Math 8 · 8.G.A.1 · G.CO.A.5

Rotations

“Turn it a quarter turn” sounds like a complete instruction. What else has to be said before everyone who follows it ends up with the same picture?

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Why this exists

A slide is fixed by how far and which way it goes. A flip is fixed by its mirror line. A turn needs more than either. Put a square card on a table and say “turn it a quarter turn”. Pin the card at its center and it spins in place. Pin it at a corner and it swings out to a new spot. Turn it to the left or to the right and it lands in two different places again. Three people following the same words can end up with three different pictures.

So a rotation is pinned down by three facts: the point it turns about, how far it turns, and which way. Once all three are named, every point of the plane has exactly one place to go, and that is what lets a rotation be drawn, computed, checked and used in a proof. On a coordinate grid, with the turn about the origin and a quarter or half turn, the whole rotation shrinks to a rule as short as swapping two numbers and changing a sign.

Definition

A rotation turns the whole plane about one fixed point. It is described by three things:

the center C, the one point that stays where it is;
the angle of rotation \theta, how far the plane turns, measured in Degrees;
the direction, counterclockwise or clockwise.

The rotation about C through \theta sends C to itself. It sends every other point P to the point P' for which

CP' = CP: the image is exactly as far from the center as the point was; and
the ray CP' is the ray CP turned through \theta in the stated direction. For \theta up to 180^\circ this says m\angle PCP' = \theta.

P' is the image of P. The same rotation is written R_{C,\theta}, so R_{C,\theta}(P) = P'.

Figure 1The rotation about C through \theta. P rides round its own circle about C until the ray CP has turned through \theta.

Because CP' always equals CP, the point P travels along the circle centered at C that passes through P, round an arc whose angle at the center is \theta.

Signs. The usual agreement is that a positive angle means counterclockwise and a negative angle means clockwise. So a rotation of -90^\circ is a quarter turn clockwise. This is the same agreement used for angles in standard position, which are measured counterclockwise from the positive x-axis.

Intuition

Lay a sheet of tracing paper over the page, pin it at C with a pencil point, and turn it. Every mark on the paper travels round its own circle about the pin. A mark close to the pin moves only a little way and a mark far out moves a long way, but every mark turns through the same angle, because the paper turns as one piece. The pin itself does not move at all. That is the definition in a picture: one angle for every point, and every distance to the center unchanged.

Two more things come free from the paper. It is not stretched, so nothing drawn on it changes size or shape: a rotation is a rigid motion. And it is never lifted and turned over, so a figure is never changed into its mirror image. A word written on the paper still reads left to right after any turn, only tilted.

Results

One rotation, several names. Turning \theta one way ends in the same place as turning 360^\circ - \theta the other way, and adding or removing a full turn changes nothing:

R_{C,\theta} = R_{C,\,\theta - 360^\circ} = R_{C,\,\theta + 360^\circ}

A half turn, 180^\circ, is the same in both directions. A full turn, 360^\circ, moves no point at all.

What a rotation keeps. A rotation keeps every length and every angle measure, so it carries every figure to a congruent one. It also keeps orientation: a triangle whose vertices run counterclockwise has an image whose matching vertices run counterclockwise too. A reflection reverses orientation, which is how a turn and a flip can be told apart.

Every line turns through the angle. If a rotation through an angle \theta between 0^\circ and 180^\circ carries a line \ell to a line \ell', then \ell and \ell' cross at the angle \theta (and at 180^\circ - \theta on the other side of the crossing). A turn of more than 180^\circ one way is a turn of less than 180^\circ the other way, so this covers every rotation. In particular, a quarter turn sends every line to a perpendicular line, and a half turn sends every line to a parallel line, or to itself when the line passes through the center.

Only the center stays put, unless the angle is a whole number of full turns.

Turns about one center add. A rotation through \alpha about C followed by a rotation through \beta about the same C is the rotation through \alpha + \beta about C, with clockwise angles counted as negative.

