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Rotations: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 One point, two specified turns

Start with $J(-8,2)$ separately in each case. Find the image about the origin and describe the direction of the turn.

Case A. The signed angle is $+90^\circ$ .

Case B. The signed angle is $-180^\circ$ .
1. Hint 1
  The sign selects the direction; the magnitude selects the amount of turning.
2. Hint 2
  Choose the origin rule separately for each case. Both cases restart at J.
Answer

Case A: $J'(-2,-8)$ , counterclockwise. Case B: $J'(8,-2)$ , clockwise.

Full solution

Case A.
A positive angle means counterclockwise.
Use $(x,y)\mapsto(-y,x)$ .

The new horizontal coordinate is the negative of the old height.
$x'=-2$

The new height is the old horizontal coordinate.
$y'=-8$

Case B.
A negative angle means clockwise.
A half-turn negates both coordinates.

The image is $J'(8,-2)$ .
A half-turn counterclockwise would have the same final image.

Answer

Case A: $J'(-2,-8)$ , counterclockwise. Case B: $J'(8,-2)$ , clockwise.

Key idea

Keep the turning direction separate from the coordinate rule for its final image.
Problem 2 Different circles, one angle

A rotation turns the plane $50^\circ$ counterclockwise about $O$ . Given $OP=4.2$ cm and $OQ=7.5$ cm, describe the circular paths of P and Q and give their image distances from O. State the image of O.

Do the different path radii mean that the distance between P and Q changes? Explain.
1. Hint 1
  A rotation keeps each point at its own distance from its center.
2. Hint 2
  Separate radius and travel along a path from distance between two points of the figure.
Answer

P follows a radius-4.2 cm circle and Q a radius-7.5 cm circle, each through $50^\circ$ counterclockwise; $OP'=4.2$ cm, $OQ'=7.5$ cm, and $O'=O$ . The distance between P and Q is unchanged.

Full solution

P stays on the circle centered at O with radius 4.2 cm.
Q stays on the circle centered at O with radius 7.5 cm.
The paths have different radii, but both turns have the same directed angle.

Each radius is preserved.
$OP'=4.2\text{ cm}$

For Q,
$OQ'=7.5\text{ cm}$

O has radius zero and stays fixed, so $O'=O$ .

The accepted distance-preservation property applies to the pair P,Q.
Thus $P'Q'=PQ$ , even though the individual circular paths differ.

Answer

P follows a radius-4.2 cm circle and Q a radius-7.5 cm circle, each through $50^\circ$ counterclockwise; $OP'=4.2$ cm, $OQ'=7.5$ cm, and $O'=O$ . The distance between P and Q is unchanged.

Key idea

A common turning angle does not require a common path radius.
Problem 3 Rotate the whole triangle

Triangle $ABC$ has vertices $A(-3,-2)$ , $B(1,-2)$ , and $C(-1,1)$ . Rotate it $270^\circ$ clockwise about the origin. Give all image coordinates and draw the image on the same coordinate axes.

State how to join the image vertices and explain why the axes do not turn.

The givens show the preimage only.

Text description of this figure

Triangle ABC is drawn on a fixed unit grid with vertices A(-3,-2), B(1,-2), C(-1,1). The image is not shown.
1. Hint 1
  Find the shorter turn that reaches the same final ray.
2. Hint 2
  Use the matching origin rule on every vertex, retaining the original correspondence.
Answer

$A'(2,-3)$ , $B'(2,1)$ , $C'(-1,-1)$ ; join $A'B'C'A'$ . The original coordinate axes stay fixed.

Full solution

A three-quarter-turn clockwise ends at the same position as a quarter-turn counterclockwise.

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

For A, negating the height and exchanging roles gives $A'(2,-3)$ .

For B, the image is $B'(2,1)$ .

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

Join $A'B'$ , $B'C'$ , and $C'A'$ because these correspond to the three original sides.
A rotation takes each straight side to its straight image side.

The coordinates describe the moving figure relative to the original axes.
Turning the paper and axes together would change the reference instead of plotting this image.

The image is plotted on the original axes.

Answer

$A'(2,-3)$ , $B'(2,1)$ , $C'(-1,-1)$ ; join $A'B'C'A'$ . The original coordinate axes stay fixed.

Key idea

Apply one center and directed turn to the entire boundary.
Problem 4 Measure positions from the actual center

Rotate $D(-1,-5)$ through $90^\circ$ counterclockwise about $C(2,-3)$ . Find $D'$ and explain the horizontal and vertical changes in position relative to C.

Why would applying $(x,y)\mapsto(-y,x)$ directly to D use the wrong rotation?
1. Hint 1
  The origin formula assumes the origin is the center.
2. Hint 2
  Find how far left or right and above or below C the point starts, then turn those directions.
Answer

$D'(4,-6)$ ; three left and two down from C become three down and two right. Direct use of $(-y,x)$ turns about the origin.

Full solution

D is three units left and two units below C.
In a counterclockwise quarter-turn, left becomes down and down becomes right.

The image is therefore two units right and three units below C.

