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consider the reflection of δabc across the line of reflection, pt. whic…

Question

consider the reflection of δabc across the line of reflection, pt. which statements must be true? check all that apply. aa = cc cq = qc pt⊥aa cc ⊥ bb aa || bb m∠trb = 90°

Explanation:

Step1: Recall reflection properties

In a reflection, the line of reflection is the perpendicular - bisector of the segments connecting pre - image and image points.

Step2: Analyze each statement

  • For $A'A = C'C$: There is no guarantee that the distances from $A$ to $A'$ and $C$ to $C'$ are equal just based on reflection. This is false.
  • For $C'Q=QC$: Since the line of reflection $\overleftrightarrow{PT}$ is the perpendicular - bisector of the segment $\overline{C'C}$, and $Q$ lies on the line of reflection, $C'Q = QC$. This is true.
  • For $\overleftrightarrow{PT}\perp\overline{A'A}$: The line of reflection is perpendicular to the line segment connecting a point and its image. So $\overleftrightarrow{PT}\perp\overline{A'A}$. This is true.
  • For $\overline{C'C}\perp\overline{B'B}$: There is no reason for $\overline{C'C}$ and $\overline{B'B}$ to be perpendicular. This is false.
  • For $\overline{A'A}\parallel\overline{B'B}$: Since the line of reflection is perpendicular to both $\overline{A'A}$ and $\overline{B'B}$, $\overline{A'A}\parallel\overline{B'B}$. This is true.
  • For $m\angle TRB = 90^{\circ}$: There is no information to suggest that $\angle TRB$ is a right - angle. This is false.

Answer:

$C'Q = QC$, $\overleftrightarrow{PT}\perp\overline{A'A}$, $\overline{A'A}\parallel\overline{B'B}$