Geometry · G.CO.A.2 · G.CO.A.4

Rigid motions defined precisely, and what they preserve

Slide, flip and turn are easy to picture. Where exactly does each one send a single point, and how can we be sure nothing gets stretched?

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▴ 6 above Rotating about a point other than the origin What rigid motions preserve Rigid motions and congruence

Rigid motions defined precisely, and what they preserve

Two reflections in a row

Why this exists

“Slide it”, “flip it” and “turn it” are good enough for moving a paper triangle across a desk. They are not good enough for a proof. A proof needs to know, for every point P of the plane, exactly which point P lands on, and that answer has to be built from objects geometry has already pinned down: segments, parallel and perpendicular lines, and circles.

What rigid motions preserve listed what survives a slide, a flip or a turn: lengths, angle measures, betweenness and parallel lines. Those facts were read off the picture. This lesson defines each motion precisely enough that the facts can be proved, and precisely enough that a claim like “this rule is a rotation” can be checked instead of eyeballed. Exact definitions are what congruence by rigid motion and transformational proof are built on.

Definition

A transformation of the plane is a rule that assigns to every point P of the plane exactly one point P', called the image of P. It is a function whose inputs and outputs are points. If the transformation is called T, we write T(P) = P'. A point with T(P) = P is a fixed point of T.

The rule acts on the whole plane, not only on the figure in front of you. The image of a triangle is just the set of images of its points.

A transformation T is a rigid motion, also called an isometry, when it keeps every distance:

P'Q' = PQ \quad\text{for every pair of points } P,\ Q

The word “every” carries the weight. A rule that keeps some distances and changes others is not a rigid motion.

The three basic rigid motions are defined next. Each definition says where an arbitrary point goes, using only a directed segment, a perpendicular bisector or a circle.

Translation

Fix a directed segment from A to B. The translation by \overrightarrow{AB}, written T_{\overrightarrow{AB}}, sends each point P to the point P' for which the trip from P to P' has the same length as the trip from A to B and points the same way.

When P is not on line AB, that says exactly that ABP'P is a parallelogram: \overline{PP'} is parallel to \overline{AB} and \overline{BP'} is parallel to \overline{AP}. When P is on line AB, the point P' is on that line too, a distance AB further along in the direction from A to B.

Figure 1The translation by \overrightarrow{AB}. Every point makes the same trip as A to B, so ABP'P is a parallelogram.

In coordinates, say the trip from A to B is a across and b up, and the trip from P to P' is c across and d up. Pointing the same way means the two trips run along parallel lines in the same sense. If neither is vertical, they share a slope m, so b = ma and d = mc, and the distance formula gives their lengths as \displaystyle |a|\sqrt{1 + m^2} and \displaystyle |c|\sqrt{1 + m^2}. Equal lengths force |c| = |a|, the same sense makes c and a the same sign, so c = a, and then d = b. If both are vertical, a = c = 0, equal lengths force |d| = |b|, and the same sense gives d = b. Either way P' is P moved by the same change as A to B:

T_{\overrightarrow{AB}}(x, y) = (x + a,\ y + b)

Describing translations this way is the business of translations.

Reflection

Fix a line \ell, the mirror. The reflection in \ell, written r_\ell, sends each point P as follows.

If P is on \ell, then P' = P.
If P is not on \ell, then P' is the point for which \ell is the perpendicular bisector of \overline{PP'}.

So \overline{PP'} crosses \ell at a right angle, at a point M that is the midpoint of \overline{PP'}. To locate P', run the perpendicular from P to \ell, meeting it at M, and carry on the same distance again past M.

Figure 2The reflection in \ell. The mirror is the perpendicular bisector of \overline{PP'}, and points on the mirror stay where they are.

In coordinates, M is the midpoint of \overline{PP'}, so each coordinate of P' is twice that of M minus that of P. The case where the mirror is an axis is reflections over the axes.

Rotation

Fix a point O, the center, and an angle \theta. The rotation about O through \theta, written R_{O,\theta}, sends each point P as follows.

