Geometry · 8.G.A.3 · G.CO.A.5

Rotating about a point other than the origin

Turning a point about the origin has a neat coordinate rule. What happens when the turn is about some other point?

You need

RotationsMath 8

Where you are 11 in Transformations

Geometry11

Transformations11

▴ 4 above Rotations Composing transformations

Rotating about a point other than the origin

What rigid motions preserve Rigid motions and congruence ▾ Show all 11 (2 more)

When you reach for it

A rotation is fixed by three things: the point it turns about, the angle, and the direction, as the lesson on rotations sets out. When the center is the origin, a quarter turn counterclockwise has a one-line rule, (x, y) \mapsto (-y, x). That rule is about the origin and nothing else. Apply it to a turn about (2, 1) and the answer is wrong, because the origin is not where the turning happens.

Most turns in the world are not about the origin. A door swings about its hinge, a wheel turns about its hub, and a figure on a grid is often turned about one of its own corners. Reach for this skill whenever the center of a rotation is a point C = (a, b) other than (0, 0) and you want the exact coordinates of the image. Reach for it backwards too, when you can see a point and where it landed and want to know what turn did it.

The procedure

What a rotation keeps

A rotation about C through an angle \theta sends a point P to the point P' with two properties:

CP' = CP: the image is the same distance from the center as the point was, so P travels along a circle centered at C;
\angle PCP' = \theta, measured in the stated direction: counterclockwise when \theta is positive, clockwise when it is negative.

The center itself does not move. Everything below is a way of finding the one point with those two properties.

Measure from the center

Neither property mentions the origin. Both are about where P sits relative to C. So the first step is to describe exactly that, by the offset from the center to the point:

(u, v) = (x - a,\ y - b)

This is the walk from C to P: u steps across (to the right when u is positive, to the left when it is negative) and v steps up (or down when v is negative). For C = (2, 1) and P = (5, 3), the offset is (3, 2). In effect C becomes a temporary origin, which is the same as translating the picture so that C sits at (0, 0).

Turn the offset

Turning the whole plane about C turns the walk from C to P with it. A step to the east becomes a step in whatever direction east was turned to, and the same for a step to the north. For the three turns that are multiples of 90^\circ this is easy to read off.

A quarter turn counterclockwise (90^\circ) sends east to north and north to west. So u steps east become u steps north, and v steps north become v steps west:

(u, v) \ \longmapsto\ (-v,\ u)

Figure 1Example 1. The walk from C to P is 3 across and 2 up. Turned a quarter turn about C, it becomes 3 up and 2 to the left.

The signs take care of themselves. A negative u means steps west, and west turns to south, which is u steps “north” with u negative. So the rule holds whichever way the offset points.

A half turn (180^\circ, either direction) sends east to west and north to south:

(u, v) \ \longmapsto\ (-u,\ -v)

A quarter turn clockwise (-90^\circ, which is the same turn as 270^\circ counterclockwise) sends east to south and north to east:

(u, v) \ \longmapsto\ (v,\ -u)

You can test the quarter turn rule with lengths and slopes. The offsets (u, v) and (-v, u) both have length \displaystyle \sqrt{u^2 + v^2}, by the distance formula, so the distance to the center is kept. Their slopes are \displaystyle \frac{v}{u} and \displaystyle \frac{u}{-v}, whose product is -1, so they are perpendicular: the angle at the center is a right angle.

Add the center back

The turned offset is a walk from C, so the image is the center plus that walk. For a quarter turn counterclockwise about (a, b),

P' = (a - v,\ b + u)

In the example, the offset (3, 2) turns to (-2, 3), and P' = (2 - 2,\ 1 + 3) = (0, 4).

The three steps are a translation, a rotation about the origin and the translation back, done in that order, which is a composition. Skipping either translation is what makes the origin rule give the wrong answer.

The half turn shortcut

For a half turn, P' = (a - u,\ b - v). The point P is the walk (u, v) from C and P' is the same walk in the opposite direction, so C is the midpoint of PP'. Written out,

P' = (2a - x,\ 2b - y)

This is the same rotation, just quicker to compute.

