.
SolutionSubstitute:
The first is the reflection in the y-axis and the second the reflection in the x-axis. A flip followed by a turn is a flip, in a different mirror.
The table predicted the kind before any algebra. Each sequence contains one reflection, so the handedness is reversed, and the origin does not move under either motion, so there is a fixed point. Reversed with a fixed point means a reflection, and the only question left was which mirror. The rule answered it: (-x, y) fixes exactly the points with x = 0, the y-axis.
Check with a point. f sends (3, 1) to (1, 3), and the quarter turn sends that to (-3, 1), the mirror image of (3, 1) in the y-axis.
A glide reflection
Let G be the glide reflection with axis y = -2 and slide \langle 6, 0 \rangle. Write its rule. Find the images of P(0, 1) and Q(2, -4) and check that their midpoints lie on the axis. Show that G has no fixed point, and find G \circ G.
SolutionThe reflection in y = -2 keeps x and sends y to -4 - y, because the midpoint of y and -4 - y is -2. The slide then adds 6 to x:
Sliding first gives (x + 6, y) and reflecting that gives (x + 6,\ -4 - y), the same rule, as it must be for a slide along the axis.
Both midpoints have y-coordinate -2, so both lie on the axis. They are different points of it, 2 apart, because P and Q start 2 apart in the direction of the axis.
A fixed point would need x + 6 = x, which no number satisfies, so G moves every point.
Doing G twice, the second G receives (x + 6, -4 - y):
The translation \langle 12, 0 \rangle. The two flips cancel and the two slides add. That is why a trail of footprints, every print a glide reflection of the one before, repeats exactly every two steps.
Naming the motion from Example 1
Name the single motions r \circ t and t \circ r from Example 1, where t = \langle -6, 2 \rangle and r is the reflection in the x-axis.
SolutionFrom the vertex lists in Example 1, the rules are
Each contains one reflection, so each reverses handedness. Neither has a fixed point, since x - 6 = x is impossible. By the table, each is a glide reflection. To find the axis of r \circ t, take the midpoint of (x, y) and its image:
Every midpoint lies on the line y = -1, so that is the axis. The reflection in y = -1 sends (x, y) to (x, -2 - y), and r \circ t is that followed by the slide \langle -6, 0 \rangle, which runs along the axis. For a check, the midpoint of A(1, 1) and its image (-5, -3) is (-2, -1), on the axis.
The same working for t \circ r gives midpoints (x - 3,\ 1): the glide reflection with axis y = 1 and the same slide \langle -6, 0 \rangle.
The slide \langle -6, 2 \rangle was at an angle to the mirror, and it split in two. The part \langle -6, 0 \rangle along the mirror stayed a slide. The part \langle 0, 2 \rangle across the mirror combined with the reflection to move the mirror by 1, half of 2, exactly as in Example 3(c). That is why the two axes are 2 apart, and why the two final triangles in the figure sit 4 apart, one straight above the other: twice the distance between the axes.
Two quarter turns about different centers
Let A = (0, 0) and B = (4, 0).
(a) Name the single motion R_{B,90^\circ} \circ R_{A,90^\circ}.
(b) Name the single motion R_{B,-90^\circ} \circ R_{A,90^\circ}.
SolutionThe rule for a turn about B measures from B, turns, and adds B back. For a quarter turn counterclockwise, the offset (x - 4, y) becomes (-y, x - 4), so
(a) Feed in R_{A,90^\circ}(x, y) = (-y, x):
The turn part is (-x, -y), a half turn, and no reflection is involved, so this is a half turn about some center. Solve for the fixed point: 4 - x = x gives x = 2, and -y - 4 = y gives y = -2. The composition is the half turn about C(2, -2).
Check with P(1, 2). The first turn sends it to P' = (-2, 1), the second to P'' = (4 - 1,\ -2 - 4) = (3, -6). The midpoint of P and P'' is (2, -2), which is what a half turn about C requires.
The angles added, 90^\circ + 90^\circ = 180^\circ, as they would for one center. The center did not stay at A or at B: it moved to a third point.
(b) A quarter turn clockwise about B sends the offset (x - 4, y) to (y, 4 - x), so R_{B,-90^\circ}(x, y) = (y + 4,\ 4 - x). Feed in (-y, x):
The translation \langle 4, 4 \rangle. The angles cancel, 90^\circ - 90^\circ = 0^\circ, and when the turning cancels, what is left cannot be a turn. It has no fixed point at all, since x + 4 = x is impossible.
Finding a sequence, then a single motion
Triangle DEF has vertices D(3, 1), E(5, 1), F(3, 4), and triangle UVW has vertices U(8, 4), V(8, 6), W(5, 4), with D matching U, E matching V and F matching W. Find a sequence of rigid motions carrying \triangle DEF onto \triangle UVW, and then a single motion that does the same.