Its horizontal coordinate is
$2+2=4$

Its vertical coordinate is
$-3-3=-6$

Thus $D'(4,-6)$ .
The direct origin rule would instead give $(5,-1)$ because it measures the turn from (0,0), rather than C.

Answer

$D'(4,-6)$ ; three left and two down from C become three down and two right. Direct use of $(-y,x)$ turns about the origin.

Key idea

Describe a point relative to the named center before using direction changes.
Problem 5 A rigid sign turns on its drawing

Coordinates on a sign drawing are in millimeters. The rigid sign turns $90^\circ$ clockwise about its hinge mark $O(2,1)$ . Before the turn, a hole is centered at $H(5,1)$ . Find the new hole center.

An edge is 18 mm long and a corner measures $76^\circ$ . Give their corresponding image measurements. The labeled triangle formed by three noncollinear marks has counterclockwise orientation; give its image orientation.
1. Hint 1
  A hinge mark is the rotation center; describe the hole from that mark.
2. Hint 2
  Turning preserves measurements within the rigid sign and does not flip its marked face.
Answer

$H'(2,-2)$ ; edge 18 mm; corner $76^\circ$ ; counterclockwise image orientation.

Full solution

H is three millimeters right of O.
A clockwise quarter-turn puts it three millimeters below O.

The horizontal position is the hinge's horizontal coordinate, 2.
The new height is
$1-3=-2$

The hole center is $H'(2,-2)$ .

Distance preservation keeps the corresponding edge length at 18 mm.

Angle preservation keeps the corner measure at 76 degrees.
The 90-degree turning angle is not added to it.

A rotation preserves the orientation of the three noncollinear marks, so their image order remains counterclockwise.

Answer

$H'(2,-2)$ ; edge 18 mm; corner $76^\circ$ ; counterclockwise image orientation.

Key idea

A turn changes placement while keeping lengths, angle measures, and orientation.
Problem 6 Use the corresponding measurements

A rotation maps $K,L,M$ to $K',L',M'$ respectively. Find the variable and common measurement in each case.

Case A. $KL=2p+9$ cm and $K'L'=5p-6$ cm.

Case B. $m\angle KLM=(4q-8)^\circ$ and $m\angle K'L'M'=(2q+26)^\circ$ .
1. Hint 1
  Both distance preservation and angle preservation are accepted rotation properties.
2. Hint 2
  Set corresponding measures equal separately in the two cases, then substitute to find each common measure.
Answer

Case A: $p=5$ , common length 19 cm. Case B: $q=17$ , common angle measure $60^\circ$ .

Full solution

Case A.
Equal corresponding lengths give
$2p+9=5p-6$

Subtract 2p and add 6.
$15=3p$

Thus $p=5$ .
Substitution gives
$2(5)+9=19$

Case B.
Equal corresponding angle measures give
$4q-8=2q+26$

Subtract 2q and add 8.
$2q=34$

Thus $q=17$ .
The common measure is
$4(17)-8=60$

The length is positive and the 60-degree angle is a valid triangle angle.

Answer

Case A: $p=5$ , common length 19 cm. Case B: $q=17$ , common angle measure $60^\circ$ .

Key idea

Pair the matching objects before equating their measures.
Problem 7 Construct a triangle turn on paper

Draw a noncollinear triangle $ABC$ on paper, using the picture as a model. Rotate the entire triangle $65^\circ$ counterclockwise about vertex A with a ruler, protractor, and compass. No exact scale for the starting triangle is required.

Label all image vertices and keep the radius and angle marks that explain your construction. Identify the fixed vertex and explain what is approximate in the finished pencil drawing.

The starting triangle is a model, not a scale requirement.

Text description of this figure

A noncollinear triangle ABC is drawn without a grid. A is the specified center. Only the original triangle is shown; the requested turn is 65 degrees counterclockwise.
1. Hint 1
  The center stays fixed. Each other vertex needs an image ray and the original radius.
2. Hint 2
  At A, draw the specified directed angle separately from AB and AC. Copy each original distance along its new ray.
Answer

A rotated triangle $A'B'C'$ with $A'=A$ ; each off-center vertex keeps its radius and turns $65^\circ$ counterclockwise. Pencil lengths and angles are approximate.

Full solution

A is the center, so $A'=A$ .
Its radius is zero; no angle based on a ray AA is needed.

Place the protractor at A.
From ray AB, draw a ray 65 degrees counterclockwise.
Copy length AB onto this ray with the compass to place B-prime.

From ray AC, draw a separate ray 65 degrees counterclockwise.
Copy length AC onto this ray to place C-prime.

These steps give $AB'=AB$ and $AC'=AC$ , with the same directed turn for both vertices.

Join $A'B'$ , $B'C'$ , and $C'A'$ .
Rotation takes each straight side to the corresponding straight image side.

The exact intended construction satisfies the rotation definition.
Protractor readings, compass settings, line thickness, and pencil placement make the physical drawing approximate.

The revealed construction retains the center, radii, and directed turn.

Answer

A rotated triangle $A'B'C'$ with $A'=A$ ; each off-center vertex keeps its radius and turns $65^\circ$ counterclockwise. Pencil lengths and angles are approximate.

