Question

which rigid transformation would map △mzk to △qzk? a rotation about point z; a reflection across the line containing \\(\overline{mz}\\); a rotation about point k; a reflection across the line containing \\(\overline{zk}\\)

Explanation:

Step1: Analyze Rigid Transformations

Rigid transformations preserve shape and size. For triangles \( \triangle MZK \) and \( \triangle QZK \), we check the relationship. \( ZK \) is a common side, \( \angle MZK = \angle QZK \) (right angles), and \( MK = QK \) (implied by symmetry about \( ZK \)). A reflection across \( ZK \) would map \( M \) to \( Q \), \( Z \) to \( Z \), \( K \) to \( K \), so \( \triangle MZK \) maps to \( \triangle QZK \). Rotation about \( Z \) or \( K \) doesn't align as \( ZK \) is the axis of symmetry here.

Step2: Evaluate Options

• Rotation about \( Z \): Doesn't align \( M \) to \( Q \) as angles/sides don't match rotation about \( Z \).
• Rotation about \( K \): \( MK \) and \( QK \) are equal, but rotation about \( K \) would not use the right angle at \( Z \) as a mirror.
• Reflection across \( MZ \): \( MZ \) is perpendicular to \( ZK \), reflection here wouldn't map \( M \) to \( Q \).
• Reflection across \( ZK \): \( ZK \) is the line where \( \triangle MZK \) and \( \triangle QZK \) are symmetric (right angles, common side \( ZK \), \( MZ = QZ \) likely), so reflection across \( ZK \) maps \( \triangle MZK \) to \( \triangle QZK \).

Answer:

a reflection across the line containing \( \overline{ZK} \)