MathsICSE Class 10 · Loci

Locus of Points Equidistant from Two Fixed Points: the Perpendicular Bisector (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

The locus of points equidistant from two fixed points A and B is the perpendicular bisector of AB. Every point P on it has PA = PB, and no point off it does. Construct it with equal arcs from A and B; where perpendicular bisectors of a triangle's sides meet is the circumcentre.

Understand Locus Equidistant from Two Points

A locus is the set of all points, and only those points, that satisfy a given condition. The CISCE syllabus lists three locus theorems (proofs not required); this page covers the third.

The theorem. The locus of a point equidistant from two fixed points $A$ and $B$ is the perpendicular bisector of $A B$ : the line through the mid-point $M$ of $A B$ at right angles to $A B$ .

If $P$ is on it, then $△ P M A ≅ △ P M B$ (SAS), so $P A = P B$ .
If $P A = P B$ , then $P$ is on it.

Construction.

1. With centre $A$ and radius more than half of $A B$ , draw arcs above and below $A B$ .
2. With centre $B$ and the same radius, draw arcs cutting the first ones at $X$ and $Y$ .
3. Join $X Y$ . This is the required locus. Keep all arcs visible; examiners look for them.

Measurements from the locus. If $P$ is on the perpendicular bisector at height $h$ above $A B$ and $A B = 2 a$ , then $P A = P B = a^{2} + h^{2}$ by Pythagoras.

Uses.

**Centres of circles through $A$ and $B$ ** all lie on the perpendicular bisector of $A B$ .
Circumcentre. The perpendicular bisectors of the three sides of a triangle meet at one point $O$ , equidistant from all three vertices; the circle with centre $O$ and radius $O A$ passes through $A$ , $B$ and $C$ . In a right-angled triangle, $O$ is the mid-point of the hypotenuse. 2024 Q10(iii) used exactly this: the point $O$ equidistant from $A$ , $B$ and $C$ , then the circle of radius $O A$ .
Combined conditions. Many questions combine this locus with another (a circle, a parallel line, an angle bisector). The required points are where the loci intersect; count them before constructing.

In coordinates. $P (x, y)$ is equidistant from $A$ and $B$ when $P A^{2} = P B^{2}$ ; this simplifies to a straight-line equation, the perpendicular bisector.

Locus Equidistant from Two Points: formulas to remember

Locus theorem: $P A = P B ⟺ P lies on the perpendicular bisector of A B$; Proof not required.
Distance from A for a point on the bisector: $P A = P B = (\frac{A B}{2})^{2} + h^{2}$; $h$ is the distance of $P$ from $A B$ .
Circumradius of a right triangle: $R = \frac{hypotenuse}{2}$; Circumcentre at the mid-point of the hypotenuse.
Coordinate condition: $P A^{2} = P B^{2}$; Gives the bisector's equation.

How to solve Locus Equidistant from Two Points questions

Name the locus, construct it with full arcs, then find where it meets any other locus.

1Identify the condition 'equidistant from $A$ and $B$ ' and name the locus: the perpendicular bisector of $A B$ .
2Construct it with equal arcs (radius more than half of $A B$ ) from both ends.
3If there is a second condition, construct that locus too.
4Mark and label the intersection point(s); say how many there are.
5Measure what is asked, or calculate it with Pythagoras as a check.
6Write a short statement: 'P is the required point; PA = … cm'.

Locus Equidistant from Two Points in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Equidistant points between two places

Suppose two friends live 8 km apart and want to meet somewhere exactly as far from one house as from the other, travelling as little as possible.

Describe all fair meeting points, and find the one that is nearest to both.

Try it first, then show the solutionSolution

All fair points lie on the perpendicular bisector of the line joining the houses.
The nearest such point is the mid-point, 4 km from each house.

Answer: Any point on the perpendicular bisector; the mid-point, 4 km from each, is nearest.

Memory hook: Fair to both = on the perpendicular bisector.

Locus meets a line

Imagine villages at $A (0, 4)$ and $B (6, 0)$ (in km) and a straight highway along the x-axis. A mobile tower must stand on the highway, equally far from both villages.

Find the tower's position and its distance from each village.

Try it first, then show the solutionSolution

Let the tower be $P (x, 0)$ . $P A^{2} = P B^{2}$ : $x^{2} + 16 = (x - 6)^{2}$ .
$16 = - 12 x + 36$ , so $x = \frac{5}{3}$ .
$P A = \frac{25}{9} + 16 = \frac{169}{9} = \frac{13}{3}$ km.

Answer: At $(\frac{5}{3}, 0)$ , about 1.67 km along the highway; $\frac{13}{3} \approx 4.33$ km from each village.

Memory hook: Where the bisector crosses the road is the fair spot on the road.

Tricks to remember Locus Equidistant from Two Points

Short memory aids. Say each one aloud once, then use it on a question the same day.

