12 multiple-choice questions, progressively harder.
Someone claims every point moves under a 25∘25^\circ25∘ rotation about OOO. Which fact refutes the claim?
Solution
Correct answer: D
The center has radius zero and is fixed, even when the rotation is not an identity. That one point refutes the claim about every point.
Which rotation about the origin makes both matches P(1,4)↦P′(4,−1)P(1,4)\mapsto P'(4,-1)P(1,4)↦P′(4,−1) and Q(−2,3)↦Q′(3,2)Q(-2,3)\mapsto Q'(3,2)Q(−2,3)↦Q′(3,2)?
Correct answer: C
The clockwise quarter-turn uses (x,y)↦(y,−x)(x,y)\mapsto(y,-x)(x,y)↦(y,−x).
(1,4)↦(4,−1)(1,4)\mapsto(4,-1)(1,4)↦(4,−1)
(−2,3)↦(3,2)(-2,3)\mapsto(3,2)(−2,3)↦(3,2)
Both pairs agree with the same rule.
For the pictured pairs, the connector bisectors y=xy=xy=x and y=−xy=-xy=−x meet at the origin. Does this candidate give one rotation making both matches?
Correct answer: B
At the origin, A is two right and A-prime two up. This needs 90 degrees counterclockwise. B is three left and B-prime three up, which needs 90 degrees clockwise.
Both pairs keep their radii, but their directed angles disagree. The intersection alone is not a consistency check.
Two nonzero connectors have the same perpendicular bisector, x = 1. What should you do to continue the two-bisector center search?
Correct answer: A
Coincident lines do not isolate a single candidate by intersection. Choose another suitable pair when available. This failure of the selected search lines does not itself prove the rotation impossible.
A 90∘90^\circ90∘ counterclockwise rotation about the origin sends P(2s−1,s+2)P(2s-1,s+2)P(2s−1,s+2) to P′(−6,7)P'(-6,7)P′(−6,7). Which value of sss satisfies both coordinates?
The image rule gives horizontal coordinate −(s+2)-(s+2)−(s+2) and vertical coordinate 2s−12s-12s−1. Match the horizontal coordinate first.
−(s+2)=−6-(s+2)=-6−(s+2)=−6
s=4s=4s=4
Check the other coordinate.
2(4)−1=72(4)-1=72(4)−1=7
Both coordinates agree.
A half-turn sends X(−5,4)X(-5,4)X(−5,4) to X′(3,−2)X'(3,-2)X′(3,−2). What image does it give Z(0,−3)Z(0,-3)Z(0,−3)?
A half-turn places its center halfway between each point and image. Halfway from -5 to 3 is -1, and halfway from 4 to -2 is 1, so the center is (-1,1).
Z is one right and four below this center. Its image is one left and four above.
x′=−1−1=−2x'=-1-1=-2x′=−1−1=−2
y′=1+4=5y'=1+4=5y′=1+4=5
Rotate the entire line x=2x=2x=2 through 180∘180^\circ180∘ about the origin. Which statement gives its image and fixed points?
A general point of the starting line is (2,y)(2,y)(2,y). The half-turn gives
(2,y)↦(−2,−y).(2,y)\mapsto(-2,-y).(2,y)↦(−2,−y).
As y ranges over all real numbers, -y does too, giving the whole line x=−2x=-2x=−2. No point with starting horizontal coordinate 2 is the origin, the only fixed point.
An unknown transformation matches a particular rotation on three tested vertices. What follows from those tests alone?
The tests describe only three inputs of an unknown rule. A chosen rotation gives an image for every point, but agreement on a finite selection does not establish that the unknown transformation uses that rule everywhere.
Rotate H(2,0)H(2,0)H(2,0) through 90∘90^\circ90∘ counterclockwise about C(−1,2)C(-1,2)C(−1,2). What is H′H'H′?
H is three right and two down from C. A counterclockwise quarter-turn changes three right to three up and two down to two right.
x′=−1+2=1x'=-1+2=1x′=−1+2=1
y′=2+3=5y'=2+3=5y′=2+3=5
A proposed rotation about OOO sends PPP and QQQ to the same image RRR. Given OP=5OP=5OP=5 cm and OQ=8OQ=8OQ=8 cm, can this happen?
Radius preservation for P requires OR=5OR=5OR=5 cm. For Q it requires OR=8OR=8OR=8 cm.
5≠85\ne85=8
One radius OR cannot have both lengths. A rotation may have different starting radii, but it keeps each one.
Which two instructions about the same center give the same final image for every point?
Negative 270 degrees means a three-quarter-turn clockwise. It ends on the same ray as a quarter-turn counterclockwise.
360−270=90360-270=90360−270=90
The same center and unchanged radius then give the same final image point.
To rotate VVV through 120∘120^\circ120∘ counterclockwise about OOO, a student has set a compass to OV=7.2OV=7.2OV=7.2 cm. Which construction is correct?
The rotation angle is between the starting and image rays based at the center O. The image must lie on the counterclockwise ray at the unchanged radius 7.2 cm. Both conditions are necessary.
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