The locus of points at a given distance from a given point is a circle. What can we say about the locus of points equidistant from two given points? (Hint: We know that any point that is equidistant from two given points A and B lies on the perpendicular bisector of AB. Does this make the perpendicular bisector the locus? For this, we have to show that all the points on the perpendicular bisector are equidistant from A and B.)
The locus of points equidistant from two given points A and B is the perpendicular bisector of the line segment AB.
For the perpendicular bisector to be the locus, we must establish two conditions. First, any point that is equidistant from A and B must lie on the perpendicular bisector of AB. Second, every point on the perpendicular bisector of AB must be equidistant from A and B.
Since both conditions are satisfied—every point on the perpendicular bisector is equidistant from A and B, and every point equidistant from A and B lies on it—the perpendicular bisector forms the locus. Consequently, the centres of all circles passing through A and B lie on this perpendicular bisector.
Explanation
The textbook context establishes that the perpendicular bisector is the locus by proving a bidirectional condition: all points equidistant from A and B lie on the perpendicular bisector, and all points on the perpendicular bisector are equidistant from A and B. This directly answers the hint provided in the question, confirming the locus.
Solution Steps
Step 1: State that the locus is the perpendicular bisector of the line segment joining the two points.
Step 2: Verify that any point equidistant from A and B lies on the perpendicular bisector.
Step 3: Verify that every point on the perpendicular bisector is equidistant from A and B.
Step 4: Conclude that since both conditions are met, the perpendicular bisector is the required locus.