Any angle

For an angle that is not a multiple of 90^\circ, the same idea needs one fact from the unit circle: one step east, turned through \theta, is the step (\cos\theta, \sin\theta). One step north is a step east already turned 90^\circ, so turned through \theta more it is (\cos(\theta + 90^\circ), \sin(\theta + 90^\circ)) = (-\sin\theta, \cos\theta). The offset is u steps east plus v steps north, and the turned offset is u turned steps plus v turned steps. Adding the center back,

\begin{aligned} x' &= a + u\cos\theta - v\sin\theta \\ y' &= b + u\sin\theta + v\cos\theta \end{aligned}

with u = x - a and v = y - b. At \theta = 90^\circ this gives (a - v, b + u) again, since \cos 90^\circ = 0 and \sin 90^\circ = 1.

Checking the answer

The two properties are the check. The distance formula should give CP' = CP. For a quarter turn the slopes of CP and CP' should multiply to -1, and the direction should be right: a point straight east of the center goes straight north under a quarter turn counterclockwise, and straight south under one clockwise.

Without coordinates

On plain paper the same two properties give a construction. Draw the segment from C to P, use a protractor at C to draw a ray at angle \theta from it in the stated direction, and set a compass to the length CP to mark P' on that ray. A polygon is turned one vertex at a time, and the images of the vertices are joined in the same order.

Worked examples

Example 1

A quarter turn counterclockwise

Rotate P(5, 3) by 90^\circ counterclockwise about C(2, 1).

Solution

The offset from the center is (5 - 2,\ 3 - 1) = (3, 2): three across and two up.

A quarter turn counterclockwise sends (u, v) to (-v, u), so the offset becomes (-2, 3): two to the left and three up.

Add the center back: P' = (2 - 2,\ 1 + 3) = (0, 4).

Check. CP^2 = 3^2 + 2^2 = 13 and CP'^2 = (-2)^2 + 3^2 = 13, so the distance is kept. The slope of CP is \displaystyle \frac{2}{3} and the slope of CP' is \displaystyle \frac{3}{-2}, and \displaystyle \frac{2}{3} \cdot \left(-\frac{3}{2}\right) = -1, so the angle at C is a right angle. P was up and to the right of C and P' is up and to the left, which is the counterclockwise side.

Applying the origin rule to P directly would give (-3, 5), which is \displaystyle \sqrt{41} from C, not \displaystyle \sqrt{13}. That point is the image under a different rotation, the one about (0, 0).

Example 2

A half turn of a triangle

Triangle PQR has vertices P(1, 2), Q(4, 2) and R(2, 5). Rotate it 180^\circ about C(3, -1).

Solution

A rotation sends segments to segments, so it is enough to rotate the three vertices and join the images in the same order.

Use the half turn shortcut (x, y) \mapsto (2a - x,\ 2b - y) with 2a = 6 and 2b = -2:

\begin{aligned} P' &= (6 - 1,\ -2 - 2) = (5, -4) \\ Q' &= (6 - 4,\ -2 - 2) = (2, -4) \\ R' &= (6 - 2,\ -2 - 5) = (4, -7) \end{aligned}

The long way agrees. The offset of P is (1 - 3,\ 2 - (-1)) = (-2, 3), a half turn makes it (2, -3), and (3 + 2,\ -1 - 3) = (5, -4).

Figure 2Example 2. A half turn about C puts C at the midpoint of every segment from a point to its image.

Check. PQ is horizontal with length 3, and so is P'Q', from (5, -4) to (2, -4). The image is upside down and Q' is to the left of P' where Q was to the right of P, which is what turning a figure halfway round does. The midpoint of PP' is \displaystyle \left(\frac{1 + 5}{2}, \frac{2 - 4}{2}\right) = (3, -1), which is C.

Example 3

A turn of 270^\circ

Rotate the segment with endpoints A(3, 0) and B(-1, 5) by 270^\circ counterclockwise about C(-1, 2).

Solution

Three quarter turns counterclockwise end in the same place as one quarter turn clockwise, so use the rule (u, v) \mapsto (v, -u).

For A the offset is (3 - (-1),\ 0 - 2) = (4, -2). It turns to (-2, -4), and A' = (-1 - 2,\ 2 - 4) = (-3, -2).