O stays where it is.
If P \neq O, then P' is the point on the circle with center O through P that is reached when the ray \overrightarrow{OP} turns through \theta about O: counterclockwise when \theta > 0, clockwise when \theta < 0.

So OP' = OP, since both are radii of one circle, and for 0^\circ < \theta \le 180^\circ the angle \angle POP' measures \theta.

Figure 3The rotation about O through \theta. P rides round its own circle until its ray has turned through \theta.

A rotation through 180^\circ is a half turn. The ray \overrightarrow{OP} turns to point the opposite way, so P, O and P' lie on one line with O between them and OP' = OP: the center is the midpoint of \overline{PP'}. Clockwise and counterclockwise give the same half turn. A rotation through 0^\circ or 360^\circ moves nothing. More on centers and directions is in rotations.

Intuition

Each definition hands the work to an object that already has the right property built in.

A translation is a parallelogram. The trip from P to P' is a copy of the trip from A to B, and since every point makes the same trip, no two points can drift apart or closer together.

A reflection is a perpendicular bisector. Folding the page along \ell lays P on top of P', and the perpendicular bisector of \overline{PP'} is exactly the locus of points equally far from P and P', which is what a fold line has to be.

A rotation is a circle and an angle. Every point rides round its own circle about O, and all of them turn by the same angle, like dots painted on a spinning plate. A point’s distance to O cannot change because a circle has one radius, and the distance between two dots cannot change because they turn together.

None of these pictures needs a grid. Coordinates are a handy way to calculate with the definitions, as the examples show, but the definitions come first and the coordinate rules are consequences.

Results

The basic motions are rigid. Every translation, every reflection and every rotation keeps distance: P'Q' = PQ for all points P and Q.

Rigid motions keep angle measure. If T is a rigid motion, then m\angle P'Q'R' = m\angle PQR for every angle \angle PQR.

Rigid motions keep betweenness. If Q lies between P and R, then Q' lies between P' and R'. So a segment is carried onto a segment of the same length, a ray onto a ray, a line onto a line, and a circle onto a circle of the same radius (Example 7).

Each basic motion is undone by one of the same kind.

\begin{aligned} T_{\overrightarrow{AB}} &\ \text{is undone by}\ T_{\overrightarrow{BA}} \\ r_\ell &\ \text{is undone by}\ r_\ell \\ R_{O,\theta} &\ \text{is undone by}\ R_{O,-\theta} \end{aligned}

Fixed points tell them apart. A translation by a segment of nonzero length fixes no point. A reflection fixes exactly the points of its mirror. A rotation through an angle that is not a multiple of 360^\circ fixes exactly one point, its center.

Why it is true

Translations keep distance. By the coordinate rule, a translation adds the same a to every x-coordinate and the same b to every y-coordinate. For P = (x_1, y_1) and Q = (x_2, y_2) the differences between the coordinates do not change:

\begin{aligned} (x_2 + a) - (x_1 + a) &= x_2 - x_1 \\ (y_2 + b) - (y_1 + b) &= y_2 - y_1 \end{aligned}

The distance formula uses nothing but these two differences, so P'Q' = PQ. Nothing beyond the definition and the distance formula was used.

Reflections keep distance. Coordinate axes can be placed anywhere, so put the x-axis along \ell. For P = (x, y) the perpendicular to \ell is the vertical line through P, meeting \ell at M = (x, 0), and going the same distance past M gives P' = (x, -y). Points of \ell have y = 0 and stay put, as the definition requires. Now for P = (x_1, y_1) and Q = (x_2, y_2),

\begin{aligned} (P'Q')^2 &= (x_2 - x_1)^2 + (-y_2 + y_1)^2 \\ &= (x_2 - x_1)^2 + (y_2 - y_1)^2 \\ &= (PQ)^2 \end{aligned}

because a number and its negative have the same square.

Rotations keep distance. Take two points P and Q. If one of them is the center, say Q = O, then P'O = PO because P' is on the circle through P centered at O.

