QUESTION IMAGE
Question
6 from unit 1, lesson 3
in this diagram, line segment cd is the perpendicular bisector of line segment ab. assume the conjecture that the set of points equidistant from a and b is the perpendicular bisector of ab is true. is point m closer to point a, closer to point b, or the same distance from both points?
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explain how you know.
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(diagram note: $overline{ab} perp overline{cd}$, with points a, m, b on $overline{ab}$ and points c, m, d on $overline{cd}$, m at the intersection with a right angle symbol)
Step1: Recall the given conjecture
The conjecture states that the set of points equidistant from \( A \) and \( B \) is the perpendicular bisector of \( AB \).
Step2: Identify the position of point \( M \)
Point \( M \) lies on \( CD \), and \( CD \) is the perpendicular bisector of \( AB \) (given: \( \overline{AB} \perp \overline{CD} \) and \( CD \) is the perpendicular bisector of \( AB \)).
Step3: Apply the conjecture to point \( M \)
Since \( M \) is on the perpendicular bisector of \( AB \), by the given conjecture, \( M \) must be equidistant from \( A \) and \( B \).
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Point \( M \) is the same distance from both points \( A \) and \( B \) because \( M \) lies on the perpendicular bisector of \( AB \), and the conjecture states that points on the perpendicular bisector of a segment are equidistant from the segment's endpoints.