Question

applications

1. tamika must walk from point a to point b. her friend tiffany must walk from point c to point d.

(a) if both friends walk the shortest path possible, at how many locations can the two meet? explain.
(b) locate all points they can meet on the diagram.
reasoning

Explanation:

(a)

Step1: Recall shortest path property

The shortest path between two points is a straight line (by the geometric postulate that the shortest distance between two points is a straight - line segment).

Step2: Analyze intersection of paths

Tamika's path is the straight line segment \(\overline{AB}\) and Tiffany's path is the straight line segment \(\overline{CD}\). Two straight - line segments (non - parallel, as per the diagram's implied configuration) intersect at at most one point. So, if we assume that the segments \(\overline{AB}\) and \(\overline{CD}\) intersect, they will intersect at exactly one point. If they are parallel, they will not intersect. But from the diagram (with points \(A\), \(D\) at the top, \(C\) at the bottom - left, and \(B\) at the bottom - right), the segments \(\overline{AB}\) and \(\overline{CD}\) are not parallel and will intersect at one point.

To locate the meeting point, we draw the straight - line segment \(\overline{AB}\) (connecting point \(A\) to point \(B\)) and the straight - line segment \(\overline{CD}\) (connecting point \(C\) to point \(D\)). The point where these two line segments intersect is the only point where they can meet.

Answer:

The two friends can meet at 1 location. This is because the shortest path between two points is a straight line, and two straight - line segments (the paths of Tamika and Tiffany) intersect at at most one point (and in this case, the segments \(\overline{AB}\) and \(\overline{CD}\) intersect at one point).

(b)