Key idea

A non-grid image is located by a directed ray and an unchanged radius for every off-center vertex.
Problem 8 Repair a quarter-turn rule

Nora rotates $K(-7,4)$ through $90^\circ$ clockwise about the origin. She writes $K'(-4,-7)$ and explains, “I exchanged the coordinates and negated the new first coordinate.”

Identify the direction her rule actually uses, give the correct clockwise image, and explain the needed sign change.
1. Hint 1
  Compare the described rule with how right, up, left, and down turn.
2. Hint 2
  A clockwise quarter-turn makes the old height the new horizontal coordinate and the negative of the old horizontal coordinate the new height.
Answer

Nora used the counterclockwise quarter-turn rule. The correct clockwise image is $K'(4,7)$ ; negate the old horizontal coordinate, which becomes the new height.

Full solution

Exchanging coordinates and negating the new first coordinate gives $(-y,x)$ .
That is the counterclockwise quarter-turn rule, so her result fits the opposite direction.

A clockwise quarter-turn instead uses $(y,-x)$ .
The new horizontal coordinate is
$x'=4$

Negate the old horizontal coordinate for the new height.
$-(-7)=7$

Thus the correct image is $K'(4,7)$ .
The starting position seven left and four up becomes seven up and four right.

Answer

Nora used the counterclockwise quarter-turn rule. The correct clockwise image is $K'(4,7)$ ; negate the old horizontal coordinate, which becomes the new height.

Key idea

Swapping coordinates is only part of a quarter-turn rule; the sign change selects the direction.
Problem 9 Test a proposed center

Proposed data are $A(4,1)\mapsto A'(1,4)$ and $B(1,3)\mapsto B'(1,-1)$ . Their two nonzero connectors have distinct perpendicular bisectors meeting at the candidate $O(1,1)$ .

Check the radii and directed turns at O. Decide whether a single rotation about this candidate makes both matches, and explain why the bisector intersection alone did not decide the question.

The candidate is given for testing.

Text description of this figure

A fixed grid shows the two stated point-image pairs and candidate O at (1,1). No derived angle arc or radius measurement is drawn.
1. Hint 1
  The definition requires an unchanged radius for each pair and one common directed angle.
2. Hint 2
  Describe the four points as right, left, above, or below O. Compare the turn needed by A with that needed by B.
Answer

The radii match within each pair, but A needs $90^\circ$ counterclockwise and B needs $180^\circ$ . No single rotation about O makes both matches.

Full solution

Relative to O, A is three right and A-prime is three above.
Both radii have length 3.
The directed turn is a counterclockwise quarter-turn.

B is two above O and B-prime two below.
Both radii have length 2.
This pair needs a half-turn, in either direction.

The turning amounts disagree:
$90\ne180$

Thus the candidate does not give one common rotation.
The intersection suggested where to test a center, but equal radii alone did not check the common directed angle.

Answer

The radii match within each pair, but A needs $90^\circ$ counterclockwise and B needs $180^\circ$ . No single rotation about O makes both matches.

Key idea

A candidate must satisfy the entire rotation definition for all supplied pairs.
Problem 10 Which center records are sufficient?

For each separate record below, decide whether the information identifies a center, gives only a candidate to check, or fails to isolate one by the two-bisector search. State any additional information or check needed.

Record A. Every point is fixed because the specified rotation is a full turn.

Record B. A rotation through an angle strictly between 0 and 360 degrees has a known fixed point F.

Record C. A nonidentity rotation is known to exist. Two selected nonzero connectors have coincident perpendicular bisectors, and other point-image pairs are available.

Record D. Arbitrary proposed data give two nonzero connectors with distinct perpendicular bisectors meeting at C. No turning angles have yet been checked.
1. Hint 1
  Treat identity and nonidentity rotations separately. A zero connector cannot be used to draw a perpendicular bisector.
2. Hint 2
  For the two-bisector method, check whether the two lines isolate one intersection.
3. Hint 3
  Once there is a candidate from arbitrary data, return to both radius and directed-angle conditions.
Answer

A: no unique center. B: center F. C: the selected bisectors do not isolate a candidate; choose another nonzero connector with a distinct bisector. D: C is only a candidate; verify both pairs keep their radii and use the same directed angle.

Full solution

Record A.
A full turn fixes every point about any chosen center.
The fixed-point data do not identify one center.

Record B.
In the stated nonidentity range, only the center is fixed.
Therefore the known fixed point F identifies the center directly.
Its zero connector needs no perpendicular-bisector construction.

Record C.
The two search lines coincide rather than meet at a single isolated point.
Keep one connector and choose another available nonzero connector with a distinct bisector.
Coincidence does not mean the promised rotation fails to exist.

Record D.
The distinct lines isolate candidate C.
Check equal start and finish radii in each pair, then check that the same angle in the same direction sends both starting rays to their image rays.
If the checks fail, C is not a center for the proposed matches.

Answer

A: no unique center. B: center F. C: the selected bisectors do not isolate a candidate; choose another nonzero connector with a distinct bisector. D: C is only a candidate; verify both pairs keep their radii and use the same directed angle.

Key idea

Decide what the data establish before treating a center-finding search as complete.

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