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⭕ Chapter 5: I'm Up and Down, and Round and Round • Think, Draw and Infer

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Exercise Set 5.1 3Think, Draw and Infer 1Exercise Set 5.2 2Exercise Set 5.3 3Exercise Set 5.4 2Exercise Set 5.5 3Exercise Set 5.6 1End-of-Chapter Exercises 12All Exercises 27
Q1

A, B and C are three collinear points. Can you find a point P such that $PA = PB = PC$? What can you say about the perpendicular bisectors of AB and BC? Draw and check. Can you show that for three collinear points A, B and C, the perpendicular bisectors of AB and BC are parallel? Is it possible for a circle to pass through collinear points? Can you draw a line that cuts a given circle in three distinct points?

Answer: No, you cannot find a point $P$ such that $PA = PB = PC$.
The perpendicular bisector of $AB$ contains all points equidistant from $A$ and $B$, and the perpendicular bisector of $BC$ contains all points equidistant from $B$ and $C$. If $A, B$, and $C$ lie on the same straight line, the perpendicular bisectors of segments $AB$ and $BC$ will both be perpendicular to that same line. Since two lines perpendicular to the same line are parallel to each other, these bisectors will never intersect. Without an intersection point, there is no single point $P$ equidistant from all three.
Consequently, no circle can pass through three collinear points, and a straight line cannot cut a circle in three distinct points.

A B C Bisector of AB Bisector of BC
Perpendicular bisectors of collinear segments AB and BC are parallel and never intersect