Equal from two points: cut it fairly
Two points → bisect the segment at right angles.
Use it when: Naming the locus quickly.
Same compass, both ends
Do not change the compass between the arcs from $A$ and $B$ .
Use it when: An accurate construction.
Three bisectors, one centre
Perpendicular bisectors of a triangle meet at the circumcentre: the centre of the circle through all three corners.
Use it when: Questions on a point equidistant from three points.

Locus Equidistant from Two Points: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Naming the locus · 1 mark · MCQBoard level

The locus of the centres of all circles passing through two fixed points $A$ and $B$ is:
(a) the line $A B$
(b) a circle with diameter $A B$
(c) the perpendicular bisector of $A B$
(d) the point mid-way between $A$ and $B$

Try it first, then show the solutionSolution

A centre $O$ of such a circle has $O A = O B$ (both radii), so it is equidistant from $A$ and $B$ .

Answer: (c)

Example 2: A distance from the bisector · 2 marks · Short answerBoard level

$A B = 10$ cm. $P$ is a point on the perpendicular bisector of $A B$ , 12 cm from $A B$ . Find $P A$ and $P B$ .

Try it first, then show the solutionSolution

The bisector meets $A B$ at its mid-point $M$ , with $A M = 5$ cm and $P M = 12$ cm.
$P A = 5^{2} + 1 2^{2} = 13$ cm, and $P B = P A = 13$ cm.

Answer: $P A = P B = 13$ cm

Example 3: Two loci, two points · 3 marks · Short answerHarder board level

Draw $A B = 8$ cm. Construct the locus of points equidistant from $A$ and $B$ . Mark the points $P$ and $Q$ on it that are 5 cm from $A$ .
(a) How far is each from $A B$ ?
(b) Find $P Q$ and name the figure $A P B Q$ .

Try it first, then show the solutionSolution

Construct the perpendicular bisector of $A B$ ; with centre $A$ and radius 5 cm, cut it at $P$ and $Q$ .
(a) $A M = 4$ cm, so $P M = 25 - 16 = 3$ cm (and $QM = 3$ cm).
(b) $P Q = 6$ cm. $A P = P B = B Q = Q A = 5$ cm, so $A P B Q$ is a rhombus.

Answer: (a) 3 cm (b) 6 cm; a rhombus

Example 4: Constructing the circumcentre · 4 marks · Long answerHarder board level

Construct $△ A B C$ with $A B = 6$ cm, $B C = 8$ cm and $C A = 7$ cm. Construct the locus of points equidistant from $A$ and $B$ , and the locus of points equidistant from $B$ and $C$ . Mark their intersection $O$ and draw the circle with centre $O$ through $A$ . Measure $O A$ .

Try it first, then show the solutionSolution

Draw $B C = 8$ cm; with centres $B$ and $C$ and radii 6 cm and 7 cm, cut arcs at $A$ .
Construct the perpendicular bisectors of $A B$ and $B C$ ; they meet at $O$ .
$O$ is equidistant from $A$ , $B$ and $C$ , so the circle through $A$ also passes through $B$ and $C$ .
$O A \approx 4.1$ cm (calculation gives 4.13 cm).

Answer: $O A \approx 4.1$ cm; the circle passes through all three vertices.

Example 5: The locus in coordinates · 3 marks · Short answerHOTS / competency

Find the point on the x-axis that is equidistant from $A (1, 2)$ and $B (5, 6)$ , and show that it lies on the line $x + y = 7$ .

Try it first, then show the solutionSolution

$P (x, 0)$ : $(x - 1)^{2} + 4 = (x - 5)^{2} + 36$ .
$- 2 x + 5 = - 10 x + 61$ , so $8 x = 56$ and $x = 7$ . $P (7, 0)$ , with $P A = P B = 40$ .
$x + y = 7 + 0 = 7$ ✓. ( $x + y = 7$ is the perpendicular bisector of $A B$ : it passes through the mid-point $(3, 4)$ and has slope $- 1$ , while $A B$ has slope 1.)

Answer: $P (7, 0)$

Example 6: Case study: a water tank for three houses · 4 marks · Case-basedHOTS / competency

Three houses stand at $A$ , $B$ and $C$ with $∠ A B C = 90°$ , $A B = 5$ km and $B C = 12$ km. A water tank must be equally far from all three.
(a) Which loci locate the tank?
(b) Where exactly is it?
(c) How far is it from each house?

Try it first, then show the solutionSolution

(a) The perpendicular bisectors of $A B$ and $B C$ (any two sides); the tank is where they meet, the circumcentre.
(b) In a right-angled triangle the circumcentre is the mid-point of the hypotenuse $A C$ .
(c) $A C = 25 + 144 = 13$ km, so the distance is $\frac{13}{2} = 6.5$ km.

Answer: (b) The mid-point of $A C$ (c) 6.5 km

Where students lose marks in Locus Equidistant from Two Points

The slips examiners see most often, and what to write instead.