Rotations about the origin. For the turns that are multiples of 90^\circ, with the origin O as center:

Rotation about O	Rule
90^\circ counterclockwise (or 270^\circ clockwise)	(x, y) \mapsto (-y,\ x)
180^\circ (either direction)	(x, y) \mapsto (-x,\ -y)
90^\circ clockwise (or 270^\circ counterclockwise)	(x, y) \mapsto (y,\ -x)

Why it is true

Lengths and angles are kept. Take two points P and Q, with P, Q and C not on one line. The rays CP and CQ both turn through \theta in the same direction, so the angle between them is the same afterwards: m\angle P'CQ' = m\angle PCQ. The definition gives CP' = CP and CQ' = CQ. So the triangles PCQ and P'CQ' have two sides and the angle between them equal, they are congruent by side-angle-side, and P'Q' = PQ. If P, C and Q do lie on one line, so do P', C and Q', because two rays that point the same way or opposite ways still do after both turn by the same amount; then PQ is the sum or the difference of CP and CQ, and P'Q' is the same sum or difference. Since every length is kept, every triangle goes to a congruent triangle by side-side-side, and so every angle measure is kept as well.

Every line turns through \theta. Take a line \ell that does not pass through C, and let F be the point of \ell nearest to C, so that CF is perpendicular to \ell. The rotation sends F to F', and because angles are kept, the image line \ell' is perpendicular to CF' at F'. Take \theta between 0^\circ and 180^\circ. Then CF and CF' do not lie along one line, so \ell and \ell', being perpendicular to them, are not parallel and cross at a point X. The four points C, F, X and F' make a quadrilateral with right angles at F and F', and its angle at C is m\angle FCF' = \theta. Its angles add to 360^\circ, so the angles at C and at X add to 180^\circ, the angle at X is 180^\circ - \theta, and the lines cross at \theta and 180^\circ - \theta. For a half turn, CF' points exactly opposite to CF, so \ell and \ell' are both perpendicular to one line through C and are parallel. (If \ell does pass through C, it is made of two rays from C, and both turn through \theta directly.)

Orientation is kept. Walk round triangle ABC from A to B to C. At each corner you turn from the direction of one side to the direction of the next. The rotation turns the direction of every side through the same angle, so at each corner the turn is the same size and the same way as before. A walk that turned left at every corner still turns left at every corner.

The rule for a quarter turn about the origin. The point P(x, y) is x steps east and y steps north of O, where a negative number of steps means the other way. Turn the whole plane a quarter turn counterclockwise about O. East turns into north and north turns into west. So the x steps east become x steps north, and the y steps north become y steps west. The image is y steps west and x steps north of O:

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

Figure 2The walk from O to P is 5 east and 2 north. A quarter turn counterclockwise makes it 5 north and 2 west, so P' is (-2, 5).

Figure 2 shows it for P(5, 2): the right triangle under OP, with legs 5 and 2, turns into a right triangle with a leg of 5 up the y-axis and a leg of 2 to the left, and P' = (-2, 5). Two checks agree with the definition. The distance to O is \displaystyle \sqrt{x^2 + y^2} before and \displaystyle \sqrt{(-y)^2 + x^2} = \sqrt{x^2 + y^2} after, by the distance formula. And when x and y are both non-zero, the slopes of OP and OP' are \displaystyle \frac{y}{x} and \displaystyle \frac{x}{-y}, whose product is -1, so the angle at O is a right angle.

A half turn is two quarter turns: (x, y) \mapsto (-y, x), and then (-y, x) \mapsto (-x, -y). It is also what reflecting in the x-axis and then in the y-axis does, since each reflection changes the sign of one coordinate.

Three quarter turns counterclockwise take (-x, -y) one step further, to (y, -x). Turning 270^\circ counterclockwise ends in the same place as turning 90^\circ clockwise, so this is also the rule for a quarter turn clockwise.

Worked examples

The lesson is taught in two parts. Examples 1 to 5 belong to the first: the center, the angle and the direction, worked from the definition with no coordinates. Examples 6 to 10 belong to the second: rotations about the origin on a coordinate grid.

Example 1

Naming the same turn

A rotation about a point C is described as 100^\circ clockwise.

