Geometry · 8.G.A.2 · G.CO.A.4

Two reflections in a row

You reflect a shape in one mirror line, then reflect the result in a second. Is there one single motion that does the same job?

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Two reflections in a row

Why this exists

A reflection is the plainest motion there is: one line, and every point jumps straight across it to the same distance on the other side. Slides and turns look like different kinds of motion altogether. A slide has a direction and a length, a turn has a center and an angle, and neither seems to have a mirror anywhere in it.

They are made of mirrors all the same. Reflect a figure in one line and then in a second, and the figure has been slid or turned, never left flipped. Which of the two you get depends only on whether the mirrors are parallel or cross, and how far the figure moves depends only on how far apart the mirrors are, or at what angle they meet.

That gives two things at once. You can say where two reflections leave a figure without tracking a single point through both. And since every slide and every turn is two reflections in a row, any fact proved for reflections, such as that they keep lengths and angles, holds at once for every slide and every turn too.

Definition

The reflection in a line \ell, written r_\ell, leaves every point of \ell where it is and sends every other point P to the point P' for which \ell is the perpendicular bisector of \overline{PP'}. The line \ell is called the mirror. Rigid motions defined precisely sets this out beside the other rigid motions.

Two reflections in a row means: reflect in \ell, then reflect the result in m. As with any composition of transformations, it is written with the first motion on the right. So r_m \circ r_\ell means first r_\ell, then r_m:

(r_m \circ r_\ell)(P) = r_m\big(r_\ell(P)\big)

Two different lines in a plane either never meet, and are parallel, or meet in exactly one point. What the two reflections do depends on one measurement in each case.

For parallel mirrors, the gap d between them, measured along a line perpendicular to both, together with which way you go to get from \ell to m.
For crossing mirrors that meet at O, the angle \theta you turn \ell through about O to land it on m, together with the direction of that turn, counterclockwise or clockwise.

Intuition

Take a shape that does not look the same in a mirror, such as a flag, and reflect it twice across parallel mirrors, as in Figure 1.

Figure 1Two flips across parallel mirrors. The flag ends up slid 2d, facing the way it started.

The first reflection flips the flag over, so it points the other way. The second flips it back, so the last flag faces exactly the way the first one did. Nothing is turned and nothing is left flipped. All that remains is a slide, straight across the mirrors.

The size of the slide comes from counting. Say the pole stands x to the left of \ell. The first reflection carries it x to reach \ell and x more beyond, so it now stands d - x to the left of m. The second carries it d - x to reach m and d - x more beyond. The whole trip is

x + x + (d - x) + (d - x) = 2d

and the x has dropped out. Every point of the flag moves the same 2d, whatever its distance from the mirrors, which is exactly what a slide does.

Now let the mirrors cross, as in Figure 2.

Figure 2Two flips across crossing mirrors. The flag ends up turned through 2\theta about O.

Again the second flip undoes the first, so the flag is not left flipped. But now each reflection swings the flag around O rather than across a strip, and the same counting, done with angles instead of distances, gives a turn of 2\theta about O.

Results

Parallel mirrors. If \ell \parallel m and the gap between them is d, then r_m \circ r_\ell is the translation through a distance 2d, perpendicular to the mirrors, in the direction from \ell toward m.

Crossing mirrors. If \ell and m cross at O, and turning \ell about O through the angle \theta lands it on m, then r_m \circ r_\ell is the rotation about O through 2\theta, in the same direction as that turn.

The order matters. Swapping the order reverses the motion. r_\ell \circ r_m is the translation through 2d from m toward \ell, or the rotation through 2\theta the other way. The two orders agree only when the mirrors are the same line, or when they are perpendicular, where both orders give the half turn about O, the rotation through 180^\circ.

One mirror twice. r_\ell \circ r_\ell moves nothing: every point is flipped across \ell and straight back.