For a half turn or a quarter turn, coordinates finish the job straight from the definition. Put the origin at O, so each point’s coordinates are its offset from the center. A half turn makes O the midpoint of \overline{PP'}, so (x, y) goes to (-x, -y). A counterclockwise quarter turn keeps the length \displaystyle \sqrt{x^2 + y^2} and turns the ray through a right angle. A ray at right angles to the offset (x, y) runs along a multiple of (-y, x), since the slopes \tfrac{y}{x} and -\tfrac{x}{y} multiply to -1, and keeping the length leaves only (-y, x) and (y, -x). The first is the counterclockwise one (right turns to up, up turns to left), so (x, y) goes to (-y, x). For the half turn,

(P'Q')^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 = (PQ)^2

and for the quarter turn,

(P'Q')^2 = (y_1 - y_2)^2 + (x_2 - x_1)^2 = (PQ)^2

A clockwise quarter turn sends (x, y) to (y, -x), and the same computation works.

For any other angle, the definition supplies three facts: OP' = OP, OQ' = OQ, and the angle between the two rays is unchanged. To see the last, measure the direction of each ray from O as an angle counterclockwise from some fixed ray. If \overrightarrow{OP} points at p^\circ and \overrightarrow{OQ} at q^\circ, then \overrightarrow{OP'} points at (p + \theta)^\circ and \overrightarrow{OQ'} at (q + \theta)^\circ. The gap between the directions is still q - p, so the angle between the rays has not changed:

m\angle P'OQ' = m\angle POQ

Figure 4Both rays turn by the same angle, so the angle between them stays \alpha.

Turning those three facts into P'Q' = PQ is exactly the job of SAS. If P, O and Q do not lie on one line, triangles POQ and P'OQ' have two pairs of equal sides and equal angles between them, so they are congruent by SAS, and P'Q' = PQ. If P, O and Q do lie on one line, the angle between the rays is 0^\circ or 180^\circ both before and after, so P', O and Q' lie on one line too, and PQ and P'Q' are both the sum of the two radii or both their difference.

Be clear about what that step rests on. In the congruence criteria SAS is a postulate, accepted without proof, so the argument stands there. The route through rigid motions and congruence runs the other way and proves SAS from rigid motions, so on that route this step would argue in a circle; there “every rotation keeps distance” is itself a starting assumption. A proof for every angle from coordinates alone needs the sine and cosine of \theta, which come with the unit circle.

Angles. Let T be any rigid motion and \angle PQR an angle, with P and R on its two rays. T keeps each of the three distances PQ, QR and PR, so triangles PQR and P'Q'R' have three pairs of equal sides. They are congruent by SSS, and corresponding angles are equal: m\angle P'Q'R' = m\angle PQR. The same caution applies. SSS is a postulate in the congruence criteria. On the rigid-motion route, “translations, reflections and rotations keep angle measure” is a second starting assumption, SSS is proved from the two assumptions, and only then does this argument carry the fact to every rigid motion.

Betweenness. A point Q lies on \overline{PR} exactly when PQ + QR = PR. For a point on the segment this is segment addition, and for a point off it the triangle inequality makes the sum larger. A rigid motion keeps all three distances, so P'Q' + Q'R' = P'R', and Q' lies on \overline{P'R'}. Running the same argument with the motion that undoes T shows that every point of \overline{P'R'} is reached, so the segment goes onto the whole segment. Rays and lines are segments extended without end, and they follow.

Undoing. Moving by the trip from B to A brings every point back. Reflecting P' in \ell asks for the point that makes \ell the perpendicular bisector of \overline{P'P}, which is P. Turning back through -\theta returns every ray to where it started.

Fixed points. A nonzero translation moves every point a distance AB > 0. A reflection sends a point off \ell to the far side of \ell, so only points on \ell stay. A rotation leaves O alone, and a point P \neq O stays only if its ray turns all the way back to itself, which takes a multiple of 360^\circ.

Worked examples

Example 1

Translating a point from the definition

A translation carries A(1, 2) to B(4, -1). Find the image P' of P(-2, 3), and check that ABP'P is a parallelogram.

