Loci (Locus & Its Constructions)

ICSE Class 10 Mathematics — Chapter 16 Master Notes

1. Concept & Definition of Locus

Origin: The word locus is derived from Latin, meaning location or place. Its plural form is loci (pronounced los-eye).

Definition: A locus is the path or curve traced out by a moving point which moves in a plane according to one or more given mathematical conditions.

To describe a locus completely, you must:

State the exact geometrical shape (e.g. straight line, circle, pair of parallel lines).
State its exact position with respect to the given fixed points or lines.
Verify that every point on the locus satisfies the condition, and every point satisfying the condition lies on the locus.

2. Fundamental Loci Theorems & Geometric Proofs

Theorem 1 (Proof): Locus of a point equidistant from two fixed points $A$ and $B$.

Statement: The locus of a point equidistant from two given fixed points $A$ and $B$ is the perpendicular bisector of the line segment joining the two points.

Proof:

Let $A$ and $B$ be two fixed points. Let $P$ be a moving point such that $PA = PB$.
Let $O$ be the mid-point of $AB$. Join $PO$.
In $\triangle AOP$ and $\triangle BOP$:
1. $PA = PB$ (Given)
2. $AO = BO$ ($O$ is midpoint of $AB$)
3. $PO = PO$ (Common side)
By SSS Congruence Criterion, $\triangle AOP \cong \triangle BOP$.
$\implies \angle AOP = \angle BOP$ (CPCTC).
Since $\angle AOP + \angle BOP = 180^\circ$ (Linear pair), $\angle AOP = \angle BOP = 90^\circ$.
$\therefore PO \perp AB$ at its mid-point $O$. Hence, $P$ lies on the perpendicular bisector of $AB$.

Theorem 2 (Proof): Locus of a point equidistant from two intersecting straight lines.

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

Proof:

Let $AB$ and $CD$ be two lines intersecting at $O$. Let $P$ be a point in the interior of $\angle AOC$ such that perpendicular distance $PL = PM$ (where $PL \perp AB$ and $PM \perp CD$).
Join $OP$. In right $\triangle POL$ and right $\triangle POM$:
1. $PL = PM$ (Given distance equality)
2. $\angle PLO = \angle PMO = 90^\circ$ (Perpendiculars)
3. $OP = OP$ (Common hypotenuse)
By RHS Congruence Criterion, $\triangle POL \cong \triangle POM$.
$\implies \angle POL = \angle POM$ (CPCTC).
$\therefore P$ lies on the angle bisector of $\angle AOC$.

3. Standard Loci Summary (8 Essential Board Results)

In ICSE Board Examinations, locus questions require applying these 8 standard geometric results:

Given Condition	Resulting Locus	Key Geometrical Property
1. Equidistant from 2 fixed points $A$ and $B$	Perpendicular bisector of segment $AB$	Line passes through midpoint of $AB$ at $90^\circ$
2. Equidistant from 2 intersecting lines	Pair of angle bisectors of the angles between them	Divides interior/exterior angles into equal halves
3. Equidistant from a fixed point $O$ (distance $r$)	Circumference of a circle	Center $= O$, Radius $= r$
4. Equidistant from 2 parallel lines $l$ and $s$	A line parallel to both $l$ and $s$	Midway between $l$ and $s$ at distance $d/2$
5. At a fixed distance $d$ from a given line $l$	A pair of lines parallel to line $l$	One line on each side of $l$ at distance $d$
6. Mid-points of all equal chords in a circle	Circumference of a concentric circle	Radius $=$ distance of equal chords from centre
7. Mid-points of all parallel chords in a circle	Diameter of the circle	Perpendicular to the given parallel chords
8. Equidistant from 2 concentric circles	Circumference of a concentric circle	Radius $= \frac{r_1 + r_2}{2}$ (midway between them)

4. Triangle Concurrency Points & Special Loci Properties

Concurrency Points in Triangles (Frequently Tested in ICSE Board):

Incentre ($I$):
Intersection of the 3 angle bisectors of a triangle.
Locus Property: Incentre $I$ is equidistant from all three sides of the triangle ($ID = IE = IF = \text{in-radius } r$).
Circumcentre ($O$):
Intersection of the 3 perpendicular bisectors of sides of a triangle.
Locus Property: Circumcentre $O$ is equidistant from all three vertices of the triangle ($OA = OB = OC = \text{circum-radius } R$).
Centroid ($G$):
Intersection of the 3 medians of a triangle. Divides each median in ratio $2 : 1$.
Orthocentre ($H$):
Intersection of the 3 altitudes of a triangle.

Master Rules for Isosceles & Equilateral Triangles:

In an Isosceles Triangle ($AB = AC$), the median from vertex $A$ to base $BC$ is simultaneously the angle bisector of $\angle A$, the perpendicular bisector of $BC$, and the altitude to $BC$. Thus, $G, I, O, H$ all lie on this single altitude line!
In an Equilateral Triangle, Centroid $=$ Incentre $=$ Circumcentre $=$ Orthocentre! All 4 points coincide into a single point!

5. Step-by-Step Solved Master Examples & Interactive Chart

Interactive Locus Visualization: Circle & Perpendicular Line Loci

Plotting Locus 1: Circle $x^2 + y^2 = 25$ (distance $r=5$ from origin) and Locus 2: Line $x = 3$. Points of intersection $P(3, 4)$ and $Q(3, -4)$.