Every slide and every turn is two reflections. The translation through a distance s in a given direction is r_m \circ r_\ell for any two lines perpendicular to that direction, a gap of \tfrac12 s apart, with m ahead of \ell in that direction. The rotation about O through an angle \varphi is r_m \circ r_\ell for any two lines through O with m turned \tfrac12\varphi from \ell, in the direction of the rotation. Only the gap or the angle between the mirrors matters, not where the pair sits, so there are infinitely many ways to choose them.

Flips cancel in pairs. A reflection reverses the direction in which a figure is lettered, clockwise to counterclockwise. Two reflections reverse it twice, so they keep it, as every slide and every turn does.

Why it is true

Parallel mirrors. Choose the axes so that both mirrors are vertical: \ell is the line x = a and m is the line x = b. The reflection in x = a sends (x, y) to (2a - x, y). That is the definition at work: the two points are level, so the segment joining them is horizontal and perpendicular to the mirror, and its midpoint has first coordinate \tfrac12\big(x + (2a - x)\big) = a, so it lies on the mirror. Do the two reflections in turn:

\begin{aligned} r_\ell(x, y) &= (2a - x,\ y) \\ r_m\big(r_\ell(x, y)\big) &= \big(2b - (2a - x),\ y\big) \\ &= \big(x + 2(b - a),\ y\big) \end{aligned}

Every point moves 2(b - a) across and 0 up. The gap is d = |b - a|, and the sign of b - a says whether m is to the right or the left of \ell, so this is a slide of 2d, perpendicular to the mirrors, from \ell toward m.

Crossing mirrors. Put the crossing point O at the center and describe each point by its distance from O and its direction, the angle from a fixed ray out of O to the ray through the point, counted counterclockwise. Let \ell have direction \lambda and m direction \mu, so the turn from \ell to m is \theta = \mu - \lambda.

First see what one reflection in a line through O does. O lies on the mirror, which is the perpendicular bisector of \overline{PP'}, and every point of a perpendicular bisector is equally far from the two ends, so OP' = OP. Let M be the midpoint of \overline{PP'}. Triangles OMP and OMP' have OM in common, MP = MP' and OP = OP', so they are congruent by SSS, and \angle POM = \angle MOP'. (If M is O itself, there are no triangles: P and P' sit on the line through O perpendicular to the mirror, a right angle either side of it, and the same conclusion holds.) The ray OP' is therefore as far past the mirror on one side as OP is on the other. If P has direction \varphi, it sits \lambda - \varphi short of the mirror \ell, so P' sits \lambda - \varphi beyond it, at direction \lambda + (\lambda - \varphi) = 2\lambda - \varphi.

Figure 3Each reflection copies an angle to the other side of its mirror. Since \alpha + \beta = \theta, the point turns through \alpha + \alpha + \beta + \beta = 2\theta.

Now do the two reflections in turn. Distances from O never change, and directions go

\begin{aligned} \text{after } r_\ell:&\quad 2\lambda - \varphi \\ \text{after } r_m:&\quad 2\mu - (2\lambda - \varphi) \\ &\quad = \varphi + 2(\mu - \lambda) = \varphi + 2\theta \end{aligned}

Every point stays the same distance from O and has its direction increased by 2\theta. That is the rotation about O through 2\theta. Figure 3 shows the same count with the angles drawn: the angle \alpha from OP up to \ell is copied to the far side of \ell, the angle \beta from OP' up to m is copied to the far side of m, and \alpha + \beta = \theta.

The rest. Swapping the order swaps \lambda and \mu (or a and b), which changes the sign of the turn or the slide. Using one mirror twice makes \theta = 0, or b - a = 0, which moves nothing. And the last result runs the first two backwards: two lines perpendicular to the direction of travel, \tfrac12 s apart, give a slide of 2 \cdot \tfrac12 s = s; two lines through O at a turn of \tfrac12\varphi give a rotation through 2 \cdot \tfrac12\varphi = \varphi.

Worked examples

Example 1

Two vertical mirrors

Triangle ABC has vertices A(1, 1), B(3, 1) and C(1, 3). Reflect it in the line x = 4, then reflect the image in the line x = 8. Find the final triangle, and the single motion that does the same job.