(a) Describe the same rotation using a counterclockwise angle between 0^\circ and 360^\circ.

(b) Is a rotation of 460^\circ counterclockwise about C the same as a rotation of 100^\circ counterclockwise about C?

Solution

(a) A full turn is 360^\circ. Going 100^\circ round one way stops at the same place as going the rest of the full turn round the other way, so the rotation is

360^\circ - 100^\circ = 260^\circ

so it is 260^\circ counterclockwise. With signs, clockwise is negative, and -100^\circ + 360^\circ = 260^\circ.

(b) Yes. 460^\circ = 360^\circ + 100^\circ: one full turn, which brings every point back to where it started, and then 100^\circ more. So 460^\circ - 360^\circ = 100^\circ, and the two are the same rotation.

(c) No. Take any point P other than C. One rotation leaves the ray CP turned 100^\circ one way, the other leaves it turned 100^\circ the other way, and these two rays are 200^\circ apart going round one way and 160^\circ apart going round the other. They are different rays, so the images are different points. The two directions agree only when \theta and 360^\circ - \theta are the same angle, which happens for the half turn, 180^\circ.

Example 2

The hands of a clock

Between 12:00 and 12:25 the two hands of a clock turn. Describe the rotation made by the minute hand and the rotation made by the hour hand: center, angle and direction.

Solution

Both hands are pinned at the center of the dial, so that is the center of both rotations, and both turn clockwise.

The minute hand makes a full turn in 60 minutes, which is \displaystyle \frac{360^\circ}{60} = 6^\circ each minute. In 25 minutes it turns 25 \times 6^\circ = 150^\circ. Its rotation is 150^\circ clockwise about the center of the dial, which is -150^\circ with signs, or 210^\circ counterclockwise.

The hour hand makes a full turn in 12 hours, which is 30^\circ each hour. Twenty-five minutes is \displaystyle \frac{25}{60} of an hour, so it turns \displaystyle 30^\circ \times \frac{25}{60} = 12.5^\circ. Its rotation is 12.5^\circ clockwise about the center of the dial.

Check. At 12:25 the minute hand points at the 5, and each number on the dial is 30^\circ from the next, so it has turned 5 \times 30^\circ = 150^\circ.

The two hands share a center and a direction but not an angle, so they make two different rotations. Naming the center and the direction is not enough; all three facts are needed.

Example 3

Every point turns through the same angle

A rotation of 50^\circ counterclockwise about C carries P to P' and Q to Q'. The lengths are CP = 2 and CQ = 4, and the ray CQ is 30^\circ counterclockwise from the ray CP. Find CQ', m\angle PCQ', m\angle P'CQ' and m\angle QCP'.

Solution

Measure every ray from C by how far counterclockwise it is from the ray CP. Then CP is at 0^\circ and CQ is at 30^\circ. The rotation turns every ray from C through 50^\circ, so CP' is at 50^\circ and CQ' is at 30^\circ + 50^\circ = 80^\circ.

CQ' = CQ = 4, because a rotation keeps the distance to the center.
m\angle PCQ' = 80^\circ, read straight off.
m\angle P'CQ' = 80^\circ - 50^\circ = 30^\circ. This equals m\angle PCQ: the angle between two rays from the center survives the turn.
m\angle QCP' = 50^\circ - 30^\circ = 20^\circ.

Figure 3Example 3. Both rays turn through 50^\circ, so the 30^\circ between them is still 30^\circ afterwards. Q travels further than P, round a bigger circle, but through the same angle.

The point Q travels further than P, because its circle is twice as big, yet both turn through exactly 50^\circ. And since CP' = CP, CQ' = CQ and m\angle P'CQ' = m\angle PCQ, the triangles PCQ and P'CQ' are congruent by side-angle-side, so P'Q' = PQ even though neither length was given.

Example 4

Turning twice about one center

(a) A figure is rotated 70^\circ counterclockwise about C and then 50^\circ clockwise about C. What single rotation has the same effect?

(b) What if the second rotation is 130^\circ clockwise instead?