Example 1 (Simultaneous Loci Construction): Construct a triangle $ABC$ with $AB = 6\text{ cm}$, $BC = 7\text{ cm}$, $CA = 6.5\text{ cm}$. Find a point $P$ inside the triangle which is equidistant from $B$ and $C$, and also equidistant from sides $AB$ and $BC$.

Solution & Construction Steps:

Point $P$ must satisfy two distinct loci conditions simultaneously:

Locus 1 (Equidistant from vertices $B$ and $C$): Point $P$ lies on the perpendicular bisector of side $BC$.
Locus 2 (Equidistant from sides $AB$ and $BC$): Point $P$ lies on the angle bisector of $\angle ABC$.

Conclusion: Point $P$ is the point of intersection of the perpendicular bisector of $BC$ and the angle bisector of $\angle ABC$.

Example 2 (Right-Angle Subtended Locus): $A$ and $B$ are two fixed points. Find the locus of a point $P$ such that $\angle APB = 90^\circ$.

Solution:

Since the angle in a semi-circle is a right angle ($90^\circ$), for any position of point $P$ such that $\angle APB = 90^\circ$, $P$ lies on the circumference of a circle drawn with segment $AB$ as diameter.

Answer: The locus of point $P$ is the circumference of a circle with diameter $AB$ (excluding endpoints $A$ and $B$).

Example 3 (Chord Equidistance): $AB$ is a chord of a circle with center $O$. Find the locus of a point in the circle which is equidistant from $A$ and $B$.

Solution:

Any point equidistant from fixed points $A$ and $B$ lies on the perpendicular bisector of $AB$.

Since the perpendicular bisector of any chord of a circle passes through its center $O$, the perpendicular bisector of chord $AB$ is a diameter of the circle.

Answer: The locus is the diameter of the circle perpendicular to chord $AB$.

Example 4 (Distance from Angle Arms): Draw an angle $\angle ABC = 120^\circ$. Find a point $P$ such that $P$ is at a distance of $3\text{ cm}$ from $AB$ and $2\text{ cm}$ from $BC$.

Solution:

Draw line $l_1$ parallel to arm $AB$ at a distance of $3\text{ cm}$ from it.
Draw line $l_2$ parallel to arm $BC$ at a distance of $2\text{ cm}$ from it.
The point of intersection of lines $l_1$ and $l_2$ is the required point $P$.

6. Quick Revision Checklist for ICSE Board Exams

Locus Target	Exact Description to Write in Board Exams
Point $P$ equidistant from $A$ and $B$	Perpendicular bisector of segment $AB$
Point $P$ equidistant from lines $AB$ and $AC$	Bisector of angle $\angle BAC$
Point $P$ at distance $r$ from fixed point $O$	Circle with center $O$ and radius $r$
Point $P$ at distance $d$ from line $AB$	Pair of lines parallel to $AB$ at distance $d$ on either side
Point $P$ such that $\angle APB = 90^\circ$	Circumference of circle with diameter $AB$
Centroid of all triangles on base $BC$ with same area	Line parallel to base $BC$ at distance equal to altitude $/ 3$

7. ICSE Board PYQs & High-Yield Practice Problems

ICSE Board PYQ 2010

BOARD Use ruler and compasses only for this question:

(i) Construct $\triangle ABC$ where $AB = 3.5\text{ cm}$, $BC = 6\text{ cm}$, and $\angle ABC = 60^\circ$.
(ii) Construct the locus of points inside the triangle equidistant from $BA$ and $BC$.
(iii) Construct the locus of points inside the triangle equidistant from $B$ and $C$.
(iv) Mark the point $P$ equidistant from $AB, BC$ and from $B, C$. Measure and record length $PB$.

ICSE Board PYQ 2008

BOARD Straight line $AB$ is $8\text{ cm}$ long. Draw and describe the locus of a point which is:

(i) Always $4\text{ cm}$ from the line $AB$.
(ii) Equidistant from $A$ and $B$.
(iii) Mark points $X$ and $Y$ which are $4\text{ cm}$ from $AB$ and equidistant from $A$ and $B$. Describe figure $AXBY$.

ICSE Board PYQ 2007

BOARD Construct a triangle $BCP$ given $BC = 5\text{ cm}$, $BP = 4\text{ cm}$, $\angle PBC = 45^\circ$. Complete rectangle $ABCD$ such that $P$ is equidistant from $AB$ and $BC$, and $P$ is equidistant from $C$ and $D$. Measure length $AB$.

ICSE Board PYQ 2003

BOARD Plot points $P(3, 2)$ and $Q(-3, -2)$. From $P$ and $Q$, draw perpendiculars $PM$ and $QN$ to the x-axis. Write down the coordinates of the point to which $M$ is mapped on reflection in: (i) x-axis, (ii) y-axis, (iii) origin.

ICSE Board PYQ 2001

BOARD Construct an isosceles triangle $ABC$ such that $AB = 6\text{ cm}$, $BC = AC = 4\text{ cm}$. Bisect $\angle C$ internally and mark point $P$ on this bisector such that $CP = 5\text{ cm}$. Find points $Q$ and $R$ which are $5\text{ cm}$ from $P$ and $5\text{ cm}$ from line $AB$.

ICSE Board PYQ 2000

BOARD Construct a triangle $ABC$ with $AB = 7\text{ cm}$, $BC = 8\text{ cm}$, $\angle ABC = 60^\circ$. Locate by construction point $P$ such that $P$ is equidistant from $B$ and $C$, and $P$ is equidistant from $AB$ and $BC$. Measure length $PB$.