Figure 4Reflect in x = 4, then in x = 8. Only the middle triangle is flipped; the last one is the first slid 8 to the right.

Solution

The reflection in x = 4 sends (x, y) to (8 - x, y): the point and its image are level, and 4 is halfway between x and 8 - x.

\begin{aligned} A(1, 1) &\mapsto A'(7, 1) \\ B(3, 1) &\mapsto B'(5, 1) \\ C(1, 3) &\mapsto C'(7, 3) \end{aligned}

The reflection in x = 8 sends (x, y) to (16 - x, y) for the same reason:

\begin{aligned} A'(7, 1) &\mapsto A''(9, 1) \\ B'(5, 1) &\mapsto B''(11, 1) \\ C'(7, 3) &\mapsto C''(9, 3) \end{aligned}

Compare the first triangle with the last: every vertex has moved 8 to the right and 0 up. The two mirrors are parallel with a gap of 4, so the result for parallel mirrors predicts a slide of 2 \cdot 4 = 8, perpendicular to the mirrors, from x = 4 toward x = 8. The single motion is the translation \langle 8, 0 \rangle.

Check. The middle triangle is lettered the other way round from ABC, since one reflection flips it. A''B''C'' is lettered the same way as ABC, as a slide must be.

In the other order the slide reverses: x = 8 first sends A(1, 1) to (15, 1), and then x = 4 sends that to (-7, 1), which is A moved 8 to the left.

Example 2

A mirror on an axis and a mirror on the diagonal

Reflect P(3, 1) in the x-axis, then in the line y = x. Then do the same to a general point (x, y), and say what single motion the pair of reflections is. What happens in the other order?

Solution

The reflection in the x-axis changes the sign of the second coordinate. The reflection in y = x swaps the two coordinates: the segment from (x, y) to (y, x) has slope -1, so it is perpendicular to y = x, and its midpoint \big(\tfrac{x + y}{2}, \tfrac{x + y}{2}\big) lies on y = x. So

P(3, 1) \mapsto (3, -1) \mapsto (-1, 3)

and for a general point

(x, y) \mapsto (x, -y) \mapsto (-y, x)

The mirrors cross at the origin, and turning the x-axis 45^\circ counterclockwise lands it on y = x. The result for crossing mirrors predicts the rotation about the origin through 2 \cdot 45^\circ = 90^\circ counterclockwise, and that rotation does send (x, y) to (-y, x).

Check. P and its image are both \displaystyle \sqrt{3^2 + 1^2} = \sqrt{10} from the origin, as a rotation about the origin requires.

In the other order, (x, y) \mapsto (y, x) \mapsto (y, -x), so P(3, 1) goes to (1, -3). That is the rotation through 90^\circ clockwise: swapping the mirrors reversed the turn.

Example 3

Perpendicular mirrors away from the origin

Reflect P(5, 3) in the line y = 1, then in the line x = 2. Show that the two reflections make a half turn, and find its center. Does the order matter here?

Solution

The reflection in y = 1 sends (x, y) to (x, 2 - y), and the reflection in x = 2 sends (x, y) to (4 - x, y), by the same halfway reasoning as in Example 1. So

P(5, 3) \mapsto (5, -1) \mapsto (-1, -1)

and for a general point

(x, y) \mapsto (x, 2 - y) \mapsto (4 - x, 2 - y)

The midpoint of (x, y) and (4 - x, 2 - y) is (2, 1) for every point, so each point is sent straight through (2, 1) to the same distance on the other side. That is the half turn about (2, 1), which is where the mirrors cross. The result predicts exactly this: the mirrors are perpendicular, so the turn is 2 \cdot 90^\circ = 180^\circ.

In the other order, (x, y) \mapsto (4 - x, y) \mapsto (4 - x, 2 - y), the same point. For perpendicular mirrors the order does not matter, because a half turn clockwise and a half turn counterclockwise are the same motion.

Example 4

Which angle to double

Lines \ell and m cross at O, making angles of 70^\circ and 110^\circ. Turning \ell counterclockwise about O, you reach m after 110^\circ. Describe r_m \circ r_\ell as a single rotation.