Solution

(a) Every ray from C turns 70^\circ one way and then 50^\circ back, so it ends up 20^\circ from where it started, counterclockwise. No distance to C ever changes. A point that keeps its distance to C while its ray turns 20^\circ counterclockwise is exactly what the rotation through 20^\circ does. With signs, 70^\circ + (-50^\circ) = 20^\circ, a rotation of 20^\circ counterclockwise about C.

(b) With signs, 70^\circ + (-130^\circ) = -60^\circ. The single rotation is 60^\circ clockwise about C.

This works only because both turns share the center C. Two turns about different centers do not simply add like this; composing transformations deals with that case.

Example 5

A turn cannot make a mirror image

Triangle ABC has its vertices in counterclockwise order. Its reflection in a line \ell is triangle A'B'C'. Is there a rotation that carries A to A', B to B' and C to C'?

Solution

No. A reflection reverses orientation, so the vertices A', B', C' run clockwise. A rotation keeps orientation: it turns the direction of every side through the same angle, so a walk round the triangle that turned left at every corner still turns left at every corner after the turn. Whatever the center and the angle, a rotation sends A, B, C to three points in counterclockwise order, and A', B', C' are not in that order.

The two triangles are congruent, since a reflection keeps lengths. But a sequence of moves carrying one onto the other with these vertices matched must include a reflection, and choosing such a sequence is the business of congruence by rigid motion.

Example 6

A quarter turn of a triangle about the origin

Triangle JKL has vertices J(2, 1), K(6, 2) and L(3, 5). Rotate it 90^\circ counterclockwise about the origin.

Solution

A rotation sends segments to segments, so it is enough to rotate the three vertices and join the images in the same order. The rule for a quarter turn counterclockwise is (x, y) \mapsto (-y, x):

\begin{aligned} J(2, 1) &\mapsto J'(-1, 2) \\ K(6, 2) &\mapsto K'(-2, 6) \\ L(3, 5) &\mapsto L'(-5, 3) \end{aligned}

Figure 4Example 6. The quarter turn carries the triangle from the first quadrant into the second, and every side of the image is at right angles to the side it came from.

Check. The side JK is the walk (4, 1) and J'K' is the walk (-1, 4); both have length \displaystyle \sqrt{17}, so the length is kept. The slope of JK is \displaystyle \frac{1}{4} and the slope of J'K' is -4, and their product is -1: the side has turned through a right angle, as every line does under a quarter turn. The triangle has moved from the first quadrant to the second, which is where a quarter turn counterclockwise carries the first quadrant.

Example 7

A half turn, and two reflections

Rotate P(3, -5) by 180^\circ about the origin. Then reflect P in the x-axis and reflect that image in the y-axis, and compare.

Solution

The half turn rule (x, y) \mapsto (-x, -y) gives P' = (-3, 5). No direction is needed: a half turn is the same both ways.

Reflecting in the x-axis changes the sign of y, giving (3, 5). Reflecting that in the y-axis changes the sign of x, giving (-3, 5). The same point. This is no accident: the two reflections together change both signs, which is exactly the half turn rule, so they agree on every point.

Check. The origin should be the midpoint of P and P', since a half turn sends P to the point the same distance away in the opposite direction: \displaystyle \left(\frac{3 + (-3)}{2}, \frac{-5 + 5}{2}\right) = (0, 0).

Example 8

Clockwise, and naming a rotation from its result

(a) Rotate Q(-2, 6) by 90^\circ clockwise about the origin.

(b) A rotation about the origin through a multiple of 90^\circ carries A(5, -3) to A'(3, 5). Which rotation is it?

Solution

(a) The rule for a quarter turn clockwise is (x, y) \mapsto (y, -x), so Q' = (6, 2).

Check. OQ^2 = 4 + 36 = 40 and OQ'^2 = 36 + 4 = 40, so the distance to the center is kept. Q is up and to the left of O; a quarter turn clockwise carries “up and to the left” to “up and to the right”, and (6, 2) is up and to the right.

(b) Try each rule on A(5, -3):

quarter turn counterclockwise, (-y, x): (3, 5);
half turn, (-x, -y): (-5, 3);
quarter turn clockwise, (y, -x): (-3, -5).

