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. l7

 

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. l6

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. l5
The locus of a point equidistant from two given parallel lines is a line parallel to the given lines and is midway between them. l4
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. l8
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. l3
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. l1
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. l2
Advertisements