For B the offset is (-1 - (-1),\ 5 - 2) = (0, 3): B is straight above the center. It turns to (3, 0), and B' = (-1 + 3,\ 2 + 0) = (2, 2). A point straight above the center lands straight to its right after a quarter turn clockwise, which is a quick check on the direction.

Check. AB^2 = 4^2 + 5^2 = 41 and A'B'^2 = 5^2 + 4^2 = 41. A rotation keeps every length, not just distances to the center, and the segment is the same length after the turn.

Example 4

The center changes the answer

Rotate P(4, 1) by 90^\circ counterclockwise (a) about the origin and (b) about C(2, 3). How are the two images related?

Solution

(a) About the origin the offset is the point itself, (4, 1), and it turns to (-1, 4).

(b) About C the offset is (4 - 2,\ 1 - 3) = (2, -2). It turns to (2, 2), so the image is (2 + 2,\ 3 + 2) = (4, 5).

The two images are different points, so the same angle and direction with a different center is a different rotation. They differ by (4, 5) - (-1, 4) = (5, 1).

That difference does not depend on P. Turning a general point (x, y) about (a, b) gives (a - (y - b),\ b + (x - a)), which rearranges to

(-y,\ x) + (a + b,\ b - a)

the origin rule followed by a translation by (a + b,\ b - a). With a = 2 and b = 3 the translation is (5, 1), as found. So a quarter turn about any center is a quarter turn about the origin followed by a slide; moving the center changes only the slide.

Example 5

Finding the center

A rotation of 90^\circ takes A(1, 4) to A'(4, 1) and B(3, 4) to B'(4, -1). Find its center and its direction.

Solution

The center C is the same distance from A as from A', and every point with that property lies on the perpendicular bisector of AA'. The same holds for BB'. So C is where the two bisectors cross.

The midpoint of AA' is \displaystyle \left(\frac{5}{2}, \frac{5}{2}\right), and AA' has slope \displaystyle \frac{1 - 4}{4 - 1} = -1. The bisector has slope 1 and passes through the midpoint, so it is y = x.

The midpoint of BB' is \displaystyle \left(\frac{7}{2}, \frac{3}{2}\right), and BB' has slope \displaystyle \frac{-1 - 4}{4 - 3} = -5. The bisector has slope \displaystyle \frac{1}{5}:

y - \frac{3}{2} = \frac{1}{5}\left(x - \frac{7}{2}\right), \quad\text{that is,}\quad y = \frac{1}{5}x + \frac{4}{5}

Where it meets y = x, \displaystyle x = \frac{1}{5}x + \frac{4}{5}, so \displaystyle \frac{4}{5}x = \frac{4}{5} and x = 1. The center is C(1, 1).

Figure 3Example 5. The center is the same distance from each point as from its image, so it lies on both perpendicular bisectors, and they cross only once.

For the direction, compare offsets. From C, the point A is (0, 3), straight up, and A' is (3, 0), straight right. North turned to east is a quarter turn clockwise.

Check with B: its offset (2, 3) under the clockwise rule becomes (3, -2), and (1 + 3,\ 1 - 2) = (4, -1), which is B'.

Example 6

A turn of 60^\circ

Rotate P(6, 2) and Q(2, 5) by 60^\circ counterclockwise about C(2, 2).

Solution

Here \displaystyle \cos 60^\circ = \frac{1}{2} and \displaystyle \sin 60^\circ = \frac{\sqrt{3}}{2}, and the general formula is needed.

For P the offset is (4, 0), four steps east. Turned through 60^\circ it is \displaystyle \left(4 \cdot \frac{1}{2},\ 4 \cdot \frac{\sqrt{3}}{2}\right) = (2, 2\sqrt{3}), so

P' = (4,\ 2 + 2\sqrt{3}) \approx (4, 5.46)

For Q the offset is (0, 3), three steps north. With u = 0 and v = 3 the formula gives \displaystyle \left(-3 \cdot \frac{\sqrt{3}}{2},\ 3 \cdot \frac{1}{2}\right), so

Q' = \left(2 - \frac{3\sqrt{3}}{2},\ \frac{7}{2}\right) \approx (-0.60, 3.5)

Figure 4Example 6. Every segment from the center turns through the same angle, whichever way it pointed to begin with.

