Question 2 of 50advanced🔍 AnalyzeLong Answer5 marks

A, B and C are three collinear points. Can you find a point P such that PA=PB=PCPA = 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 bisector 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?

Correct Answer

(1) No, we cannot find a point P such that PA = PB = PC for three collinear points.

(2) The perpendicular bisectors of AB and BC are parallel to each other.

(3) Since A, B, and C are collinear, AB and BC lie on the same straight line. The perpendicular bisectors of AB and BC are both perpendicular to this same line. Two lines perpendicular to the same line are parallel, hence the perpendicular bisectors are parallel.

(4) No, it is not possible for a circle to pass through three collinear points. A unique circle can only pass through three non-collinear points.

(5) No, you cannot draw a line that cuts a given circle in three distinct points. If a line cut a circle in three points, those three collinear points would lie on the circle, which is impossible.

Exercise: Think, Draw and Infer | Q: 1 | (Chapter: 7)
For More Understanding

Explanation

The textbook states that there is a unique circle passing through three non-collinear points, whose center is the intersection of their perpendicular bisectors. For collinear points, the perpendicular bisectors do not intersect (they are parallel), so there is no center and no circle. Because no circle can pass through three collinear points, a line cannot intersect a circle at three distinct points.

Solution Steps

  1. Step 1: Determine if point P exists. A point P equidistant from A, B, and C would be the center of a circle passing through them. Since no circle passes through collinear points, no such P exists.

  2. Step 2: Analyze the perpendicular bisectors. For collinear points A, B, and C, segments AB and BC lie on the same line. Their perpendicular bisectors are both perpendicular to this line.

  3. Step 3: Show the bisectors are parallel. Since both perpendicular bisectors are perpendicular to the same straight line, they must be parallel to each other and will never intersect at a center point.

  4. Step 4: Conclude on the circle and line intersection. Since three collinear points cannot lie on a circle, a single straight line cannot intersect a circle at three distinct points.