Revision: Mathematics >> Playing with Constructions Maths Secondary School (English Medium) (5 to 8) Class 6 CBSE

Statement:
The locus of a point which is equidistant from two fixed points is the perpendicular bisector of the line segment joining the two fixed points.

Given:

• A and B are two fixed points
• P is a point such that PA = PB

To Prove:

• P lies on the perpendicular bisector of AB

Construction:

• Join AB
• Let M be the midpoint of AB
• Join PM

Proof:

1. In ΔPMA and ΔPMB,
   PA = PB (given), MA = MB (M is the midpoint of AB), PM = PM (common).

2. ∴ ΔPMA ≅ ΔPMB (SSS).

3. ∴ ∠PMA = ∠PMB (c.p.c.t.).

4. ∠PMA + ∠PMB = 180° (straight line AB).

5. ∴ ∠PMA = ∠PMB = 90°.

6. Hence, PM ⟂ AB passes through midpoint M.

Thus, P lies on the perpendicular bisector of AB.

Statement:

Every point on the perpendicular bisector of a line segment is equidistant from the two fixed points.

Given:

• A and B are two fixed points
• MQ is the perpendicular bisector of AB
• P is any point on MQ

To Prove:

PA = PB

Construction:

• Join PA and PB

Proof:

1. In ΔPMA and ΔPMB,
   MA = MB (M is the midpoint of AB), PM = PM (common),
   ∠PMA = ∠PMB = 90° (MQ ⟂ AB).

2. ∴ ΔPMA ≅ ΔPMB (SAS).

3. ∴ PA = PB (c.p.c.t.).

Conclusion:

Hence, every point on the perpendicular bisector of AB is equidistant from A and B.