Check. \displaystyle CP'^2 = 2^2 + (2\sqrt{3})^2 = 4 + 12 = 16, so CP' = 4 = CP. And \displaystyle CQ'^2 = \frac{27}{4} + \frac{9}{4} = 9, so CQ' = 3 = CQ. The coordinates are irrational, which is normal for an angle that is not a multiple of 90^\circ; the exact forms are the answer and the decimals are for plotting.

Example 7

Building a square from one side

The points A(1, 1) and B(4, 2) are neighboring vertices of a square ABCD, labeled counterclockwise. Find C and D.

Solution

Walking counterclockwise round the square, the side AD is the side AB turned a quarter turn counterclockwise about A. So D is the image of B under that rotation.

The offset of B from A is (3, 1). It turns to (-1, 3), so D = (1 - 1,\ 1 + 3) = (0, 4).

In the same way, BC is BA turned a quarter turn clockwise about B. The offset of A from B is (-3, -1), and the clockwise rule (u, v) \mapsto (v, -u) turns it to (-1, 3), so C = (4 - 1,\ 2 + 3) = (3, 5).

Check. The four sides, as walks, are AB = (3, 1), BC = (-1, 3), CD = (-3, -1) and DA = (1, -3). Each has length \displaystyle \sqrt{10}, and each is the one before it turned a quarter turn counterclockwise, so every corner is a right angle. Labeled clockwise instead, the square sits on the other side of AB, with C(5, -1) and D(2, -2).

Boundaries

The coordinate rules are for multiples of 90^\circ. Swapping and negating the offset is exact only for quarter and half turns. Any other angle needs the cosine and sine formula, and the image usually has irrational coordinates.

The direction must be stated. A positive angle means counterclockwise. “Rotate 90^\circ” with no direction names two different rotations. Among turns of less than a full circle, only the half turn is the same both ways.

The center is the only point that does not move, for any angle that is not a whole number of full turns. A whole number of full turns moves nothing at all.

Turning the vertices is enough only because a rotation keeps straight lines straight. A rotation is a rigid motion, so it keeps lengths, angles and betweenness, and the image of a polygon is the polygon on the images of its vertices.

Undoing a rotation means the same center and the opposite angle. To find where a point came from, rotate its image back about the same center.

Finding the center needs the right two points when the angle is not known. One point and its image fix only a line of possible centers, the perpendicular bisector. (When the angle and its direction are known, one pair is enough: only one point on that line turns P to P' through that angle.) A second pair usually crosses it at one point, but not always: if the two bisectors are the same line, a third pair is needed, and if they are parallel and different, no rotation does the job at all. If a point is its own image, that point is the center.

Two rotations about different centers are not one rotation about either. Their combination is a rotation through the sum of the angles about a new center, or a translation when the angles add to a full turn. That belongs to composing transformations.

Where this goes next

A rotation about any point is a rotation about the origin with a slide attached, as Example 4 showed. That makes it the natural partner of composing transformations, and it is one of the moves that decides whether two figures are congruent. Two reflections in crossing mirrors give exactly such a rotation, centered where the mirrors cross, and a figure that a turn about its own center carries onto itself has rotational symmetry.

The cosine and sine formula is the start of finding the matrix of a transformation: the turned east step and the turned north step are the columns of a matrix acting on (u, v), and the center is added afterwards. In the complex plane the same move is one multiplication, since turning z through \theta about c gives c + (z - c)(\cos\theta + i\sin\theta), which is where the polar form of a complex number leads.

Exercises

Quarter and half turns

Find the image of each point.

P(4, 2), rotated 90^\circ counterclockwise about C(1, 1).
P(-2, 5), rotated 90^\circ clockwise about C(1, 3).
P(6, -1), rotated 180^\circ about C(2, 3).
The origin, rotated 90^\circ counterclockwise about C(3, -2).
P(-4, -1), rotated 270^\circ counterclockwise about C(-2, 2).