Only the first gives A'. The rotation is 90^\circ counterclockwise about the origin, which is the same as 270^\circ clockwise.

Example 9

The image of a line

Find an equation of the image of the line y = 3x - 2 under a rotation of 90^\circ counterclockwise about the origin.

Solution

A rotation keeps straight lines straight, so the image is a line, and two of its points fix it. Take two points of the original line and rotate them with (x, y) \mapsto (-y, x):

(0, -2) \mapsto (2, 0) \qquad\qquad (1, 1) \mapsto (-1, 1)

The image line through (2, 0) and (-1, 1) has slope \displaystyle \frac{1 - 0}{-1 - 2} = -\frac{1}{3}, so

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

Check. The slopes 3 and \displaystyle -\frac{1}{3} multiply to -1, so the image is perpendicular to the original, as a quarter turn requires. A third point agrees: (2, 4) is on the original line, it goes to (-4, 2), and \displaystyle -\frac{1}{3}(-4) + \frac{2}{3} = 2.

Example 10

A turn that is not a multiple of 90^\circ

Rotate P(3, 3) by 45^\circ counterclockwise about the origin. Then find where P goes under a rotation of 135^\circ counterclockwise about the origin instead.

Solution

None of the swap rules applies, so go back to the definition: the image is just as far from O, and the ray has turned through the angle.

The distance is \displaystyle OP = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}. The point P lies on the line y = x, so the ray OP is 45^\circ counterclockwise from the positive x-axis. Turned 45^\circ more, it is 90^\circ from the positive x-axis: it points straight up the y-axis. The image is the point on the positive y-axis at distance \displaystyle 3\sqrt{2} from O:

P' = (0,\ 3\sqrt{2}) \approx (0, 4.24)

Figure 5Example 10. P starts at 45^\circ from the x-axis. Turned 45^\circ more, it points straight up, still \displaystyle 3\sqrt{2} from O.

Under 135^\circ instead, the ray ends at 45^\circ + 135^\circ = 180^\circ, along the negative x-axis, so the image is \displaystyle (-3\sqrt{2},\ 0).

Check. A quarter turn would send P to (-3, 3), and 45^\circ is half of a quarter turn, so P' should sit halfway round the arc from (3, 3) to (-3, 3). By symmetry about the y-axis that halfway point is on the y-axis, where \displaystyle (0, 3\sqrt{2}) is. The coordinates are irrational, which is usual for angles that are not multiples of 90^\circ; the general formula uses the cosine and sine of the angle.

Boundaries

The direction has to be stated. “Rotate 90^\circ” with no direction names two different rotations. The only turns that are the same both ways are the half turn and whole numbers of full turns.

The center has to be stated. The same angle about a different center is a different rotation. A quarter turn counterclockwise sends (4, 0) to (0, 4) when the center is the origin, but leaves it where it is when the center is (4, 0) itself.

The swap rules are for the origin and for multiples of 90^\circ. About any other center, the rules give the wrong point; turning about another center is rotating about a point other than the origin. For other angles there is no swap rule, and the image usually has irrational coordinates, as in Example 10.

The center has no angle of its own. The angle \angle PCP' makes sense only when P is not C, which is why the definition sends C to itself separately. When the angle is not a whole number of full turns, every other point moves, so the center is the only point that stays put.

A rotation keeps orientation. It can never carry a figure onto its mirror image with each point matched to its own reflection. That needs a reflection.

Turns add only about one center. Two rotations about the same center make the rotation through the sum of the angles. About different centers they do not, and the result is found by composing the transformations.

This is a turn of the plane. In space a turn happens about a line, the axis, not about a point, and its direction only makes sense when you say from which end of the axis you are looking.

Where this goes next

Turning about the origin is the base case. Rotating about any other point measures the walk from the center instead of from the origin and then uses these same rules, and for an angle that is not a multiple of 90^\circ it brings in the cosine and the sine.

