Question

rosalyn tried to define a rotation about point c.

• for the center point c, the image c is at the same point as c.
• for a point p that is not at c, a counterclockwise rotation of θ about point c creates point p, such that
• m∠pcp = θ
• point c is the midpoint of pp

which counterexample shows that rosalyns definition does not define a rotation?
choose 1 answer:
a (image: 180° rotation with p, c, p colinear, c midpoint)
b (image: p, c, p with angle θ, cp=cp)
c (image: p, c, p with cp≠cp)
d rosalyn correctly defined a rotation.

Explanation:

To determine the counterexample, we analyze Rosalyn's definition. A rotation about a point \( C \) should preserve the distance from \( C \) to \( P \) (i.e., \( CP = CP' \)). Let's examine each option:

• Option A: \( C \) is the midpoint of \( PP' \), and \( \angle PCP' = 180^\circ \). Here, \( CP = CP' \) (since \( C \) is the midpoint), and the angle is \( 180^\circ \), which is a valid rotation (a 180° rotation).
• Option B: \( \angle PCP' = \theta \), and \( CP = CP' \) (marked with congruent segments), so this fits a rotation.
• Option C: The segment \( CP' \) is longer than \( CP \) (different segment markings), so \( CP

eq CP' \). Rosalyn's definition requires \( C \) to be the midpoint (implying \( CP = CP' \)) and \( m\angle PCP' = \theta \), but here \( CP
eq CP' \), violating the rotation property (distance from center to point and its image must be equal).

• Option D: Incorrect, as Option C shows a violation.

Answer:

C. The diagram where \( CP
eq CP' \) (different segment markings for \( CP \) and \( CP' \))