Question

Carlos is playing pool. He has 1 ball left to make in a pocket and then he will win. He knows that a pool table is a rectangle and there are 6 pockets that you can hit the balls into (indicated by points A, B, C, F, E, and D on the diagram below). In his head, he drew a line segment between points F and C and then pictured its perpendicular bisector, GH¯¯¯¯¯¯¯¯ . The ball he needs to hit is at point I, on the perpendicular bisector. He is trying to decide which pocket is closer, F or C. a. Which pocket is closer? (1 point) Responses Pocket C is closer to point I than Pocket F Pocket C is closer to point I than Pocket F They are the same distance from point I They are the same distance from point I Pocket F is closer to Point I than Pocket C Pocket F is closer to Point I than Pocket C Question 2 b. How do you know? (Either give the name of the theorem

Answer 1

To determine which pocket is closer to point I on the perpendicular bisector GH of the segment FC, we rely on the property of perpendicular bisectors in geometry. The theorem that is relevant here is:

Perpendicular Bisector Theorem: Any point on the perpendicular bisector of a segment is equidistant from the endpoints of that segment.

Since point I lies on the perpendicular bisector GH of segment FC, it must be equidistant from points F and C. This means the distance from point I to pocket F is the same as the distance from point I to pocket C.

The correct responses are: a. They are the same distance from point I. b. The reasoning is based on the Perpendicular Bisector Theorem.

Answer 2

What other point on the diagram would also be equidistanct from points F and C? Explain how you know.

Answer 3

Any point that lies on the line segment FC itself (between points F and C) will also be equidistant from points F and C. Here’s why:

1. Midpoint of FC: The midpoint of segment FC, which we can denote as M, is a specific point that is equidistant from both F and C. The distance from M to F is equal to the distance from M to C, by definition of a midpoint.

2. Forming triangles: For any point P on the line segment FC, you can draw two segments, PF and PC. Since P lies directly between F and C on the same straight line, the lengths of PF and PC will depend on where P is located along the segment. However, because both points are on the same line segment, point P's distance to F and C will always be proportional.

Thus, while the perpendicular bisector uniquely identifies all points that are equally distant from F and C, any point along the line segment connecting F and C will also exhibit this property—although the specific distances will vary depending on the exact location of point P on that line segment.

Therefore, any point on the line segment FC, including but not limited to its midpoint, is equidistant from points F and C.