Turning figures

Triangle KLM has vertices K(2, 1), L(5, 1) and M(2, 3). Rotate it 90^\circ counterclockwise about K.
A segment has endpoints (1, 4) and (3, 0). Rotate it 180^\circ about (2, 2). What do you notice?
A rectangle has vertices (1, 1), (4, 1), (4, 3) and (1, 3). Rotate it 90^\circ clockwise about (4, 3).

Working backwards

A rotation of 90^\circ counterclockwise about C(2, -1) takes a point P to P'(0, 2). Find P.
A rotation of 180^\circ takes (3, 7) to (-1, 1). Find its center.
A rotation of 90^\circ takes A(0, 0) to A'(-1, 5) and B(4, 1) to B'(0, 1). Find its center and its direction.
Which of Q(6, 1), R(5, 3) and S(-2, 5) could be the image of P(4, 5) under some rotation about C(1, 1)?

Any angle

Give exact coordinates.

P(5, 1), rotated 30^\circ counterclockwise about C(1, 1).
P(3, 4), rotated 45^\circ counterclockwise about C(1, 2).
P(-1, 3), rotated 120^\circ counterclockwise about C(1, 3).
P(2, 5), rotated 60^\circ clockwise about C(2, 1).

Problems

The points A(2, 1) and B(5, 3) are neighboring vertices of a square ABCD, labeled counterclockwise. Find C and D.
The segment from A(1, 2) to B(5, 2) is one side of an equilateral triangle whose third vertex lies above it. Find the third vertex.
The hub of a wheel is at (0, 15), in meters, with the ground along the x-axis. A seat starts at (10, 15) and the wheel turns 150^\circ counterclockwise. Where is the seat now, and how high above the ground is it?
Find an equation of the image of the line y = 2x under a rotation of 90^\circ counterclockwise about C(1, 1).

Reasoning

Explain why, for a rotation through an angle that is not a whole number of full turns, the center is the only point that does not move.
Show that a half turn about C(a, b) sends P(x, y) to (2a - x,\ 2b - y), and that C is the midpoint of P and its image.
Show that the offsets (u, v) and (-v, u) have the same length, and that when neither is horizontal or vertical their slopes multiply to -1.
Show that a rotation of 90^\circ counterclockwise about C(a, b) is the same as the rotation of 90^\circ counterclockwise about the origin followed by the translation by (a + b,\ b - a).
Explain why one point and its image are not enough to find the center of a rotation whose angle is not known, and describe a case where two points and their images are not enough either.

