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Question
here are triangles stu and stj.
\\( \overline { u s } \cong \overline { j s } \\)
\\( \overline { t u } \cong \overline { t j } \\)
reflect triangle stu across line st. without any additional justification, which of these is a valid reason why the image of u will coincide with j?
a the image of u and j are on the same side of line st and make the same angle with it at t.
b the image of u and j are the same distance along the same ray from t.
c the image of u and j coincide after reflection because we defined our transformation that way.
d line st is the perpendicular bisector of the segment connecting u and j, because the perpendicular bisector is determined by 2 points that are both equidistant from the endpoints of a segment.
- Option A: Reflecting over \( ST \), the image of \( U \) and \( J \) may not be on the same side of \( ST \), and the angle reasoning is not valid for coincidence.
- Option B: The "same ray from \( T \)" is incorrect as reflection over \( ST \) would place the image on the opposite ray (or same only if \( U \) is on \( ST \), which it's not).
- Option C: The transformation (reflection) is not defined to make \( U \)'s image coincide with \( J \); it's based on geometric properties, not arbitrary definition.
- Option D: Given \( \overline{US} \cong \overline{JS} \) and \( \overline{TU} \cong \overline{TJ} \), points \( S \) and \( T \) are equidistant from \( U \) and \( J \). By the perpendicular bisector theorem, a line through two points equidistant from the endpoints of a segment is the perpendicular bisector of that segment. So \( ST \) is the perpendicular bisector of \( UJ \), meaning reflecting \( U \) over \( ST \) (the perpendicular bisector) maps \( U \) to \( J \) (since reflection over a perpendicular bisector of a segment swaps its endpoints).
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D. Line \( ST \) is the perpendicular bisector of the segment connecting \( U \) and \( J \), because the perpendicular bisector is determined by 2 points that are both equidistant from the endpoints of a segment.