Question

select the counterexample that shows that the conjecture is false.
conjecture: a line ( s ) divides ( overline{mn} ) into two line segments. so, the line ( s ) is the segment bisector of ( overline{mn} ).

• if another line also divides ( overline{mn} ), then line ( s ) is not the only segment bisector of ( overline{mn} ).
• line ( s ) must intersect ( overline{mn} ) at a right angle in order to be the segment bisector of ( overline{mn} ).
• if line ( s ) intersects ( overline{mn} ) anywhere besides the midpoint, it is not the segment bisector of ( overline{mn} ).
• if line ( s ) intersects ( overline{mn} ) at the midpoint, it is the perpendicular bisector of ( overline{mn} ).

Explanation:

To find a counterexample, we need a case where a line divides \(\overline{MN}\) into two segments but is not a segment bisector. A segment bisector must intersect at the midpoint. The option "If line \( s \) intersects \(\overline{MN}\) anywhere besides the midpoint, it is not the segment bisector of \(\overline{MN}\)" shows that if \( s \) divides \(\overline{MN}\) (into two segments) but not at the midpoint, it's not a bisector, contradicting the conjecture. Other options either talk about uniqueness, right angles (perpendicular bisector, not just bisector), or incorrect definitions.

Answer:

If line \( s \) intersects \(\overline{MN}\) anywhere besides the midpoint, it is not the segment bisector of \(\overline{MN}\)