Answers

(0, 4). The offset (3, 1) turns to (-1, 3), and (1 - 1,\ 1 + 3) = (0, 4).
(3, 6). The offset (-3, 2) turns clockwise to (2, 3), and (1 + 2,\ 3 + 3) = (3, 6).
(-2, 7). The half turn shortcut gives (2 \cdot 2 - 6,\ 2 \cdot 3 - (-1)) = (-2, 7).
(1, -5). The offset (-3, 2) turns to (-2, -3), and (3 - 2,\ -2 - 3) = (1, -5).
(-5, 4). A turn of 270^\circ counterclockwise is a quarter turn clockwise. The offset (-2, -3) turns to (-3, 2), and (-2 - 3,\ 2 + 2) = (-5, 4).
K'(2, 1), L'(2, 4), M'(0, 1). K is the center, so it stays. L has offset (3, 0), which turns to (0, 3); M has offset (0, 2), which turns to (-2, 0).
The endpoints swap: (1, 4) \mapsto (3, 0) and (3, 0) \mapsto (1, 4), so the segment lands on itself. The center (2, 2) is the segment’s midpoint, and a half turn about a segment’s midpoint always does this.
(2, 6), (2, 3), (4, 3), (4, 6) in order. The offsets (-3, -2), (0, -2), (0, 0), (-3, 0) turn clockwise to (-2, 3), (-2, 0), (0, 0), (0, 3). The 3 by 2 rectangle becomes a 2 by 3 one.
P(5, 1). Undo the turn by rotating P' a quarter turn clockwise about C: the offset (-2, 3) turns to (3, 2), and (2 + 3,\ -1 + 2) = (5, 1).
(1, 4), the midpoint \displaystyle \left(\frac{3 - 1}{2}, \frac{7 + 1}{2}\right), since a half turn puts the center at the midpoint of a point and its image.
Center (2, 3), clockwise. The bisector of BB' is x = 2; the bisector of AA' passes through \displaystyle \left(-\frac{1}{2}, \frac{5}{2}\right) with slope \displaystyle \frac{1}{5}, and at x = 2 it gives y = 3. From (2, 3) the offset of A is (-2, -3) and of A' is (-3, 2), and the clockwise rule sends the first to the second.
Q and S, not R. A rotation about C keeps the distance to C, which is 5 for P. CQ = 5 and CS = 5, but \displaystyle CR = \sqrt{20}.
\displaystyle (1 + 2\sqrt{3},\ 3). The offset (4, 0) turns to \displaystyle (4\cos 30^\circ, 4\sin 30^\circ) = (2\sqrt{3}, 2).
\displaystyle (1,\ 2 + 2\sqrt{2}). The offset (2, 2) points at 45^\circ, so after the turn it points straight up, with its length \displaystyle 2\sqrt{2} unchanged.
\displaystyle (2,\ 3 - \sqrt{3}). The offset (-2, 0) turns to \displaystyle (-2\cos 120^\circ, -2\sin 120^\circ) = (1, -\sqrt{3}).
\displaystyle (2 + 2\sqrt{3},\ 3). With \theta = -60^\circ, the offset (0, 4) turns to \displaystyle (-4\sin(-60^\circ), 4\cos(-60^\circ)) = (2\sqrt{3}, 2).
C(3, 6), D(0, 4). D is B turned a quarter turn counterclockwise about A: the offset (3, 2) becomes (-2, 3). C is A turned a quarter turn clockwise about B: the offset (-3, -2) becomes (-2, 3), and (5 - 2,\ 3 + 3) = (3, 6).
\displaystyle (3,\ 2 + 2\sqrt{3}). Every angle of an equilateral triangle is 60^\circ, so the third vertex is B turned 60^\circ counterclockwise about A: the offset (4, 0) becomes \displaystyle (2, 2\sqrt{3}).
At \displaystyle (-5\sqrt{3},\ 20), which is 20 meters above the ground and about 8.66 meters left of the hub. The offset (10, 0) turns to \displaystyle (10\cos 150^\circ, 10\sin 150^\circ) = (-5\sqrt{3}, 5).
\displaystyle y = -\frac{1}{2}x + 1. The points (0, 0) and (1, 2) of the line have offsets (-1, -1) and (0, 1), which turn to (1, -1) and (-1, 0), giving images (2, 0) and (0, 1). The image line passes through both. Its slope \displaystyle -\frac{1}{2} is perpendicular to 2, as a quarter turn requires.
A point P other than C is at a positive distance from C, and after the turn the segment CP' makes the angle \theta with CP. Since \theta is not a whole number of full turns, CP' points in a different direction from CP, so P' \ne P. The center is at distance 0 and has no direction to turn, so it stays where it is.
The offset (x - a,\ y - b) turns to (a - x,\ b - y), and adding the center gives (2a - x,\ 2b - y). The midpoint of (x, y) and (2a - x,\ 2b - y) is \displaystyle \left(\frac{2a}{2}, \frac{2b}{2}\right) = (a, b).
Both lengths are \displaystyle \sqrt{u^2 + v^2}, since (-v)^2 = v^2. When u and v are both non-zero the slopes are \displaystyle \frac{v}{u} and \displaystyle \frac{u}{-v}, and their product is \displaystyle -\frac{uv}{uv} = -1.
About C, the offset (x - a,\ y - b) turns to (b - y,\ x - a), so the image is (a + b - y,\ b - a + x). The rotation about the origin gives (-y, x), and adding (a + b,\ b - a) gives the same point, for every (x, y).
The center is equally far from a point and its image, so it can be anywhere on their perpendicular bisector, a whole line of candidates. A second pair gives a second bisector, and the center is where the two cross. That fails when the bisectors are the same line, which happens when the two points lie on one line through the center: then the segments from each point to its image are parallel, and both bisectors pass through the center at right angles to them.