Rotations

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Learning goals

Describe rotations using centers, directed angles, circular paths, and fixed points
Draw rotated images with angle tools and origin coordinate rules
Explain distance and angle preservation and the retention of orientation
Assess candidate centers and check a common directed angle for paired data

One center and one directed angle

A turn keeps each point the same distance from its center. For example, turn a point $4$ cm from a pin through $90^\circ$ counterclockwise. Its image is still $4$ cm from the pin, one quarter of the way around its circle. A point $7$ cm from that pin follows a different circle but turns through the same angle.

A rotation about a center $O$ sends every point other than $O$ around a circle centered at $O$ , through one specified angle in one specified direction. Write $P\mapsto P'$ as in Translations. The radius condition is $OP'=OP$ , and the directed turn from ray $OP$ to ray $OP'$ is the given turn.

Counterclockwise turns run opposite to clock hands; clockwise turns run with them. When we use signed angles, counterclockwise is positive and clockwise is negative. Thus $+90^\circ$ means $90^\circ$ counterclockwise, while $-90^\circ$ means $90^\circ$ clockwise. An ordinary angle measure alone does not choose a turning direction.

Every off-center point turns through the same angle on a circle of its own radius.

The center itself has radius zero. Define $O'=O$ ; there is no ray $OO$ to which we could assign a turning angle. For a turn strictly between $0^\circ$ and $360^\circ$ , every other point moves. A $0^\circ$ turn or a full $360^\circ$ turn gives the identity rotation, leaving every point fixed. Full extra turns do not change the final image, although they describe a longer motion along the path.

A quarter-turn is $90^\circ$ , a half-turn is $180^\circ$ , and a three-quarter-turn is $270^\circ$ . A half-turn has the same final image in either direction. A three-quarter-turn counterclockwise has the same final image as a quarter-turn clockwise because the two routes reach the same ray.

Worked example 1 Draw a turn without a grid

Problem. Rotate $P$ through $35^\circ$ clockwise about $O$ , where $OP=3.8$ cm. Describe how to locate $P'$ with a ruler, protractor, and compass. Give $OP'$ and the image of $O$ .

The givens do not yet include the constructed image.

Solution. Place the protractor at $O$ , with its baseline on ray $OP$ . Draw a new ray $35^\circ$ clockwise from $OP$ . The direction matters: drawing the ray on the other side would give a different image.

Set the compass to length $OP$ . Mark $P'$ on the new ray at that same distance from $O$ :

$OP'=3.8\text{ cm}.$

The center stays fixed, so $O'=O$ . A pencil drawing approximates the specified angle and radius; the rotation definition gives their exact intended values.

The constructed image uses the clockwise ray and the unchanged radius.

Check your understanding

Which information specifies a 70-degree rotation exactly?

A Only its center B Only the distance a point travels C Its center and a choice of clockwise or counterclockwise direction D Only one image point

Turn the figure while the axes stay fixed

A coordinate rule records a turn relative to the original axes. Keep the paper still and turn the cutout on it. Rotating the paper and its axes together would change your viewpoint rather than show the figure’s new coordinates on those fixed axes.

Start with $(3,2)$ and turn it $90^\circ$ counterclockwise about the origin. Its position is three units right and two up. Under the turn, “right” becomes “up,” and “up” becomes “left.” The image is two left and three up, or $(-2,3)$ .

This reasoning works for signed coordinates too: left becomes down, and down becomes right. Horizontal and vertical roles exchange, with the old vertical contribution becoming a negative horizontal contribution. Thus

$90^\circ\text{ counterclockwise}:\quad(x,y)\mapsto(-y,x).$

For a half-turn, right becomes left and up becomes down, so both signs reverse. For $90^\circ$ clockwise, right becomes down and up becomes right. These direction changes give the following origin rules.

Turn about the origin	Image of $(x,y)$
$90^\circ$ counterclockwise, or $270^\circ$ clockwise	$(-y,x)$
$180^\circ$ in either direction	$(-x,-y)$
$90^\circ$ clockwise, or $270^\circ$ counterclockwise	$(y,-x)$
$0^\circ$ or $360^\circ$	$(x,y)$

These formulas use the origin as center. For another center, reason from that center’s position. Relative to center $(1,2)$ , point $(4,3)$ is three right and one up. A quarter-turn counterclockwise puts it one left and three up from the center, at $(0,5)$ . Applying the origin formula to $(4,3)$ would turn about the wrong point.

Worked example 2 Rotate every vertex on fixed axes

Problem. Triangle $ABC$ has $A(2,1)$ , $B(4,1)$ , and $C(3,3)$ . Rotate it $90^\circ$ counterclockwise about the origin. Find all three image vertices and draw the image with the axes unchanged.

Solution. Use $(x,y)\mapsto(-y,x)$ for each vertex.

For $A$ , the old height $1$ becomes horizontal coordinate $-1$ , and the old horizontal coordinate $2$ becomes height $2$ . Thus $A'(-1,2)$ .

For $B$ , the image is $B'(-1,4)$ .

For $C$ , the image is $C'(-3,3)$ .

Plot those points on the original axes and join $A'B'C'$ in corresponding order. A rotation takes straight segments to straight segments, so these joined sides give the image boundary.

Solid blue is the preimage; dashed red is the image. The axes have not turned.

Check your understanding

Rotate T(-2,5) through 270 degrees counterclockwise about the origin. What is T-prime?