Solution

The angle to double is the turn from the first mirror to the second, with its direction. Turning counterclockwise, that turn is 110^\circ, so the result gives a rotation about O through 2 \cdot 110^\circ = 220^\circ counterclockwise.

But \ell also lands on m by turning the other way: clockwise, it reaches m after 70^\circ, because a line turned through 180^\circ is the same line and 110 + 70 = 180. Read that way, the result gives a rotation through 2 \cdot 70^\circ = 140^\circ clockwise.

The two answers agree: turning 220^\circ counterclockwise ends in the same place as turning 360^\circ - 220^\circ = 140^\circ clockwise. So r_m \circ r_\ell is the rotation about O through 140^\circ clockwise, the shorter way to say it. Doubling the 70^\circ angle counterclockwise would be wrong: the direction belongs to the turn from \ell to m, not to whichever angle at the crossing is smaller.

Example 5

A rotation taken apart into mirrors

Write the rotation through 90^\circ counterclockwise about C(1, 2) as two reflections, and test the answer on P(3, 2).

Solution

By the last result, any two lines through C will do, provided the second is turned \tfrac12 \cdot 90^\circ = 45^\circ counterclockwise from the first. Choose the first to make the arithmetic easy: \ell is the vertical line x = 1, which has direction 90^\circ. Turning it 45^\circ counterclockwise gives direction 135^\circ, a line of slope -1 through (1, 2):

\begin{aligned} y - 2 &= -(x - 1) \\ y &= -x + 3 \end{aligned}

So the rotation is r_m \circ r_\ell with \ell the line x = 1 and m the line y = -x + 3.

Test. The reflection in x = 1 sends (x, y) to (2 - x, y), so P(3, 2) \mapsto (-1, 2). The reflection in y = -x + 3 sends (x, y) to (3 - y, 3 - x): the segment joining them has slope 1, perpendicular to the mirror, and its midpoint \big(\tfrac{x - y + 3}{2}, \tfrac{y - x + 3}{2}\big) satisfies y = -x + 3. So (-1, 2) \mapsto (3 - 2,\ 3 + 1) = (1, 4).

Directly, P is 2 to the right of C. A quarter turn counterclockwise about C puts it 2 above C, at (1, 4). The two reflections agree with the rotation.

Example 6

Two half turns make a slide

Let A = (1, 1) and B = (4, 3). Show that the half turn about A followed by the half turn about B is a translation, find it, and check it on P(2, 5).

Solution

Write each half turn as two reflections, choosing the mirrors so that one of them is shared. Let n be the line AB, let p be the line through A perpendicular to n, and let q be the line through B perpendicular to n. A half turn about a point is two reflections in perpendicular lines through it, in either order, so

\begin{aligned} \text{half turn about } A &= r_n \circ r_p \\ \text{half turn about } B &= r_q \circ r_n \end{aligned}

The half turn about A comes first, so it sits on the right:

\begin{aligned} &(r_q \circ r_n) \circ (r_n \circ r_p) \\ &= r_q \circ (r_n \circ r_n) \circ r_p \\ &= r_q \circ r_p \end{aligned}

Regrouping is allowed because composition of transformations is associative, and r_n \circ r_n moves nothing, so it drops out. What is left is two reflections in p and q, which are both perpendicular to n and so are parallel, with a gap of AB between them. By the result for parallel mirrors, this is the translation through 2 \cdot AB, along n, from A toward B: twice the step from A to B, which is 2\langle 3, 2 \rangle = \langle 6, 4 \rangle.

Check. The half turn about A sends a point to the far end of a segment with midpoint A, so P(2, 5) \mapsto (2 \cdot 1 - 2,\ 2 \cdot 1 - 5) = (0, -3). The half turn about B then gives (2 \cdot 4 - 0,\ 2 \cdot 3 + 3) = (8, 9). And P moved by \langle 6, 4 \rangle is (2 + 6, 5 + 4) = (8, 9).