Solution

The trip from A to B changes x by 4 - 1 = 3 and y by -1 - 2 = -3. Every point makes the same trip, so

P' = (-2 + 3,\ 3 - 3) = (1, 0)

To check the parallelogram, use the fact that a quadrilateral whose diagonals bisect each other is a parallelogram. The diagonals of ABP'P are \overline{AP'} and \overline{BP}:

\begin{aligned} \text{midpoint of } \overline{AP'} &= \left(\tfrac{1 + 1}{2},\ \tfrac{2 + 0}{2}\right) = (1, 1) \\ \text{midpoint of } \overline{BP} &= \left(\tfrac{4 - 2}{2},\ \tfrac{-1 + 3}{2}\right) = (1, 1) \end{aligned}

They share a midpoint, and the four points are not on one line (line AB is y = -x + 3, and P is not on it since -(-2) + 3 = 5 \neq 3), so ABP'P is a parallelogram.

Check. The trip has the right length: \displaystyle PP' = \sqrt{3^2 + 3^2} = 3\sqrt{2}, and \displaystyle AB = \sqrt{3^2 + (-3)^2} = 3\sqrt{2} as well.

Example 2

Reflecting in a slanted line

Reflect P(5, 0) in the line \ell: y = x + 1.

Solution

P is not on \ell, since 0 \neq 5 + 1, so the definition asks for the point P' that makes \ell the perpendicular bisector of \overline{PP'}.

First, \overline{PP'} is perpendicular to \ell. The mirror has slope 1, so the segment has slope -1, because slopes of perpendicular lines multiply to -1. The line through P with that slope is

y = -(x - 5) = -x + 5

Second, the midpoint M of \overline{PP'} lies on \ell. It also lies on this perpendicular, so it is where the two lines meet:

\begin{aligned} x + 1 &= -x + 5 \\ 2x &= 4 \\ x &= 2 \end{aligned}

and y = 2 + 1 = 3, so M = (2, 3). P' is as far past M as P is before it, so each coordinate of P' is twice that of M minus that of P:

P' = (2 \cdot 2 - 5,\ 2 \cdot 3 - 0) = (-1, 6)

Check. Both conditions of the definition hold. The midpoint of \overline{PP'} is \left(\tfrac{5 - 1}{2}, \tfrac{0 + 6}{2}\right) = (2, 3), which is on \ell because 2 + 1 = 3. The slope of \overline{PP'} is \tfrac{6 - 0}{-1 - 5} = -1, perpendicular to \ell.

Example 3

A half turn

Find the image of P(5, 3) under the rotation about C(2, -1) through 180^\circ.

Solution

By the definition, P' is on the circle centered at C through P, and the ray \overrightarrow{CP} turns through 180^\circ to reach \overrightarrow{CP'}. A turn of 180^\circ points the ray the opposite way, so P, C and P' lie on one line with C between them, and CP' = CP. That makes C the midpoint of \overline{PP'}:

P' = (2 \cdot 2 - 5,\ 2 \cdot (-1) - 3) = (-1, -5)

Check. \displaystyle CP = \sqrt{3^2 + 4^2} = 5 and \displaystyle CP' = \sqrt{(-3)^2 + (-4)^2} = 5. The two points are opposite ends of a diameter of the circle of radius 5 about C.

Example 4

Finding the mirror from a point and its image

A reflection sends P(1, 2) to P'(5, 6). Find the line of reflection, find the image Q' of the origin Q(0, 0), and check that P'Q' = PQ.

Solution

Since P' \neq P, the point P is not on the mirror, and the definition says the mirror is the perpendicular bisector of \overline{PP'}. A segment has only one perpendicular bisector, so the mirror is pinned down by this single pair.

The midpoint of \overline{PP'} is \left(\tfrac{1 + 5}{2}, \tfrac{2 + 6}{2}\right) = (3, 4). The slope of \overline{PP'} is \tfrac{6 - 2}{5 - 1} = 1, so the mirror has slope -1:

y - 4 = -(x - 3), \qquad\text{that is}\qquad y = -x + 7

Now reflect Q. It is not on the mirror, since 0 \neq 7. The perpendicular through Q has slope 1, so it is y = x, and it meets the mirror where x = -x + 7, at M = \left(\tfrac{7}{2}, \tfrac{7}{2}\right). Twice M minus Q gives Q' = (7, 7).

