Question

describing the properties of a
consider the reflection of (\triangle abc) across the line of
reflection, (overleftrightarrow{pt}).
which statements must be true? check all that apply.
(square aa = cc)
(square cq = qc)
(square overleftrightarrow{pt}perpoverline{aa})
(square overline{cc}perpoverline{bb})
(square overline{aa}paralleloverline{bb})
(square mangle trb = 90^{circ})

Explanation:

Step1: Properties of reflection

In a reflection, the line of reflection is the perpendicular bisector of the segments joining corresponding points. So, \(C'Q = QC\) (since \(PQ\) is the line of reflection and \(Q\) is on the line of reflection, it bisects \(C'C\)). Also, the line of reflection \(PT\) is perpendicular to the segments joining corresponding points. So, \(m\angle TRB=90^{\circ}\) (as \(PT\) is the line of reflection and \(B\) and \(B'\) are corresponding points, the angle between \(PT\) (or \(TR\) which is part of \(PT\)) and \(BB'\) (or \(RB\) which is part of \(BB'\)) is \(90^{\circ}\)).

Step2: Analyzing other options

• \(A'A = C'C\): There is no guarantee from the reflection property that the lengths of \(A'A\) and \(C'C\) are equal. Corresponding segments are congruent in the reflected figure but \(A'A\) and \(C'C\) are not necessarily equal.
• \(\overleftrightarrow{PT}\perp\overline{A'A}\): The line of reflection \(PT\) is the perpendicular bisector of \(A'A\).
• \(\overline{C'C}\perp\overline{B'B}\): There is no property of reflection that makes \(C'C\) and \(B'B\) perpendicular.
• \(\overline{A'A}\parallel\overline{B'B}\): Since both \(A'A\) and \(B'B\) are perpendicular to the line of reflection \(PT\), by the property that two lines perpendicular to the same line are parallel, \(\overline{A'A}\parallel\overline{B'B}\)

Answer:

\(C'Q = QC\), \(m\angle TRB = 90^{\circ}\), \(\overline{A'A}\parallel\overline{B'B}\)