Loci ICSE Class 10: Locus Theorems, Constructions and Board Questions

What Loci is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Mathematics. Loci is in the 3. Geometry (b) Loci part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Mathematics syllabus, examination year 2027

Key concepts

What a locus is

A locus is a set of points defined by a condition. To describe a locus fully, give its shape and position (and size): for example 'a circle with centre O and radius 3 cm', not just 'a circle'.

Two things must be true of a locus:
1. Every point on it satisfies the condition.
2. Every point that satisfies the condition is on it.

Where two loci intersect, the points satisfy both conditions. This is how most board questions find a special point.

Worked example

Describe the locus of the tip of a minute hand 7 cm long as it moves for one hour.

Try it first, then show the solutionSolution

1. The tip is always 7 cm from the centre of the clock (a fixed point).
2. So the locus is a circle with the centre of the clock as centre and radius 7 cm.

Answer: A circle, centre at the clock's centre, radius 7 cm

Locus 1: fixed distance from a fixed point

The locus of a point at a fixed distance $r$ from a fixed point $O$ is the circle with centre $O$ and radius $r$.

To construct: open the compass to $r$, put the point on $O$ and draw the full circle (or the arc needed). Label it, e.g. 'locus of points 4 cm from O'.

In 2024, many candidates did not see that 'points at a fixed distance OA from O' is a circle through $A$ with centre $O$.

Worked example

$O$ is a point 5 cm from a straight line $l$. How many points on $l$ are 6 cm from $O$? How many are 4 cm from $O$?

Try it first, then show the solutionSolution

1. The points 6 cm from $O$ form a circle of radius 6 cm. It cuts $l$ at two points, because $6>5$.
2. The circle of radius 4 cm does not reach $l$, because $4<5$: no points.

Answer: Two points; none

Locus 2: equidistant from two points

The locus of points equidistant from two fixed points $A$ and $B$ is the perpendicular bisector of $AB$.

To construct: with centre $A$ and radius more than half of $AB$, draw arcs above and below $AB$; repeat from $B$ with the same radius; join the two crossing points.

Any point $P$ on it satisfies $PA=PB$. If $M$ is the mid-point of $AB$, then $P{A}^2=P{M}^2+A{M}^2$ (Pythagoras), which lets you check measurements.

Worked example

$AB=6$ cm. $P$ lies on the perpendicular bisector of $AB$, 4 cm from $AB$. Find $PA$.

Try it first, then show the solutionSolution

1. $AM=3$ cm and $PM=4$ cm, with a right angle at $M$.
2. $PA=\sqrt{3^2+4^2}=5$ cm, and $PB=5$ cm too.

Answer: PA = PB = 5 cm

Locus 3: equidistant from two intersecting lines

The locus of points equidistant from two intersecting lines is the pair of lines that bisect the angles between them. The two bisectors are perpendicular to each other.

If the point must be inside a given angle (such as $\angle ABC$, equidistant from $BA$ and $BC$), the locus is the bisector of that angle.

To construct: with centre at the vertex, draw an arc cutting both arms; from these two points draw equal arcs that cross; join the vertex to the crossing point. Distance from a line is measured along the perpendicular.

Worked example

$P$ lies on the bisector of $\angle XOY=60^{\circ }$ with $OP=6$ cm. Find the distance of $P$ from each arm.

Try it first, then show the solutionSolution

1. The bisector makes $30^{\circ }$ with each arm.
2. Perpendicular distance $=6\sin 30^{\circ }=3$ cm, the same for both arms.

Answer: 3 cm from each arm

Circumcentre: where perpendicular bisectors meet

The perpendicular bisectors of the three sides of a triangle meet at one point, the circumcentre $O$. It is equidistant from all three vertices, so the circle with centre $O$ and radius $OA$ passes through $A$, $B$ and $C$: the circumcircle.

Two bisectors are enough to locate it. Its position depends on the triangle:

• acute triangle: inside;
• right triangle: at the mid-point of the hypotenuse (so $R=\frac{\text{hypotenuse}}{2}$);
• obtuse triangle: outside.

Worked example

A triangle has sides 5 cm, 12 cm and 13 cm. Where is its circumcentre, and what is the circumradius?

Try it first, then show the solutionSolution

1. $5^2+12^2=169=13^2$, so the triangle is right-angled.
2. The circumcentre is the mid-point of the hypotenuse; $R=13÷2=6.5$ cm.

Answer: Mid-point of the 13 cm side; R = 6.5 cm

Incentre: where angle bisectors meet

The bisectors of the three angles of a triangle meet at the incentre $I$, which is equidistant from the three sides. The circle with centre $I$ touching all three sides is the incircle; its radius is the perpendicular distance from $I$ to any side.

To construct: bisect two angles, mark $I$, drop a perpendicular from $I$ to one side to get the radius, then draw the circle. For a right triangle with legs $a$, $b$ and hypotenuse $c$, the inradius is $r=\frac{a+b-c}{2}$, a good check on your measurement.

Worked example

Find the inradius of a right triangle with sides 6 cm, 8 cm and 10 cm.

Try it first, then show the solutionSolution

1. $r=\frac{6+8-10}{2}=2$ cm.
2. So a careful construction should give an incircle of radius about 2 cm.

Answer: 2 cm

The construction drill

For every locus construction:
1. Draw a rough sketch first and decide which locus each condition gives.
2. Construct the given figure accurately (lengths to the mm; $60^{\circ }$, $90^{\circ }$, $120^{\circ }$ and their halves with compasses, not a protractor, when the question says ruler and compass).
3. Construct each locus with all arcs visible; extend lines far enough to meet.
4. Label each locus and the intersection point(s) as asked; respect conditions like 'on the same side of AB as C'.
5. Measure and state the required length, to the nearest mm.