The mirrors never had to be found as equations. Choosing them to share a line made the middle pair cancel, which is the main way two reflections are used in a proof.

Boundaries

Two reflections are never a reflection. A reflection flips the way a figure is lettered, and two flips cancel, so a pair of reflections always gives a slide, a turn, or nothing. A single reflection, or a reflection followed by a slide along its mirror, needs an odd number of reflections.

The order matters unless the mirrors are perpendicular or the same. r_m \circ r_\ell and r_\ell \circ r_m are the same motion run backwards.

Double the turn from the first mirror to the second, with its direction. A line turned through 180^\circ is the same line, so the turn from \ell to m can be read either way round, as in Example 4. Both readings give the same rotation, because the doubled angles differ by 360^\circ. Doubling the smaller angle at the crossing without regard to direction can give the wrong turn.

Only the end agrees, not the journey. The single slide or turn lands every point where the two reflections land it. It does not pass through the flipped middle image, which no slide or turn ever reaches.

The mirrors are not determined by the motion. A given slide or turn comes from infinitely many pairs of mirrors, so asking for “the” two mirrors of a rotation has no single answer until one of them is chosen.

In space the mirrors are planes. Two reflections in parallel planes make a slide, and two reflections in planes that cross make a rotation about their line of intersection. The lines in this lesson are what you see in one plane cut straight across both.

Where this goes next

The two results turn questions about slides and turns into questions about mirrors, and back. They explain why a translation and a rotation, including a rotation about a point other than the origin, keep every length and every angle: each one is two reflections, and a reflection keeps both. That is the route the precise definitions of the rigid motions take, and it is why congruence by rigid motion can be checked with mirrors alone.

Symmetry is the same fact seen in a figure. A figure with two lines of symmetry crossing at an angle \theta is carried onto itself by each reflection, so it is also carried onto itself by the turn through 2\theta: two mirror lines at 45^\circ in a square give its quarter-turn symmetry.

Example 6 points at the larger game. Writing motions as mirrors and canceling a shared mirror in the middle shows that two half turns make a slide and that two turns about one center add up, and it is the main tool in a transformational proof. Going one mirror further, three reflections can make the motions two cannot: a reflection, or a reflection followed by a slide along its mirror.

Exercises

Parallel mirrors

Reflect P(2, 5) in the line x = 3, then in the line x = 6. Find the final image and the single translation that does the same job.
Reflect Q(-1, 4) in the line y = 1, then in the line y = -2. Find the final image and the single translation.
Reflect P(2, 5) in the line x = 6 first, then in the line x = 3. Find the final image and the single translation, and compare with Exercise 1.
Reflecting in the line x = 2 and then in the line x = k gives the translation \langle -8, 0 \rangle. Find k.
Reflect R(0, 0) in the line y = x, then in the line y = x + 4. Find the final image and the single translation, and check that its length is twice the gap between the lines.

Crossing mirrors

Lines \ell and m cross at O, and turning \ell through 25^\circ counterclockwise about O lands it on m. Describe r_m \circ r_\ell and r_\ell \circ r_m.
Reflect (4, 1) in the y-axis, then in the x-axis. Find the image and name the single motion.
Reflect (2, 5) in the line y = x, then in the y-axis. Find the image and name the single motion.
Reflect (3, 0) in the y-axis, then in the line y = x. Find the image and name the single motion.
Reflect (5, 7) in the line x = 1, then in the line y = 4. Find the image and name the single motion.
Lines \ell and m cross at O, making angles of 60^\circ and 120^\circ. Turning \ell counterclockwise about O, you reach m after 120^\circ. Describe r_m \circ r_\ell in two ways, once as a counterclockwise turn and once as a clockwise one.

