Revision: Geometry >> Loci Maths (English Medium) ICSE Class 10 CISCE

Locus is the path traced by a moving point, which moves so as to satisfy a certain given condition/conditions. 

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.

Statement:
The locus of a point equidistant from two intersecting lines is the pair of angle bisectors of the angles formed by the given lines.

Given:

• Lines AB and CD intersect at O
• P is a point inside ∠BOD
• PM ⟂ OB, PN ⟂ OD
• PM = PN

To Prove:

• P lies on the bisector of ∠BOD

Construction:

• Join OP

Proof:

1. In ΔOMP and ΔONP, we have
   (i) PM = PN (given)
   (ii) ∠OMP = ∠ONP = 90° (PM ⟂ OB, PN ⟂ OD)
   (iii) OP = OP (common)

2. ∴ ΔOMP ≅ ΔONP (RHS congruence)

3. ∴ ∠MOP = ∠PON (c.p.c.t.)

4. Hence, OP bisects ∠BOD.

Conclusion:
Therefore, P lies on the bisector of ∠BOD.

Statement:
Every point on the angle bisector of two intersecting lines is equidistant from the lines.

Given:

• Lines AB and CD intersect at O
• OE is the bisector of ∠BOD
• P is a point on OE
• PM ⟂ OB and PN ⟂ OD

To Prove:

PM = PN

Proof:

1. In ΔOMP and ΔONP, we have
   (i) ∠MOP = ∠PON (OE bisects ∠BOD)
   (ii) ∠OMP = ∠ONP = 90° (PM ⟂ OB, PN ⟂ OD)
   (iii) OP = OP (common)

2. ∴ ΔOMP ≅ ΔONP (ASA congruence)

3. ∴ PM = PN (c.p.c.t.)

Conclusion:

Hence, P is equidistant from OB and OD, and therefore from the intersecting lines AB and CD.

• Equidistant from two fixed points → lies on the perpendicular bisector.

• Equidistant from two intersecting lines → lies on the angle bisectors.

• Fixed distance from a point → forms a circle.

• Equidistant from two parallel lines → lies on a line midway and parallel to them.

• Fixed distance from a line → forms two parallel lines.

• Midpoints of equal chords → lie on a concentric circle.

• Midpoints of parallel chords → lie on a diameter perpendicular to the chords.

• Equidistant from two concentric circles → forms a circle midway between them.