Rotations sit beside slides and flips as the three rigid motions, and what all three preserve is what makes them the tool for deciding when two figures are congruent. Two rotations, or a rotation and another motion, done one after the other are the subject of composing transformations, and two reflections in crossing mirrors turn out to be a single rotation about the point where the mirrors cross, through twice the angle between them.

A figure that some turn of less than a full circle carries onto itself has rotational symmetry. A graph that the half turn about the origin carries onto itself is symmetric about the origin, which is exactly the half turn rule (x, y) \mapsto (-x, -y) applied to every point of the graph.

Further on, the quarter turn rule (x, y) \mapsto (-y, x) is the first example of the matrix of a transformation: where the rotation sends one step east and one step north decides where it sends every point.

Exercises

Exercises 1 to 15 go with the first part of the lesson (Examples 1 to 5), Exercises 16 to 29 with the second (Examples 6 to 10), and the reasoning group draws on both.

Part 1: naming a rotation

Describe a rotation of 75^\circ clockwise using a counterclockwise angle between 0^\circ and 360^\circ.
Describe a rotation of 300^\circ counterclockwise using a clockwise angle between 0^\circ and 360^\circ.
All about the same center, which of these are the same rotation as 90^\circ counterclockwise? (a) 270^\circ clockwise; (b) 450^\circ counterclockwise; (c) 90^\circ clockwise; (d) a rotation of -270^\circ.
What does a rotation of 720^\circ counterclockwise do to each point of the plane?

Part 1: clocks, wheels and distances

Describe the rotation the minute hand of a clock makes between 3:00 and 3:40.
Describe the rotation the hour hand of a clock makes between 1:00 and 5:00.
A wheel makes 2.5 full turns counterclockwise about its hub. Which rotation through less than a full turn has the same effect on the wheel?
A rotation about C carries P to P', and CP = 6. Find PP' when the angle of rotation is (a) 90^\circ, (b) 60^\circ, (c) 180^\circ.

Part 1: what a rotation keeps

A rotation of 40^\circ counterclockwise about C carries P to P' and Q to Q'. The ray CQ is 25^\circ counterclockwise from the ray CP. Find m\angle PCQ', m\angle P'CQ' and m\angle QCP'.
Triangle DEF has DE = 7, EF = 5 and m\angle DEF = 48^\circ. It is rotated 120^\circ clockwise about a point O. Find D'E', E'F' and m\angle D'E'F'.
A rotation of 30^\circ about C carries the line \ell to the line m, and \ell does not pass through C. At what angle do \ell and m cross?
What can you say about a line and its image under a rotation of 90^\circ? Under a rotation of 180^\circ?

Part 1: turning twice

A figure is rotated 80^\circ counterclockwise about C and then 35^\circ clockwise about C. Give one rotation with the same effect.
A rotation of 100^\circ counterclockwise about C is done four times in a row. Give one rotation, through less than a full turn, with the same effect.
Which rotation undoes a rotation of 110^\circ counterclockwise about C?

Part 2: points about the origin

Rotate each point about the origin.

(3, 7), 90^\circ counterclockwise.
(-4, 2), 180^\circ.
(6, -1), 90^\circ clockwise.
(-5, -3), 270^\circ counterclockwise.

Part 2: figures about the origin

Triangle STU has vertices S(2, 0), T(5, 1) and U(3, 4). Rotate it 90^\circ clockwise about the origin.
The segment from (1, 3) to (4, -2) is rotated 180^\circ about the origin. Find its image, and show that the image is parallel to the segment and has the same length.
The square with vertices (1, 2), (4, 2), (4, 5) and (1, 5) is rotated 90^\circ counterclockwise about the origin. Find the image.

Part 2: which rotation?

In Exercises 23 to 25 the rotation is about the origin, through a multiple of 90^\circ. Name it.

(2, -5) \mapsto (5, 2)
(-3, 4) \mapsto (3, -4)
(6, 1) \mapsto (1, -6)
Can any rotation about the origin carry (3, 4) to (5, 1)? Explain.

Part 2: beyond the swap rules

Find an equation of the image of the line y = x + 3 under a rotation of 90^\circ counterclockwise about the origin.
Rotate (4, -4) by 45^\circ counterclockwise about the origin.
Rotate (6, 0) by 60^\circ clockwise about the origin. Use the triangle whose vertices are the origin, the point and its image.

