Pearson Geometry Common Core, 2011
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Pearson Geometry Common Core, 2011 View details
6. Locus: A Set of Points
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Exercise 24 Page 809

Begin by considering three points that are not collinear. How do you find a point equidistant from these points? Then, consider that these points are collinear.

Yes, she is correct.

Practice makes perfect

We will begin by recalling our classmate's statement.

It is impossible to find a point equidistant from three collinear points.

Before we decide whether she is correct, we will examine things step by step. Let's first think about the points that are equidistant from two points, A and B, in a plane.

As we can see, the locus of two points in a plane forms a line. This line is the perpendicular bisector of AB.

Note that any point equidistant from A and B lies on the perpendicular bisector of AB. Now, let's consider another point C in the same plane with A and B such that A, B, and C are not collinear.

If we are looking for a point equidistant from A, B, and C, it is the intersection of the perpendicular bisectors of AB and BC.

Finally, to understand whether our classmate is correct, let's assume A, B, and C are collinear points.

As we can see, when the points are collinear, the perpendicular bisectors do not intersect. Therefore, our classmate is correct. We cannot find a point equidistant from three collinear points.