Mistake: Arcs with radius less than half of AB
The arcs from $A$ and $B$ will not meet.
Wrong working
Radius 2 cm for $A B = 6$ cm.
Right working
Any radius more than 3 cm, the same from both ends.
Do this: Open the compass to clearly more than half the segment.
Usually costs: The construction marks.
Mistake: Not identifying the circle in a combined question
Points at a fixed distance from $O$ form a circle with centre $O$ .
Wrong working
Part (c) of 2024 Q10(iii) left as a single arc or line.
Right working
Draw the full circle with centre $O$ and radius $O A$ .
Do this: Name each locus in words before constructing it.
Usually costs: Reported in 2024 Q10(iii).
Mistake: Rubbing out construction arcs
The arcs show the method.
Wrong working
A clean perpendicular line with no arcs.
Right working
Arcs from both $A$ and $B$ left visible.
Do this: Never erase construction lines.
Usually costs: Method marks (the 2024 report asks for all construction arcs to be shown).

Exam tips

2024 Q10(iii) (4 marks): perpendicular bisectors of $B C$ and $A B$ met at $O$ ; then the locus at distance $O A$ from $O$ was a circle through $A$ , $B$ and $C$ .
Label every locus on your diagram, for example 'locus of points equidistant from A and B'.
Draw lightly but clearly; extend the bisector beyond the segment so intersections are visible.
If asked to measure, give the length to the nearest mm and check it with a quick calculation when possible.

Practice questions on Locus Equidistant from Two Points

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If $P A = P B$ , then $P$ lies on:
(a) the line $A B$
(b) the circle with centre $A$ through $B$
(c) the perpendicular bisector of $A B$
(d) the bisector of angle $A P B$ only
Hint
Equidistant from two points.
Answer
(c)
Q2 · 2 marks · Short answerBoard level
$A B = 6$ cm. How many points are equidistant from $A$ and $B$ and 3 cm from $A$ ? Describe them.
Hint
Compare 3 cm with half of $A B$ .
Answer
One: the mid-point of $A B$ , since the circle of radius 3 cm just touches the perpendicular bisector there.
Q3 · 2 marks · Short answerHarder board level
$A B = 12$ cm and $P$ is on the perpendicular bisector of $A B$ with $P A = 10$ cm. Find the distance of $P$ from $A B$ .
Hint
$1 0^{2} - 6^{2}$ .
Answer
8 cm
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The circumcentre of a right-angled triangle is the mid-point of its hypotenuse.
Reason (R): The angle in a semicircle is a right angle.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
The right-angle vertex lies on the circle with the hypotenuse as diameter.
Answer
(c)
Q5 · 4 marks · Long answerHarder board level
Draw $A B = 7$ cm. Construct the locus of points equidistant from $A$ and $B$ , and the locus of points 4 cm from the line $A B$ . How many points satisfy both conditions? Find the distance of each from $A$ .
Hint
The second locus is a pair of lines parallel to $A B$ .
Answer
Two points (one on each side of $A B$ ); each is $3. 5^{2} + 4^{2} = 28.25 \approx 5.3$ cm from $A$ .
Q6 · 3 marks · Short answerHarder board level
Find the equation of the locus of points equidistant from $A (- 2, 3)$ and $B (4, - 1)$ .
Hint
Mid-point $(1, 1)$ ; slope of $A B$ is $- \frac{2}{3}$ .
Answer
Perpendicular bisector: slope $\frac{3}{2}$ through $(1, 1)$ : $3 x - 2 y - 1 = 0$ .
Q7 · 3 marks · Short answerHOTS / competency
In $△ A B C$ , $A B = 5$ cm and $∠ A B C = 60°$ . The perpendicular bisector of $A B$ meets $B C$ at $P$ . Without measuring, find $B P$ .
Hint
$P A = P B$ and $∠ B = 60°$ .
Answer
$P A = P B$ makes $∠ P A B = ∠ P B A = 60°$ , so $△ A B P$ is equilateral and $B P = 5$ cm.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Locus of points equidistant from $A$ and $B$ ?Show answer
The perpendicular bisector of $A B$ .
2.What is the circumcentre?Show answer
The meeting point of the perpendicular bisectors of a triangle's sides; it is equidistant from the vertices.
3.Where is the circumcentre of a right triangle?Show answer
At the mid-point of the hypotenuse.
4. $A B = 8$ cm, $P$ on the bisector 3 cm from $A B$ . $P A$ ?Show answer
5 cm
5.Radius rule for the construction arcs?Show answer
Same radius from both ends, more than half of $A B$ .

Locus Equidistant from Two Points: frequently asked questions

What is the locus of points equidistant from two fixed points?

The perpendicular bisector of the segment joining them. Every point on it is the same distance from both points, and no other point is.

How do I construct the perpendicular bisector?

With the same radius (more than half of AB), draw arcs from A and from B on both sides of AB. Join the two points where the arcs cross.

Do I need to prove the locus theorems for ICSE?

No. The CISCE syllabus states the three locus theorems and says proofs are not required; you need to use them and do the constructions.