Check.

\begin{aligned} PQ &= \sqrt{1^2 + 2^2} = \sqrt{5} \\ P'Q' &= \sqrt{(7 - 5)^2 + (7 - 6)^2} = \sqrt{5} \end{aligned}

The distance survived the reflection, as the Results promise.

Example 5

A rule that is not a rigid motion

The rule F(x, y) = (x + y,\ y) slides each point sideways by its own height. Is F a rigid motion?

Solution

No. Take P(0, 0) and Q(0, 1), which are 1 apart. Their images are F(P) = (0, 0) and F(Q) = (1, 1), and

P'Q' = \sqrt{1^2 + 1^2} = \sqrt{2} \neq 1

One pair whose distance changes is enough, because the definition demands every pair. It is no defense that some distances survive: (0, 0) and (1, 0) are both fixed by F, so the distance between them stays 1.

Figure 5The rule F(x, y) = (x + y,\ y) keeps the bottom side but stretches the left side from 1 to \displaystyle \sqrt{2}.

Figure 5 shows the whole unit square. The bottom side keeps its length, the left side stretches to \displaystyle \sqrt{2}, and the right angle at the origin closes to 45^\circ. Rigid motions keep angles, so the angle alone already shows F is not one.

Example 6

A rule that is a rotation in disguise

Show that G(x, y) = (3 - y,\ x) is a rigid motion, and that it is a rotation. Find its center and its angle.

Solution

Distance. For P = (x_1, y_1) and Q = (x_2, y_2) the images are (3 - y_1,\ x_1) and (3 - y_2,\ x_2), so

\begin{aligned} (P'Q')^2 &= \big((3 - y_2) - (3 - y_1)\big)^2 + (x_2 - x_1)^2 \\ &= (y_1 - y_2)^2 + (x_2 - x_1)^2 \\ &= (PQ)^2 \end{aligned}

and G is rigid.

Center. A rotation through less than a full turn fixes only its center, so look for a fixed point. G(x, y) = (x, y) means 3 - y = x and x = y, so 2x = 3 and the only candidate is C = \left(\tfrac{3}{2}, \tfrac{3}{2}\right).

Angle. Measure every point from C. Write P = \left(\tfrac{3}{2} + a,\ \tfrac{3}{2} + b\right), so P sits a across and b up from C. Then

G(P) = \left(\tfrac{3}{2} - b,\ \tfrac{3}{2} + a\right)

so P' sits -b across and a up from C. The offset (a, b) has become (-b, a). Its length \displaystyle \sqrt{a^2 + b^2} is unchanged, so CP' = CP and P' is on the circle about C through P. And (a, b) \mapsto (-b, a) is exactly a counterclockwise quarter turn of the offset: pointing right, (1, 0), becomes pointing up, (0, 1); pointing up becomes pointing left, (-1, 0); and the slopes \tfrac{b}{a} and \tfrac{a}{-b} multiply to -1, so \overline{CP} and \overline{CP'} are always perpendicular. So G = R_{C,90^\circ} with C = \left(\tfrac{3}{2}, \tfrac{3}{2}\right).

Check. The point \left(\tfrac{5}{2}, \tfrac{3}{2}\right), one unit right of C, goes to \left(\tfrac{3}{2}, \tfrac{5}{2}\right), one unit above C.

Example 7

The image of a circle

Let T be a translation, a reflection or a rotation, and let c be the circle with center C and radius r. Show that the image of c is exactly the circle with center C' and radius r.

Solution

Two things have to be shown: every image point lies on the new circle, and every point of the new circle is an image.

Images lie on the new circle. If P is on c, then CP = r. T keeps distance, so C'P' = CP = r, and P' is on the circle with center C' and radius r.

Every point of the new circle is reached. Let X be any point with C'X = r. Let S be the motion of the same kind that undoes T; it is rigid too, and S(C') = C. Put P = S(X). Then

CP = S(C')\,S(X) = C'X = r

so P is on c, and T(P) = T(S(X)) = X.

