Line R T is a perpendicular bisector of Line segment U V.

Question

Line R T is a perpendicular bisector of line segment U V. The lines intersect at point S. Line segment U S and S V are congruent.

What is true of any triangle created by points U, V, and any point on Line R T other than S?

It will be a right triangle.
It will be an acute triangle.
It will be an equilateral triangle.
It will be an isosceles triangle.

Answer 1

Given the conditions stated in your problem, line segment \( RT \) is the perpendicular bisector of segment \( UV \), meaning that \( S \) is the midpoint of segment \( UV \), and that \( RT \) intersects \( UV \) at a right angle. Let’s analyze what can be said about a triangle formed by points \( U, V, \) and any point \( P \) on line \( RT \) that is not point \( S \).

1. Congruence: Since \( S \) is the midpoint, it follows that \( US \) is congruent to \( SV \).

2. Angles: Since \( RT \) is the perpendicular bisector of \( UV \), the angle \( USP \) (where \( P \) is a point on \( RT \)) will be a right angle (90 degrees), because \( RT \) intersects \( UV \) perpendicularly at \( S \).

When considering triangle \( UVP \):

• One angle (angle \( USP \)) is \( 90^\circ \).
• Thus, triangle \( UVP \) must be a right triangle as it contains a right angle.

Conclusion:

The correct answer is:

It will be a right triangle.