Class 10: Loci (Locus and Its Construction) – Lecture Notes – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Definition:

In geometry, a locus (plural: loci) (Latin word for “place”, “location”) is a set of all points (commonly, a line, a line segment, a curve or a surface), whose location satisfies or is determined by one or more specified conditions.

In another terms…Locus can be defined as the path traced by a point, which moves so as to satisfy certain given conditions such as equidistant from two given lines, equidistant from a given point etc.

Theorems based on symmetry:

Theorem 2: The locus of a point equidistant from two intersecting lines is the bisector of the angles between the lines.

Given: Two straight lines $AB \ and \ CD$ intersecting at $O$ . Point $P$ is such that it is equidistant from $AB \ and \ CD$ (the two given lines)

To Prove: Locus of $P$ is the bisector of $\angle AOC$

(i) $\Rightarrow P$ lies on the bisector of $\angle AOC$

(ii) $\Rightarrow$ Each point on the bisector of $\angle AOC$ is equidistant from $AB \ and \ CD$

Proof: Consider $\triangle POL$ and $\triangle POM$

(i) $PL = PM$ (Given)

$\angle PLO = \angle PMO$ (right angles triangle)

$PO$ is common

Therefore $\triangle POL \cong \triangle POM$

$\Rightarrow \angle POL = \angle POM$

Therefore $P$ lies on the angle bisector of $\angle AOC$

(ii) Conversely, if $Q$ be any point on the angle bisector $OP$

Consider $\triangle QOS \ and \ \triangle QOR$

$\angle SOQ = \angle ROQ$ (given)

$OQ$ is common

$\angle SQO = \angle RQO$

Therefore $\triangle QOS \cong \triangle QOR$

Hence $QS = QR$

$\Rightarrow Q \ is \ equidistant \ from \ AB \ and \ CD$

Theorem 3: The set of points equidistant from two points is a perpendicular bisector to the line segment connecting the two points.

Given: Two fixed points $A \ and \ B$ . $P$ is a point equidistant from $A \ and \ B \ i.e. \ PA = PB$ at all times.

To Prove: Locus of moving point $P$ is perpendicular bisector of line $AB$

(i) $P$ lies on perpendicular bisector of $AB$ and conversely

(ii) Every point on this perpendicular bisector is equidistant from points $A \ and \ B$

Proof:

(i) Consider $\triangle AOP \ and \ \triangle BOP$

$PA=PB$ (Given)

$AO = OB$ (Given)

$PO$ is common

Therefore $\triangle AOP \cong \triangle BOP$ (S.S.S postulate)

Therefore corresponding angles are equal

Hence $\angle AOP = \angle BOP$

Since $\angle AOP + \angle BOP = 180 \Rightarrow \angle AOP = 90^o$

Hence proved that $P$ lies on perpendicular bisector of $AB$

(ii) Given $PO$ is the perpendicular bisector of $AB$

Consider $\triangle AOQ \ and \ \triangle BOQ$

$AO=OB$

$\angle AOQ = \angle BOQ$

$OQ$ is common

Therefore $\triangle AOQ \cong \triangle BOQ$

Hence $AQ=BQ$

Therefore proved that every point on a perpendicular bisector is equidistant from fixed points $A \ and \ B$ .

Notes:

The locus of a point in a plane at a fixed distance from a given point is the circumference of a circle with the fixed given point as the center of the circle and the distance as the radius.
The locus of a point equidistant from two given parallel lines is a line parallel to the given lines and is midway between them.
The locus of a point, which is at a given distance from a given line, is a pair of lines parallel to the given line and at the given distance from it.
The locus of all mid-points of all equal chords, in a circle, is the circumference of the circle concentric with the given circle and having radius equal to the distance of equal chords from the center.
The locus of mid-point of all parallel chords in a circle is the diameter of the circle which is perpendicular to the given parallel chords.
The locus of a point equidistant from two concentric circles is the circumference of the circle concentric with the given circle and midway between them.