T'(5,2)

T'(-5,-2)

T'(2,-5)

T'(5,-2)

What a turn preserves

Turn a stiff transparent sheet with a traced $11$ cm segment and a $64^\circ$ angle. The segment is not stretched, and the two rays do not open or close. We accept a rotation preserves every distance and every angle measure directly from this turning picture. It is a rigid motion. We also use that it takes straight segments to straight segments. Measurements of a few drawings provide approximate evidence, rather than a proof for the whole plane.

A turn also keeps the sheet’s marked face upward. The clockwise or counterclockwise order of corresponding vertices around a noncollinear triangle stays the same. This is orientation preservation. In the paired triangles of Example 2, following $A,B,C$ and then following $A',B',C'$ goes counterclockwise in both figures. In contrast, Reflections reverse that order by turning the marked face over. Neither motion changes corresponding angle measures.

Worked example 3 Separate turning angle from preserved measurements

Problem. A $120^\circ$ rotation maps noncollinear $D,E,F$ to $D',E',F'$ respectively. Given $DE=4u-3$ cm, $D'E'=25$ cm, and $m\angle DEF=112^\circ$ , find $u$ and the image angle at $E'$ . The order $D,E,F$ is clockwise. State the corresponding image order.

Solution. Corresponding side lengths are equal, so

$4u-3=25.$

Adding $3$ and dividing by $4$ gives $u=7$ .

Angle preservation gives $m\angle D'E'F'=112^\circ$ . The $120^\circ$ turning angle describes how the whole figure moves; it is not added to the angle inside it.

The order $D',E',F'$ is clockwise because a rotation preserves orientation.

Points at different radii can travel different distances along their circles. Preserved distance compares two points before and after the turn, such as $DE$ and $D'E'$ . It does not say that every point travels an equal distance.

Check your understanding

A 40-degree rotation sends A, B, C to A-prime, B-prime, C-prime. The noncollinear order A, B, C is counterclockwise and angle ABC measures 57 degrees. What is guaranteed?

A Clockwise image order and a 57-degree angle B Counterclockwise image order and a 97-degree angle C Clockwise image order and a 40-degree angle D Counterclockwise image order and a 57-degree angle

Investigate a missing center

An image can leave clues about its center. When a nonidentity rotation is known to exist, try this center-finding investigation: draw two nonzero point-image connectors, then draw each connector’s perpendicular bisector. These are lines through the connectors’ midpoints at right angles. Use a ruler and protractor for this exploration.

Two distinct bisectors have at most one intersection. If they intersect, their intersection is a candidate center; if they are parallel, the search provides no candidate. If the two chosen bisectors coincide, choose another point-image pair when one is available. A fixed point has a zero connector, so it supplies no perpendicular bisector. If a rotation is known to be nonidentity, a known fixed point already identifies its center directly. For an identity rotation, every point is fixed and these data do not single out a center.

This search is exploratory here; the general perpendicular-bisector justification returns in lesson 9.1. An intersection alone does not establish a rotation for arbitrary data. At a candidate center, check equal start and finish radii for each pair and check that the same directed angle maps both pairs. If the angles disagree, one rotation cannot make both matches.

Worked example 4 Locate a candidate and verify the turn

Problem. Investigate whether one rotation can send $A(-1,-1)$ to $A'(5,-1)$ and $B(2,2)$ to $B'(2,-4)$ . Use the two-bisector search to find a candidate center, then verify both pairs from the rotation definition.

The given pairs provide two nonzero connectors. The center has not been marked.

Solution. The horizontal connector $AA'$ has midpoint $(2,-1)$ . Its perpendicular bisector is the vertical line $x=2$ .

The vertical connector $BB'$ also has midpoint $(2,-1)$ . Its perpendicular bisector is the horizontal line $y=-1$ . The two distinct lines meet at candidate $O(2,-1)$ .

Relative to $O$ , $A$ is three left and $A'$ is three right. Both radii have length $3$ , and a $180^\circ$ turn matches them.

Point $B$ is three above $O$ and $B'$ is three below. Again both radii have length $3$ , and the same $180^\circ$ turn matches them. Thus the half-turn about $O(2,-1)$ makes both specified matches. Either direction gives this half-turn’s final image.

Two distinct bisectors give a candidate; equal radii and the same half-turn verify these pairs.

One nonzero pair without a specified turn generally leaves many possible centers. Two pairs with distinct bisectors give at most one candidate: parallel bisectors have no intersection. When an intersection exists, the common directed-angle check is still necessary. Agreement for the supplied pairs verifies those matches; the chosen rotation then specifies the image of every other point.

Check your understanding

Two nonzero point-image connectors have distinct perpendicular bisectors meeting at O. What additional check is needed for arbitrary proposed data?

A Only that O is on a coordinate axis B Equal radii for each pair and the same directed turning angle for both C Only that the two connectors have equal lengths D No check; the intersection already proves a rotation

Status	What we may use
Earlier foundations and results	Retain the angle and line results and the accepted slide and fold foundations linked in Reflections
Defined here	A common center and directed turn; unchanged radius for each point; the center fixed; identity turns fix every point
Accepted rotation foundations	Every distance and angle measure is preserved; straight segments map to straight segments
Derived from coordinate directions and the turn picture	Origin rules, equivalent quarter and three-quarter turn images, and retention of triangle orientation
Exploratory and observed	Circular paths and measurements illustrate the turn; the two-bisector search suggests a center, which must be checked against the paired data

Rotations

Learning goals

One center and one directed angle

Turn the figure while the axes stay fixed

What a turn preserves

Investigate a missing center

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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