So the image is the whole circle, nothing more and nothing less. The second half is not a formality: without it we would only know the image sits somewhere on the new circle, perhaps as a single arc.

Boundaries

Every pair, not some pairs. The rule in Example 5 keeps every horizontal distance and still is not rigid. A single pair whose distance changes settles the question.

Keeping angles is not enough. The rule (x, y) \mapsto (2x, 2y) doubles every distance and keeps every angle (Exercise 23). It is a dilation, the starting point of similarity transformations, and it is not rigid.

A rotation needs a direction. R_{O,90^\circ} and R_{O,-90^\circ} are different motions; only the half turn is the same both ways. Angles that differ by 360^\circ give the same transformation, so R_{O,450^\circ} is R_{O,90^\circ}.

A translation needs a directed segment. T_{\overrightarrow{AB}} and T_{\overrightarrow{BA}} move points in opposite directions. A segment with no direction does not decide which.

Rigid does not mean the same way round. Translations and rotations keep the order of labels: a triangle lettered counterclockwise has a counterclockwise image. A reflection reverses it. Distances and angles survive every rigid motion; the handedness of a figure survives only some.

Three kinds are not the whole list. The rule (x, y) \mapsto (x + 4,\ -y) is rigid but is none of the three (Exercise 17). It is a reflection followed by a translation along the mirror, a glide reflection.

Stated for later, not proved here. Every rigid motion of the plane turns out to be a translation, a rotation, a reflection or a glide reflection, and each can be made by doing at most three reflections in a row. Nothing in this lesson uses that fact. Proving it takes composing transformations, starting from two reflections in a row.

What the proofs lean on. Translations, reflections, half turns and quarter turns are shown to keep distance from their definitions and the distance formula alone. A rotation through any other angle, and keeping angle measure, lean on SAS and SSS as postulates. A development that proves SAS and SSS from rigid motions takes those two facts as its starting assumptions instead. The definitions of the three motions are the same either way.

Where this goes next

With exact definitions, two figures can be compared by asking for a motion rather than by laying one on the other. Two figures are congruent exactly when some sequence of rigid motions carries one onto the other, and rigid motions and congruence makes that the definition. Transformational proof then proves facts about figures by naming the motion that moves one part onto another, using the Results here as its reasons.

Doing one motion after another is composition, and the most surprising case is that two reflections make a translation or a rotation. A motion that carries a figure onto itself is a symmetry, which is what line symmetry and rotational symmetry study.

The coordinate rules met along the way are worked out in full in translations, reflections over the axes and rotating about any point. Drop the requirement that distance is kept, and keep only angles, and you reach similarity transformations.

Exercises

Reading the definitions

Which points does the reflection r_\ell leave where they are? Explain from the definition.
The rotation about O through 90^\circ sends P to P', and OP = 6. Find OP', m\angle POP' and PP'.
A translation sends A to B, where AB = 8. A point P has AP = 3. How far does P move, and how far is P' from B?

Translations

Find the image of P(3, -1) under the translation by \overrightarrow{AB}, where A(0, 0) and B(2, 5).
A translation sends A(-1, 4) to B(3, 1). Find the images of P(0, 0) and Q(2, -3), and check that P'Q' = PQ.
A translation sends P(2, 7) to P'(-3, 4). Which point does it send to the origin?

Reflections

Reflect P(4, 1) in the vertical line x = 1.
Reflect P(-1, 3) in the horizontal line y = -2.
Reflect P(6, 2) in the line y = x.
Reflect P(0, 7) in the line y = 2x + 2.
A reflection sends P(1, 1) to P'(7, 5). Find an equation of its mirror.

Rotations

Find the image of P(7, 5) under the half turn about C(3, 2).
Find the image of P(4, 2) under the rotation about C(1, 1) through 90^\circ counterclockwise, and check that CP' = CP.
Find the image of P(1, 4) under the rotation about C(-2, 0) through 90^\circ clockwise.
A rotation sends P(4, 1) to P'(2, 3) and Q(2, 4) to Q'(-1, 1). Find its center and its angle.