SolutionCompare directions from the matching corners. \overrightarrow{DE} runs 2 east and \overrightarrow{UV} runs 2 north. \overrightarrow{DF} runs 3 north and \overrightarrow{UW} runs 3 west. East to north and north to west are both a quarter turn counterclockwise, and both pairs agree, so the handedness is kept and no reflection is needed. Start with R_{O,90^\circ}, (x, y) \mapsto (-y, x):
The turned triangle now faces the right way, and all that is left is a slide. D is at (-1, 3) and has to reach U(8, 4), so the slide is \langle 9, 1 \rangle. Check the other two: (-1, 5) goes to (8, 6) = V and (-4, 3) goes to (5, 4) = W. The sequence is the quarter turn about the origin, then the translation \langle 9, 1 \rangle.
As one rule, (x, y) \mapsto (9 - y,\ x + 1). The turn part is a quarter turn, so this is a quarter turn about its fixed point. Solve 9 - y = x and x + 1 = y: substituting, 9 - (x + 1) = x, so x = 4 and y = 5. A single quarter turn counterclockwise about C(4, 5) carries \triangle DEF onto \triangle UVW. Check on D: the offset of D from C is (-1, -4); turned, it is (4, -1); added back to C, it gives (8, 4) = U.
The sequence was not the only one. Turning about D first and then sliding by \langle 5, 3 \rangle works too. The single motion, though, is unique: there is only one rigid motion that sends D, E and F to U, V and W.
Boundaries
The order is read from the right. g \circ f means f first. Two motions in the other order are a different composition, and in general a different motion. The cases where order does not matter (two translations, two turns about one center, a reflection and a slide along its mirror) are special, not the rule.
Undoing reverses the order. (g \circ f)^{-1} = f^{-1} \circ g^{-1}, never g^{-1} \circ f^{-1} unless f and g happen to commute.
The angles add, but the center stays put only when the turns share it. Two rotations about the same center make the rotation about that center through the sum of the angles. About different centers the angles still add, but the center moves, as in Example 8. When the sum is a whole number of full turns, such as 90^\circ - 90^\circ or 270^\circ + 90^\circ, the result is a translation, possibly the identity, and has no center at all.
A glide reflection needs the slide along the mirror. A slide of length zero leaves a plain reflection. A slide at an angle to the mirror still gives a glide reflection, but not with the mirror you started with: the part of the slide across the mirror moves the axis, as in Example 7. So the axis has to be found, from midpoints, rather than assumed.
The four-kind table is for rigid motions. A composition that includes a dilation with scale factor other than 1 changes lengths, so it is none of the four. A fixed point alone proves nothing either: (x, y) \mapsto (2x, 2y) fixes the origin and is not a rotation. The table may be used only once the motion is known to be rigid, as every composition of slides, flips and turns is.
A sequence is not unique, the motion is. Many sequences can carry one figure onto another, as in Example 9, and any of them is a correct answer. The single motion they all amount to is one motion, because a rigid motion of the plane is fixed by where it sends three points that are not on one line.
Handedness counts reflections only. Turns and slides never change it, however many there are. A sequence with three reflections gives a mirror image; a sequence with two does not.
Where this goes next
The next step is two reflections in a row, which composes the simplest motion with itself and finds that two flips make a turn or a slide. From there, every rigid motion turns out to be at most three reflections, which is what makes the four-kind table complete.
Composition is how congruence is defined: two figures are congruent when some sequence of slides, flips and turns carries one onto the other, and a transformational proof exhibits that sequence. Add a dilation to the sequence and the figures are similar instead.
The symmetries of a figure compose too. Two motions that each map a figure onto itself, a reflection and a turn, compose to a third one that does, and the glide reflection is the symmetry of every trail of footprints and of many border patterns.
In algebra, the same idea is composing functions and undoing them with inverse functions, with the same right-to-left reading and the same reversed order for an inverse. And when each motion that keeps the origin is written as a matrix, composing them is multiplying the matrices, which is one reason matrix multiplication does not commute.
Exercises
Part 1: following a point
- The point (2, 5) is slid by \langle -3, 1 \rangle and then reflected in the y-axis. Where does it end?
- The point (2, 5) is reflected in the y-axis and then slid by \langle -3, 1 \rangle. Where does it end?
- Triangle JKL has vertices J(0, 1), K(3, 1), L(3, 2). It is given a half turn about the origin, R_{O,180^\circ}, and then slid by \langle 2, 4 \rangle. Find the vertices of the image.