Reasoning

Use the half turn rule to show that a half turn about the origin carries the line y = 2x onto itself, and find the image of the line y = 2x + 1. How are the two answers different, and why?
Explain why, for a rotation of 50^\circ about C, the center is the only point that does not move.
Show that the rule (x, y) \mapsto (-y, x) keeps every point’s distance from the origin, and that the segments from the origin to (x, y) and to its image are perpendicular.
Apply the rule (x, y) \mapsto (-y, x) four times in a row to (x, y). What do you get, and why did it have to happen?
You are told the center C of a rotation and that it carries one point P, not C, to P'. Is that enough to know the whole rotation? What must be true of CP and CP'?
The rule (x, y) \mapsto (y, x) keeps every point’s distance from the origin. Show that it is not a rotation about the origin.

Answers

285^\circ counterclockwise. 360^\circ - 75^\circ = 285^\circ.
60^\circ clockwise. 360^\circ - 300^\circ = 60^\circ.
(a), (b) and (d). Clockwise is negative, so (a) is -270^\circ, which is (d), and -270^\circ + 360^\circ = 90^\circ. For (b), 450^\circ - 360^\circ = 90^\circ. But (c) is -90^\circ, which is -90^\circ + 360^\circ = 270^\circ counterclockwise, not 90^\circ.
Every point ends where it started. 720^\circ = 2 \times 360^\circ, two full turns, so every ray from the center comes back to itself.
240^\circ clockwise about the center of the dial, the same as 120^\circ counterclockwise. The minute hand turns 6^\circ a minute, and 40 \times 6^\circ = 240^\circ.
120^\circ clockwise about the center of the dial. Four hours at 30^\circ an hour is 4 \times 30^\circ = 120^\circ.
A half turn, 180^\circ. The wheel turns 2.5 \times 360^\circ = 900^\circ, and removing two full turns leaves 900^\circ - 720^\circ = 180^\circ. Each point of the rim ends opposite where it started.
(a) \displaystyle 6\sqrt{2}; (b) 6; (c) 12. In (a), triangle CPP' has a right angle at C and legs 6 and 6, so PP'^2 = 36 + 36 = 72. In (b), CP = CP', so the two angles at P and P' are equal, and they add to 180^\circ - 60^\circ = 120^\circ; each is 60^\circ, the triangle is equilateral, and PP' = 6. In (c), C is the midpoint of PP', so PP' = 6 + 6.
m\angle PCQ' = 65^\circ, m\angle P'CQ' = 25^\circ, m\angle QCP' = 15^\circ. Measured from the ray CP, the ray CQ is at 25^\circ, CP' at 40^\circ and CQ' at 25^\circ + 40^\circ.
D'E' = 7, E'F' = 5 and m\angle D'E'F' = 48^\circ. A rotation keeps every length and every angle measure, whatever its center and angle.
At 30^\circ (and 150^\circ on the other side of the crossing). Every line turns through the angle of the rotation.
Under 90^\circ the image is perpendicular to the line. Under 180^\circ the image is parallel to the line, or is the line itself when the line passes through the center: a half turn sends every direction to its opposite, which runs along the same line.
45^\circ counterclockwise about C. With signs, 80^\circ + (-35^\circ) = 45^\circ.
40^\circ counterclockwise about C. Four turns of 100^\circ make 400^\circ, and 400^\circ - 360^\circ = 40^\circ.
110^\circ clockwise about the same center C, which is the same as 250^\circ counterclockwise. The two together turn 110^\circ + (-110^\circ) = 0^\circ.
(-7, 3). The rule is (x, y) \mapsto (-y, x).
(4, -2). A half turn changes both signs.
(-1, -6). The rule for a quarter turn clockwise is (x, y) \mapsto (y, -x).
(-3, 5). A turn of 270^\circ counterclockwise is a quarter turn clockwise, so (x, y) \mapsto (y, -x) gives (-3, -(-5)).
S'(0, -2), T'(1, -5), U'(4, -3), using (x, y) \mapsto (y, -x).
