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.

Question

a. Which pocket is closer?
(1 point)
Pocket C is closer to point I than Pocket F
Pocket F is closer to Point I than Pocket C
They are the same distance from point I
b. How do you know? (Either give the name of the theorem or write out what it says) (2 points)
c. What other point on the diagram would also be equidistanct from points F and C? Explain how you know. (2 points)

Answer 1

To solve the problem, let’s analyze the situation step by step.

a. Which pocket is closer?

Since point I lies on the perpendicular bisector GH of segment FC, it is equidistant from both points F and C.

Answer: They are the same distance from point I.

b. How do you know?

The theorem that supports this conclusion is called the Perpendicular Bisector Theorem. It states that any point on the perpendicular bisector of a segment is equidistant from the endpoints of that segment. In this case, point I is on the perpendicular bisector GH of segment FC, and therefore it is equidistant from both points F and C.

Answer: The Perpendicular Bisector Theorem states that any point on the perpendicular bisector of a segment is equidistant from the segment's endpoints (points F and C).

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

Other points that are equidistant from F and C include any point on the line segment GH (the perpendicular bisector itself). This is because, according to the Perpendicular Bisector Theorem, every point on the perpendicular bisector of segment FC is at the same distance from both endpoints F and C.

Answer: Any point on the line segment GH (the perpendicular bisector of FC) is also equidistant from points F and C, as all points on the perpendicular bisector are equidistant from the endpoints of the segment.