Rigid or not

Is F(x, y) = (2x,\ y) a rigid motion? Explain.
Show that G(x, y) = (x + 4,\ -y) is a rigid motion. Is it a translation, a reflection or a rotation?
Is H(x, y) = (x,\ |y|) a rigid motion? Explain.
Triangle ABC has AB = 6, BC = 8 and CA = 10. A rotation sends it to triangle A'B'C'. Find the perimeter of A'B'C' and m\angle A'B'C'.

Reasoning

Explain why a translation by a segment of nonzero length has no fixed point.
Show that reflecting in \ell twice returns every point to where it started.
A rigid motion fixes three points A, B and C that are not on one line. Prove that it fixes every point of the plane.
The rule (x, y) \mapsto (2x, 2y) doubles every distance. Is it a rigid motion? Does it keep angle measure? What does that say about defining rigid motions by angles alone?
Show that a half turn about O carries every line through O onto itself. Does it fix every point of such a line?
Show that a translation, reflection or rotation carries parallel lines to parallel lines.

Answers

Exactly the points of \ell. The definition leaves every point of \ell where it is. A point P off \ell goes to the P' for which \ell is the perpendicular bisector of \overline{PP'}, so P and P' are on opposite sides of \ell and P' \neq P.
OP' = 6, m\angle POP' = 90^\circ and \displaystyle PP' = 6\sqrt{2}. P' is on the circle centered at O through P, so OP' = OP = 6. The ray turns through 90^\circ, so \angle POP' is a right angle, and triangle POP' is an isosceles right triangle: \displaystyle PP' = \sqrt{6^2 + 6^2} = 6\sqrt{2}.
P moves 8 units, and BP' = 3. By the definition the trip from P to P' is as long as the trip from A to B. The translation sends A to B and keeps distance, so BP' = A'P' = AP = 3.
(5, 4). The trip from A to B is 2 across and 5 up, so P' = (3 + 2,\ -1 + 5).
P' = (4, -3) and Q' = (6, -6). The trip is 4 across and -3 up. Then \displaystyle PQ = \sqrt{2^2 + 3^2} = \sqrt{13} and \displaystyle P'Q' = \sqrt{(6 - 4)^2 + (-6 + 3)^2} = \sqrt{13}.
(5, 3). The trip is -5 across and -3 up, so the point that lands on (0, 0) started at (0 + 5,\ 0 + 3).
(-2, 1). The perpendicular to x = 1 through P is the horizontal line y = 1, meeting the mirror at M(1, 1), so P' = (2 \cdot 1 - 4,\ 1).
(-1, -7). The perpendicular is the vertical line x = -1, meeting the mirror at M(-1, -2), so P' = (-1,\ 2 \cdot (-2) - 3).
(2, 6). The perpendicular through P has slope -1: y = -x + 8. It meets y = x at M(4, 4), so P' = (8 - 6,\ 8 - 2).
(4, 5). The mirror has slope 2, so the perpendicular through P has slope -\tfrac{1}{2}: y = -\tfrac{1}{2}x + 7. It meets the mirror where 2x + 2 = -\tfrac{1}{2}x + 7, so \tfrac{5}{2}x = 5, x = 2 and M = (2, 6). Then P' = (4 - 0,\ 12 - 7).
y = -\tfrac{3}{2}x + 9, which is 3x + 2y = 18. The mirror is the perpendicular bisector of \overline{PP'}. Its midpoint is (4, 3) and its slope is \tfrac{4}{6} = \tfrac{2}{3}, so the mirror has slope -\tfrac{3}{2}: y - 3 = -\tfrac{3}{2}(x - 4).
(-1, -1). In a half turn the center is the midpoint of \overline{PP'}, so P' = (2 \cdot 3 - 7,\ 2 \cdot 2 - 5). Both points are 5 from C.
P' = (0, 4). From C, P is 3 across and 1 up. A counterclockwise quarter turn sends the offset (a, b) to (-b, a), here (-1, 3), so P' = (1 - 1,\ 1 + 3). Check: \displaystyle CP = \sqrt{3^2 + 1^2} = \sqrt{10} and \displaystyle CP' = \sqrt{(-1)^2 + 3^2} = \sqrt{10}, and the slopes of \overline{CP} and \overline{CP'}, \tfrac{1}{3} and -3, multiply to -1.