The image runs from (-1, -3) to (-4, 2). The slope of the segment is \displaystyle \frac{-2 - 3}{4 - 1} = -\frac{5}{3} and the slope of the image is \displaystyle \frac{2 - (-3)}{-4 - (-1)} = -\frac{5}{3}, so they are parallel. Both lengths are \displaystyle \sqrt{3^2 + 5^2} = \sqrt{34}.
(-2, 1), (-2, 4), (-5, 4) and (-5, 1), in order. It is still a 3 by 3 square, now in the second quadrant.
90^\circ counterclockwise. (-y, x) gives (5, 2); the half turn would give (-2, 5) and the clockwise quarter turn (-5, -2).
180^\circ. Both signs have changed.
90^\circ clockwise. (y, -x) gives (1, -6).
No. A rotation about the origin keeps the distance to the origin, but (3, 4) is \displaystyle \sqrt{9 + 16} = 5 from it and (5, 1) is \displaystyle \sqrt{25 + 1} = \sqrt{26} from it.
y = -x - 3. The points (0, 3) and (1, 4) of the line go to (-3, 0) and (-4, 1). The slope through these is \displaystyle \frac{1 - 0}{-4 - (-3)} = -1, and the line of slope -1 through (-3, 0) is y = -(x + 3).
\displaystyle (4\sqrt{2}, 0). The distance is \displaystyle \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}. The point lies on y = -x in the fourth quadrant, so its ray is 45^\circ clockwise from the positive x-axis; 45^\circ counterclockwise brings it back onto the positive x-axis.
\displaystyle (3, -3\sqrt{3}). The image is 6 from the origin and the angle at the origin is 60^\circ, so the triangle is isosceles with a 60^\circ angle between its equal sides, which makes it equilateral. A clockwise turn puts its third vertex below the midpoint (3, 0) of the side on the x-axis, at depth \displaystyle \sqrt{6^2 - 3^2} = \sqrt{27} = 3\sqrt{3}.
The half turn sends (x, y) to (-x, -y). A point (t, 2t) of y = 2x goes to (-t, -2t), whose second coordinate is again twice its first, so it lies on y = 2x: the line is carried onto itself. A point (t, 2t + 1) of y = 2x + 1 goes to (-t, -2t - 1); with x = -t its second coordinate is 2x - 1, so the image is y = 2x - 1. That is a different line with the same slope, so it is parallel. The first line passes through the center and the second does not, and a half turn sends a line to itself exactly when it passes through the center.
Any point P other than C is a positive distance from C, and its image lies on the ray from C that is 50^\circ away from the ray CP. That is a different ray, so P' \ne P. The center is sent to itself by the definition.
The distance from the origin to (-y, x) is \displaystyle \sqrt{(-y)^2 + x^2} = \sqrt{x^2 + y^2}, the same as for (x, y). When x and y are both non-zero the two slopes are \displaystyle \frac{y}{x} and \displaystyle \frac{x}{-y}, and their product is \displaystyle -\frac{xy}{xy} = -1. When one coordinate is 0, one segment lies along an axis and the other along the other axis, which are perpendicular too.
The point comes back to itself. In turn, (x, y) \mapsto (-y, x), then (-y, x) \mapsto (-x, -y), then (-x, -y) \mapsto (y, -x), and finally (y, -x) \mapsto (x, y). Four quarter turns about one center add to 360^\circ, a full turn, and a full turn moves nothing.
Yes, as long as CP' = CP. Then the angle from the ray CP round to the ray CP', in the counterclockwise direction, is an angle of rotation that works, and every other point must turn through that same angle about the same center. If CP' \ne CP, no rotation about C carries P to P' at all.
Every point of the line y = x, such as (1, 1) and (2, 2), is sent to itself, so the rule has a whole line of points that stay put. A rotation that moves anything has only one such point, its center. And the rule does move some points, for example (1, 0) \mapsto (0, 1), so it is not the rotation that moves nothing either. It is the reflection in the line y = x.