Building a motion from mirrors

Write the translation \langle 8, 0 \rangle as two reflections in vertical lines, one of which is x = 1. There are two answers; find both.
Write the rotation through 60^\circ counterclockwise about the origin as the reflection in the y-axis followed by the reflection in a second line. Give the second line’s equation.
Write the half turn about (2, -1) as two reflections in lines whose equations you give.
r_m \circ r_\ell is the rotation about O through 100^\circ counterclockwise. Through what angle, and which way, does \ell turn about O to land on m?
Write the translation \langle 3, 4 \rangle as two reflections, the first in the line 3x + 4y = 0. Give the equation of the second mirror.
Do the half turn about A(0, 0), then the half turn about B(2, -1). Find the image of P(5, 5) and the single translation the two half turns make.

Reasoning

Explain why two reflections in a row can never be a single reflection.
Lines \ell and m are different, and r_m \circ r_\ell = r_\ell \circ r_m. Show that \ell and m are perpendicular.
Mirrors \ell and m cross at 45^\circ. A flag is reflected in \ell, then m, then \ell, then m, and so on, alternating. After how many reflections is it back where it started for the first time? Explain.
Show, using mirrors, that the rotation about O through \alpha followed by the rotation about the same O through \beta is the rotation about O through \alpha + \beta.
Lines \ell and m are different. Which points does r_m \circ r_\ell leave where they are? Answer for crossing mirrors and for parallel mirrors.