(2, -3). From C, P is 3 across and 4 up. A clockwise quarter turn sends the offset (a, b) to (b, -a) (up becomes right), here (4, -3), so P' = (-2 + 4,\ 0 - 3). Both points are 5 from C.
Center (2, 1), angle 90^\circ counterclockwise. The center is equally far from P and P', so it is on the perpendicular bisector of \overline{PP'}: midpoint (3, 2), slope of \overline{PP'} is -1, so y = x - 1. Likewise for \overline{QQ'}: midpoint \left(\tfrac{1}{2}, \tfrac{5}{2}\right), slope 1, so y = -x + 3. These meet where x - 1 = -x + 3, at (2, 1). From there P is offset (2, 0) and P' is offset (0, 2): right becomes up, a counterclockwise quarter turn. Q is offset (0, 3) and Q' is (-3, 0), the same turn.
No. (0, 0) and (1, 0) are 1 apart, but their images (0, 0) and (2, 0) are 2 apart.
Rigid, and none of the three. For two points the differences become (x_2 + 4) - (x_1 + 4) = x_2 - x_1 and (-y_2) - (-y_1) = -(y_2 - y_1), and squaring removes the sign, so the distance formula gives the same answer and every distance is kept. It is not a translation: one sending (0, 0) to (4, 0) would send (0, 1) to (4, 1), but G sends it to (4, -1). It has no fixed point, since x + 4 = x is impossible, so it is not a reflection (which fixes its mirror) or a rotation (which fixes its center). It is the reflection in the x-axis followed by a translation 4 to the right, a glide reflection.
No. (0, 1) and (0, -1) are 2 apart, but both go to (0, 1), which is 0 apart. H keeps distances between points on the same side of the x-axis, but not between every pair.
Perimeter 24 and m\angle A'B'C' = 90^\circ. The rotation keeps each side, so the perimeter is 6 + 8 + 10 = 24. Since 6^2 + 8^2 = 100 = 10^2, the converse of the Pythagorean theorem makes \angle ABC a right angle, and rigid motions keep angle measure.
Every point P moves to a P' with PP' = AB. If AB > 0, then P' \neq P for every P.
A point on \ell stays put both times. For P off \ell, the first reflection gives P' with \ell the perpendicular bisector of \overline{PP'}. The second asks for the point X with \ell the perpendicular bisector of \overline{P'X}. X = P works, and the definition gives only one such point, so the second reflection returns P' to P.
Suppose some point P goes to P^* \neq P. The motion fixes A and keeps distance, so AP^* = AP, and A is equally far from P and P^*. By the perpendicular bisector theorem A lies on the perpendicular bisector of \overline{PP^*}. The same holds for B and C, so A, B and C all lie on one line, which contradicts the given. So every point is fixed.
It is not rigid: (0, 0) and (1, 0), 1 apart, go to points 2 apart. It does keep angle measure: a triangle and its image have all three sides in the ratio 2, so they are similar by SSS similarity and their angles are equal. So a rule can keep every angle and still not be rigid, and a definition by angles alone would let in dilations.
Let m be a line through O and P \neq O a point on it. The half turn makes O the midpoint of \overline{PP'}, so P' is on line OP, which is m. Every point of m is also an image of a point of m, since the half turn is undone by itself, so m goes onto all of m. It does not fix the points of m: only O stays, and every other point swaps to the far side of O.
Let T be the motion and \ell \parallel m. By the Results, T carries \ell and m onto lines \ell' and m'. Suppose \ell' and m' shared a point X. The motion S that undoes T carries \ell' back onto \ell and m' back onto m, so S(X) lies on both \ell and m. Parallel lines share no point, so that is impossible, and \ell' \parallel m'.