Constructing Parallel and Perpendicular Lines Quick Check 2 of 52 of 5 Items Question Select the choice that indicates AB←→ is the perpendicular bisector of CD←→ if they intersect at P .(1 point) Responses CD¯¯¯¯¯¯¯¯⊥AB¯¯¯¯¯¯¯¯ Modifying above upper C upper D with bar perpendicular to Modifying above upper A upper B with bar AP¯¯¯¯¯¯¯¯≅BP¯¯¯¯¯¯¯¯ Modifying above upper A upper P with bar congruent to Modifying above upper B upper P with bar AC¯¯¯¯¯¯¯¯⊥BD¯¯¯¯¯¯¯¯ and AP¯¯¯¯¯¯¯¯≅BP¯¯¯¯¯¯¯¯ Modifying above upper A upper C with bar perpendicular to Modifying above upper B upper D with bar and Modifying above upper A upper P with bar congruent to Modifying above upper B upper P with bar AB¯¯¯¯¯¯¯¯⊥CD¯¯¯¯¯¯¯¯ and CP¯¯¯¯¯¯¯¯≅DP¯¯¯¯¯¯¯¯ Modifying above upper A upper B with bar perpendicular to Modifying above upper C upper D with bar and Modifying above upper C upper P with bar congruent to Modifying above upper D upper P with bar Skip to navigation page 14 of 15 Formula keypad has been closed. Press Control + Backslash to open it again.Formula keypad has been closed. Press Control + Backslash to open it again.Formula keypad has been closed. Press Control + Backslash to open it again.Formula keypad has been closed. Press Control + Backslash to open it again.Formula keypad has been closed. Press Control + Backslash to open it again.

1 answer

To indicate that line segment \( \overleftrightarrow{AB} \) is the perpendicular bisector of segment \( \overleftrightarrow{CD} \) when they intersect at point \( P \), we need to mention two critical characteristics:

  1. Perpendicularity: The two lines intersect at right angles.
  2. Bisection: The point of intersection divides the segment \( \overline{CD} \) into two equal parts.

Considering the options you provided, the correct choice would be:

\( \overline{AB} \perp \overline{CD} \) and \( CP \cong DP \).

This indicates that \( AB \) is perpendicular to \( CD \) and that the segments \( CP \) and \( DP \) are congruent, satisfying the conditions of being a perpendicular bisector.

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