Part 1: composing rules
- Let f(x, y) = (x + 2,\ y - 5) and g(x, y) = (y, x). Find rules for g \circ f and f \circ g.
- A figure is slid by \langle 5, 3 \rangle and then by \langle -1, -8 \rangle. What single translation does the same, and what translation undoes the pair?
Part 1: does the order matter?
- Let r be the reflection in the line y = x and t the translation \langle 3, 3 \rangle. Find r \circ t and t \circ r. Does the order matter?
- Let r be the reflection in the y-axis and t the translation \langle 6, 0 \rangle. Find t \circ r and r \circ t, and describe each as a single reflection.
- Let R = R_{O,90^\circ} and t = \langle 1, 0 \rangle. Find rules for R \circ t and t \circ R, and say where each sends the origin.
Part 1: undoing
- A point is turned by R_{O,90^\circ} and then slid by \langle -4, 2 \rangle, and it ends at (1, 7). Where did it start?
- Find a rule for the inverse of g \circ f in Exercise 4.
Part 2: one motion for two
Name each composition as a single translation, rotation or reflection.
- R_{O,90^\circ} followed by the reflection in the x-axis.
- R_{O,180^\circ} followed by the reflection in the y-axis.
- R_{O,90^\circ} followed by the translation \langle 6, 0 \rangle.
- R_{O,180^\circ} followed by the translation \langle 6, -2 \rangle.
Part 2: glide reflections
- G is the glide reflection with axis y = 2 and slide \langle -3, 0 \rangle. Find the images of A(1, 5) and B(-2, 0), and check that each midpoint lies on the axis.
- Write a rule for the glide reflection G in Exercise 15, and show that G \circ G is a translation.
- Show that (x, y) \mapsto (x + 2,\ 6 - y) is a glide reflection, and find its axis and slide.
- Show that (x, y) \mapsto (4 - x,\ y + 5) is a glide reflection, and find its axis and slide.
- The translation \langle 4, 6 \rangle is followed by the reflection in the x-axis. Find the rule, and name the single motion with its axis and slide.
Part 2: turns about different centers
- Name the single motion R_{B,-90^\circ} \circ R_{O,-90^\circ}, where B = (0, 2).
- Name the single motion R_{B,-90^\circ} \circ R_{O,90^\circ}, where B = (2, 0).
Part 2: finding a sequence
- Find a sequence of rigid motions carrying \triangle ABC, with A(0, 0), B(3, 0), C(0, 2), onto \triangle DEF, with D(-4, 2), E(-4, 5), F(-2, 2), so that A, B, C go to D, E, F. Then name the single motion.
- Find a sequence of rigid motions carrying the same \triangle ABC onto \triangle PQR, with P(4, 2), Q(1, 2), R(4, 0), so that A, B, C go to P, Q, R. Then name a single motion that does it.
Reasoning
- Explain why a composition of two rigid motions is a rigid motion.
- Explain why the inverse of g \circ f is f^{-1} \circ g^{-1} and not g^{-1} \circ f^{-1}.
- Explain why a glide reflection with a non-zero slide is not a translation, a rotation or a reflection.
- A figure is reflected, turned, reflected, slid and reflected, in that order. Is the final figure a mirror image of the first? Which kinds of single motion could the sequence be?
- Show that a reflection followed by a slide along its mirror gives the same motion as the slide followed by the reflection.
Answers
- (1, 6). The slide gives (-1, 6), and the reflection changes the sign of x.
- (-5, 6). The reflection gives (-2, 5), and the slide gives (-5, 6). Compare Exercise 1.
- J'(2, 3), K'(-1, 3), L'(-1, 2). The half turn changes both signs, giving (0, -1), (-3, -1), (-3, -2), and the slide adds 2 and 4.
- (g \circ f)(x, y) = (y - 5,\ x + 2) and (f \circ g)(x, y) = (y + 2,\ x - 5). For g \circ f, swap the coordinates of (x + 2, y - 5); for f \circ g, slide the swapped pair (y, x).
- \langle 4, -5 \rangle, undone by \langle -4, 5 \rangle. The vectors add: 5 - 1 = 4 and 3 - 8 = -5.
- Both are (x, y) \mapsto (y + 3,\ x + 3), so the order does not matter. The slide runs along the mirror y = x; this is a glide reflection.
- t \circ r is (x, y) \mapsto (6 - x,\ y), the reflection in x = 3. r \circ t is (x, y) \mapsto (-x - 6,\ y), the reflection in x = -3. Midpoints of x and 6 - x are 3; of x and -x - 6, -3.
- (R \circ t)(x, y) = (-y,\ x + 1), sending the origin to (0, 1). (t \circ R)(x, y) = (1 - y,\ x), sending the origin to (1, 0). The order matters.