Answers

(8, 5), and the translation \langle 6, 0 \rangle. In x = 3: (2 \cdot 3 - 2, 5) = (4, 5). In x = 6: (12 - 4, 5) = (8, 5). The gap is 3, so the slide is 2 \cdot 3 = 6, from x = 3 toward x = 6.
(-1, -2), and the translation \langle 0, -6 \rangle. In y = 1: (-1, 2 - 4) = (-1, -2). That point lies on the second mirror y = -2, so the second reflection leaves it there. The gap is 3 and the second mirror is below the first, so the slide is 6 down.
(-4, 5), and the translation \langle -6, 0 \rangle. In x = 6: (12 - 2, 5) = (10, 5). In x = 3: (6 - 10, 5) = (-4, 5). These are the mirrors of Exercise 1 in the other order, so the slide is reversed.
k = -2. The slide is 2(k - 2) across, so 2(k - 2) = -8 and k - 2 = -4.
(-4, 4), and the translation \langle -4, 4 \rangle. R lies on y = x, so the first reflection leaves it there. The reflection in y = x + 4 sends (x, y) to (y - 4, x + 4), since the segment joining them has slope -1 and its midpoint lies on the line; so (0, 0) \mapsto (-4, 4). The gap between the lines is \displaystyle \frac{4}{\sqrt 2} = 2\sqrt 2, and the slide has length \displaystyle \sqrt{16 + 16} = 4\sqrt 2, twice the gap, at right angles to the mirrors.
r_m \circ r_\ell is the rotation about O through 50^\circ counterclockwise, since 2 \cdot 25 = 50. r_\ell \circ r_m is the rotation about O through 50^\circ clockwise, because swapping the order reverses the turn.
(-4, -1), the half turn about the origin. (4, 1) \mapsto (-4, 1) \mapsto (-4, -1). The axes are perpendicular, so the turn is 2 \cdot 90^\circ = 180^\circ.
(-5, 2), the rotation through 90^\circ counterclockwise about the origin. (2, 5) \mapsto (5, 2) \mapsto (-5, 2). Turning y = x through 45^\circ counterclockwise lands it on the y-axis.
(0, -3), the rotation through 90^\circ clockwise about the origin. (3, 0) \mapsto (-3, 0) \mapsto (0, -3). Turning the y-axis through 45^\circ clockwise lands it on y = x.
(-3, 1), the half turn about (1, 4). (5, 7) \mapsto (2 - 5, 7) = (-3, 7), then (-3, 8 - 7) = (-3, 1). The mirrors are perpendicular and cross at (1, 4), and the midpoint of (5, 7) and (-3, 1) is (1, 4).
The rotation about O through 240^\circ counterclockwise, which is the same as 120^\circ clockwise. Counterclockwise the turn from \ell to m is 120^\circ, giving 2 \cdot 120 = 240. Clockwise it is 60^\circ, giving 2 \cdot 60 = 120, and 360 - 240 = 120, so the two readings agree.
x = 1 then x = 5, or x = -3 then x = 1. The gap must be 4, half of 8, with the second mirror to the right of the first: 2(5 - 1) = 8 and 2(1 - (-3)) = 8.
\displaystyle y = -\sqrt{3}\,x. The second mirror passes through the origin turned 30^\circ counterclockwise from the y-axis, so its direction is 90^\circ + 30^\circ = 120^\circ and its slope is \displaystyle \tan 120^\circ = -\sqrt 3. Check with (0, 1): the first reflection leaves it, the second sends it to direction 2 \cdot 120^\circ - 90^\circ = 150^\circ, which is (0, 1) turned through 60^\circ.
The lines x = 2 and y = -1, in either order. They are perpendicular and cross at (2, -1): (x, y) \mapsto (4 - x, y) \mapsto (4 - x, -2 - y), and the midpoint of (x, y) and (4 - x, -2 - y) is (2, -1). Any two perpendicular lines through (2, -1) also work.
50^\circ counterclockwise, since the rotation is twice the turn from \ell to m and 2t = 100 gives t = 50. Turning 130^\circ clockwise lands \ell on the same line, so that answer is equally right.
\displaystyle 3x + 4y = \frac{25}{2}, that is 6x + 8y = 25. Both mirrors must be perpendicular to \langle 3, 4 \rangle, so both have the form 3x + 4y = c, with a gap of half its length 5, which is \displaystyle \frac52. The line 3x + 4y = c is \displaystyle \frac{c}{5} from the origin on the side \langle 3, 4 \rangle points to, so \displaystyle \frac{c}{5} = \frac52. Check: the origin is on the first mirror, and the second sends it to (3, 4).
(9, 3), and the translation \langle 4, -2 \rangle. The half turn about A sends (5, 5) to (-5, -5), and the half turn about B sends that to (2 \cdot 2 + 5,\ 2 \cdot (-1) + 5) = (9, 3). As in Example 6, the translation is twice the step from A to B: 2\langle 2, -1 \rangle = \langle 4, -2 \rangle.
A reflection reverses the direction in which a figure is lettered, clockwise to counterclockwise. Two reflections reverse it twice, so they leave it as it was. A single reflection would reverse it, so the pair cannot be one.
If the lines were parallel, the two orders would be slides of 2d in opposite directions, equal only if d = 0, which would make the lines the same. So they cross at some O, and the two orders are rotations about O through 2\theta and -2\theta, where \theta is the turn from \ell to m, between 0^\circ and 180^\circ. They are equal only if 4\theta is a whole turn, and with 0 < 4\theta < 720 that means 4\theta = 360, so \theta = 90^\circ and the lines are perpendicular.
After 8 reflections. Each pair, \ell then m, is a rotation through 2 \cdot 45^\circ = 90^\circ about the crossing point, so k pairs turn the flag through 90k degrees, which is first a whole turn at k = 4. (If the turn from \ell to m is 45^\circ the other way, each pair turns 90^\circ the other way, and the count is the same.) After an odd number of reflections the flag is flipped, so it cannot be back where it started.
Choose any line \ell through O. Let m be the line through O turned \tfrac12\alpha from \ell, and n the line through O turned \tfrac12\beta from m. Then the first rotation is r_m \circ r_\ell and the second is r_n \circ r_m, since a rotation’s first mirror may be any line through its center. Doing them in order gives (r_n \circ r_m) \circ (r_m \circ r_\ell), which by associativity is r_n \circ (r_m \circ r_m) \circ r_\ell, and that is r_n \circ r_\ell, as r_m \circ r_m moves nothing. n is turned \tfrac12\alpha + \tfrac12\beta from \ell, so this is the rotation through \alpha + \beta.
Crossing at O: only O. The motion is the rotation about O through 2\theta with 0^\circ < 2\theta < 360^\circ, which keeps every point’s distance from O but changes the direction of every point other than O. Parallel: no point at all, since a slide of 2d with d > 0 moves every point. A point of \ell is left alone by r_\ell, but r_m then moves it unless it is also on m.