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ICSE Class 10 Mathematics Chapter 18 Loci Digital Edition
For Class 10 Mathematics, this chapter in ICSE Class 10 Maths Chapter 18 Loci provides a detailed overview of important concepts. We highly recommend using this text alongside the ICSE Solutions for Class 10 Mathematics to learn the exercise questions provided at the end of the chapter.
Chapter 18 Loci ICSE Book Class Class 10 PDF (2026-27)
Chapter 18
Loci
Points To Remember
1. Locus is the path traced out by a moving point which moves according to some given geometrical conditions.
Thus, (i) Every point which satisfies the given geometrical conditions will lie on the locus. And, (ii) Every point lying on the locus will satisfy the given geometrical conditions.
Remark: The plural of locus is loci, read as 'losai'.
Example 1. A circle with centre O and radius r cm is the locus of a point which moves in a plane in such a way that its distance from the fixed point O is always equal to r cm.
Example 2. Let a point P move in such a way that its distance from a fixed line AB is always equal to d cm.
Clearly, the locus of the moving point is a pair of straight lines CD and EF, each parallel to AB at a distance of d cm from it.
Points Equidistant From Two Given Points
Theorem 1. The locus of a point which is equidistant from two given fixed points, is the perpendicular bisector of the line segment joining the given fixed points.
Proof: We shall prove the theorem in two parts (a) and (b) given below.
Part (a): Every point which is equidistant from two fixed points A and B, lies on the perpendicular bisector of AB.
Given: Two fixed points A and B and P is a point, such that PA = PB.
To Prove: P lies on the perpendicular bisector of AB.
Construction: Join AB. Find its middle point M and Join MP.
| Statement | Reason |
|---|---|
| 1. In ΔPMA and ΔPMB, PA = PB MA = MB PM = PM ∴ ΔPMA ≅ ΔPMB ⇒ ∠AMP = ∠BMP ...I ∠AMP + ∠BMP = 180° ...II ∠AMP = ∠BMP = 90° 4. PM is the right bisector of AB | Given. By construction, M is the mid-point of AB. Common. SSS-congruency axiom. c.p.c.t. AMB is a straight line. From I and II. PM is the right bisector of AB |
Hence, P lies on the perpendicular bisector of AB.
Part (b): Conversely, every point on the perpendicular bisector of AB is equidistant from A and B.
Given: Two fixed points A and B, MQ is the perpendicular bisector of AB and P is any point on MQ.
To Prove: PA = PB.
Construction: Join PA and PB.
Proof.
| Statement | Reason |
|---|---|
| 1. In ΔPMA and ΔPMB, MA = MB PM = PM ∠PMA = ∠PMB ∴ ΔPMA ≅ ΔPMB 2. So, PA = PB Hence, PA = PB. | M is the mid-point of AB (given). Common. Each equal to 90° (given). R.H.S. congruency axiom. c.p.c.t. |
Hence, the locus of a point which is equidistant from two fixed points A and B, is the perpendicular bisector of AB.
Point Equidistant From Two Intersecting Lines
Theorem 2. The locus of a point which is equidistant from two intersecting lines is the pair of lines bisecting the angles formed by the given lines.
Proof: We shall prove the theorem in two parts; Part (a) and Part (b).
Part (a): Every point which is equidistant from two intersecting lines, lies on the bisector of the angle between the given lines.
Given: Two straight lines, AB and CD, intersecting at a point O and P is a point in the interior of ∠BOD such that PM - AB, PN - OD and PM = PN.
To Prove: P lies on the bisector of ∠BOD.
Construction: Join OP.
Proof:
| Statement | Reason |
|---|---|
| 1. In ΔOMP and ΔONP, PM = PN ∠OMP = ∠ONP OP = OP 2. ΔOMP ≅ ΔONP 3. ∠MOP = ∠NOP | Given Each equal to 90° (Given) Common. RHS-axiom of congruency c.p.c.t. |
∴ P lies on the bisector of ∠BOD
Thus, P lies on the bisector of ∠BOD
Similarly, if P' is a point such that P'M' - OA and P'N' - OD and P'M' = P'N', then P' lies on the bisector of ∠AOD.
Part (b): Conversely, every point on the angle bisector of two intersecting lines, is equidistant from the lines.
Given: Two lines AB and CD intersecting at a Point O; OE is the bisector of ∠BOD and P is a point on OE ; PM - OB and PN - OD.
To Prove: PM = PN
Proof.
| Statement | Reason |
|---|---|
| 1. In ΔOMP and ΔONP, we have ∠MOP = ∠NOP ∠OMP = ∠ONP OP = OP 2. ∴ ΔOMP ≅ ΔONP 3. PM = PN | Given, as OE is the bisector of ∠BOD. Each equal to 90°. Common. ASA-axiom of congruency. c.p.c.t. |
∴ P is equidistant from OB and OD, and therefore from AB and CD.
Similarly, if P' is a point on the bisector of ∠AOD and P'M' - LOA, P'N' - OD, then P'M' = P'N'.
Some Examples Of Loci
In each figure, we shall show the locus by dotted lines.
1. The locus of a point which is equidistant from two given points A and B is the perpendicular bisector of AB.
2. The locus of a point equidistant from two intersecting lines AB and BC is the bisector of ∠ABC.
3. If A and B are two fixed points, then the locus of a point P such that ∠APB = 90° is the circle with AB as diameter. (∴ Angle in a semi-circle is always a right angle)
4. The locus of a point P inside a circle such that PA = PB, where A and B are two fixed points on the circle is a diameter of the circle bisecting the chord AB.
5. The locus of mid-points of all chords parallel to a chord AB of a circle is the diameter of the circle which is the right bisector of AB.
6. The locus of mid-points of all equal chords of a circle concentric with the given circle with radius equal to distance of each chord from the centre.
7. The locus of centres of circles touching a given line AB at a given point T on it is the straight line perpendicular to AB at T.
Teacher's Note
Understanding loci helps us in navigation systems and GPS technology, where we track the path of moving objects in real-time by applying geometrical principles similar to those learned here.
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ICSE Book Class 10 Mathematics Chapter 18 Loci
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