- (5, -5). Remove the slide: (1 + 4,\ 7 - 2) = (5, 5). Turn back clockwise with (x, y) \mapsto (y, -x): (5, -5).
- (x, y) \mapsto (y - 2,\ x + 5). Undo g first (it is its own inverse), then undo f by sliding \langle -2, 5 \rangle: (y, x) \mapsto (y - 2,\ x + 5).
- The reflection in the line y = -x. The quarter turn gives (-y, x) and the flip then gives (-y, -x), so the rule is (x, y) \mapsto (-y, -x).
- The reflection in the x-axis. The half turn gives (-x, -y) and the flip then gives (x, -y).
- The quarter turn counterclockwise about (3, 3). The rule is (6 - y,\ x); solving 6 - y = x, x = y gives (3, 3).
- The half turn about (3, -1). The rule is (6 - x,\ -2 - y), and 6 - x = x, -2 - y = y give (3, -1).
- A'(-2, -1) and B'(-5, 4). The reflection in y = 2 sends y to 4 - y, then 3 is subtracted from x. Midpoints \displaystyle \left(-\frac{1}{2}, 2\right) and \displaystyle \left(-\frac{7}{2}, 2\right), both on y = 2.
- G(x, y) = (x - 3,\ 4 - y), and G\big(G(x, y)\big) = (x - 6,\ y), the translation \langle -6, 0 \rangle.
- Axis y = 3, slide \langle 2, 0 \rangle. Midpoints are (x + 1, 3), all on y = 3; the reflection in y = 3 is (x, 6 - y), and the rule adds 2 to x. No point is fixed, since x + 2 \ne x.
- Axis x = 2, slide \langle 0, 5 \rangle. Midpoints are \displaystyle \left(2,\ y + \frac{5}{2}\right), all on x = 2; the reflection in x = 2 is (4 - x, y), and the rule adds 5 to y.
- (x, y) \mapsto (x + 4,\ -y - 6), the glide reflection with axis y = -3 and slide \langle 4, 0 \rangle. Midpoints are (x + 2, -3); the part \langle 0, 6 \rangle across the mirror moved it down by 3.
- The half turn about (-1, 1). R_{O,-90^\circ} is (y, -x) and R_{B,-90^\circ} is (y - 2,\ 2 - x), so the composition is (-x - 2,\ 2 - y), with fixed point (-1, 1).
- The translation \langle 2, 2 \rangle. R_{B,-90^\circ} is (y + 2,\ 2 - x); feeding in (-y, x) gives (x + 2,\ y + 2). The angles cancel.
- Reflect in the line y = x, then slide by \langle -4, 2 \rangle (one answer of many). The handedness is reversed (\overrightarrow{AB} east goes to north, \overrightarrow{AC} north goes to east), so a reflection is needed. (x, y) \mapsto (y, x) sends A, B, C to (0, 0), (0, 3), (2, 0), and the slide finishes the job. The whole, (x, y) \mapsto (y - 4,\ x + 2), is a glide reflection with axis y = x + 3 and slide \langle -1, -1 \rangle: the midpoints \displaystyle \left(\frac{x + y - 4}{2},\ \frac{x + y + 2}{2}\right) all lie on that line, and the reflection in it, (y - 3,\ x + 3), differs from the rule by \langle -1, -1 \rangle.
- A half turn about the origin, then a slide by \langle 4, 2 \rangle, which is the single half turn about (2, 1). East goes to west and north to south, so the handedness is kept. The half turn gives (0, 0), (-3, 0), (0, -2); the rule (4 - x,\ 2 - y) fixes (2, 1).
- If f and g keep distances, then the distance between g(f(P)) and g(f(Q)) equals the distance between f(P) and f(Q), because g keeps distances, and that equals PQ, because f does. So g \circ f keeps every distance.
- g \circ f does f first and g last, so g is the outermost motion and has to be removed first. Applied to g(f(P)), g^{-1} removes g and leaves f(P), and then f^{-1} leaves P. The other order would apply f^{-1} to g(f(P)), which only cancels if f and g commute.
- It reverses handedness, since it contains one reflection, so it is not a translation or a rotation, which keep it. With its axis as the x-axis it is (x + a, -y) with a \ne 0, which has no fixed point, while a reflection fixes every point of its mirror.
- Yes. There are three reflections, an odd number, so the handedness is reversed. By the table the sequence is a single reflection or a glide reflection.
- Take the mirror as the x-axis and the slide as \langle a, 0 \rangle. Reflecting first gives (x, -y) and then (x + a, -y). Sliding first gives (x + a, y) and then (x + a, -y). The rules are the same, so the motions are.