Worked example

Which loci would you construct to find a point equidistant from $A$ and $B$ and also equidistant from the lines $AB$ and $AC$?

Try it first, then show the solutionSolution

1. Equidistant from $A$ and $B$: perpendicular bisector of $AB$.
2. Equidistant from $AB$ and $AC$: bisector of $\angle BAC$.
3. The required point is where these two lines meet.

Answer: Perpendicular bisector of AB and the bisector of ∠BAC; their intersection

Maths in real life: Loci

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.

Topics in this chapter

• Locus Equidistant from Two Points
• Locus at a Fixed Distance from a Point
• Locus Equidistant from Two Intersecting Lines

Loci formula sheet

Fixed distance r from point O

$\text{locus}=\text{circle}(O,r)$

Give centre and radius.

Equidistant from A and B

$PA=PB⟺P\text{ on the perpendicular bisector of }AB$

Construct with equal arcs from A and B.

Equidistant from two intersecting lines

$\text{locus}=\text{bisectors of the angles between them}$

Inside a given angle: just that angle's bisector.

Check with Pythagoras

$PA=\sqrt{P{M}^2+A{M}^2}$

M = mid-point of AB, P on its perpendicular bisector.

Circumradius of a right triangle

$R=\frac{c}{2}$

c = hypotenuse; centre at its mid-point.

Inradius of a right triangle

$r=\frac{a+b-c}{2}$

Use to check a measured incircle.

How the board asks Loci

The question patterns to expect, and the method that works for each.

Tricks to remember Loci

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

• Point–point–bisect

  Equidistant from two POINTS → PERPENDICULAR bisector. Equidistant from two LINES → ANGLE bisector.

  Use it when: Choosing the right construction.

• Circum = corners, In = inside edges

  Circumcentre is the same distance from the corners; incentre is the same distance from the sides.

  Use it when: Circumcircle and incircle questions.

• Fixed distance, think compass

  If the rule says 'always r cm from a point', you only need the compass: it is a circle.

  Use it when: Recognising circle loci.

• Right triangle shortcut

  Circumcentre of a right triangle sits on the hypotenuse, halfway along.

  Use it when: Checking a circumcircle construction.

Loci: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Loci

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

• ✗Mistake: Not recognising a circle locus

  CISCE examiners' comments, 2024: many candidates could not identify that the locus of points at a fixed distance OA from O is a circle with centre O and radius OA.

  ✓Do this: 'Fixed distance from a fixed point' always means a circle: put the compass on the point.

  Usually costs: 1 mark.

• ✗Mistake: Not bisecting the angle

  CISCE examiners' comments, 2024: for points equidistant from BA and BC, some candidates did not bisect ∠ABC.

  ✓Do this: Equidistant from two lines → bisect the angle between them.

  Usually costs: 1 mark.

• ✗Mistake: Inaccurate base figure

  CISCE examiners' comments, 2024: candidates made mistakes constructing triangle ABC, and some could not construct ∠ABC = 90° with ruler and compass.

  ✓Do this: Practise 60°, 90°, 120°, 45° and 30° with compasses until they are automatic.

  Usually costs: The base mark, and accuracy of every later length.

• ✗Mistake: Construction arcs rubbed out or missing

  The CISCE examiners' suggestions for 2024 ask students to show all traces of construction clearly, and the 2025 report lists locus constructions among the topics candidates found difficult.

  ✓Do this: Leave every arc visible; draw lightly but do not erase.

  Usually costs: Marks for each construction step.

• ✗Mistake: Describing a locus only by its shape

  A locus answer such as 'a circle' or 'a line' does not fix which circle or line.

  Wrong working

  Locus: a circle

  Right working

  Locus: a circle with centre O and radius 4 cm

  ✓Do this: Always give centre and radius, or which segment or angle is bisected.

  Usually costs: 1 mark.

Practice questions on Loci

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

1. Q1 · 1 mark · MCQBoard level

   The locus of a point that is always 3 cm from a fixed point $O$ is:
   (a) a line 3 cm from $O$
   (b) a circle with centre $O$ and radius 3 cm
   (c) a circle with diameter 3 cm
   (d) the perpendicular bisector of a 3 cm segment

   Hint

   Fixed point, fixed distance.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   The point inside a triangle that is equidistant from its three sides is the:
   (a) centroid
   (b) circumcentre
   (c) incentre
   (d) mid-point of the longest side

   Hint

   Sides, not vertices.

   Answer

   (c)

3. Q3 · 1 mark · MCQHarder board level

   The circumcentre of a triangle with sides 7 cm, 24 cm and 25 cm lies:
   (a) inside the triangle, 12.5 cm from each vertex
   (b) at the mid-point of the 25 cm side, 12.5 cm from each vertex
   (c) outside the triangle
   (d) at the vertex of the right angle

   Hint

   Check for a right angle.

   Answer

   (b): $7^2+24^2=625=25^2$.

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): The locus of points equidistant from two intersecting lines consists of two lines.
   Reason (R): Two intersecting lines form two pairs of vertically opposite angles, and the bisectors of these give two perpendicular lines.
   (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

   Think of an X shape.

   Answer

   (c)

5. Q5 · 4 marks · Short answerHarder board level

   $AB=10$ cm. Find the number of points that are equidistant from $A$ and $B$ and 13 cm from $A$, and the distance between them.

   Hint

   Right triangle with hypotenuse 13 and one leg 5.

   Answer

   Two points on the perpendicular bisector, each $\sqrt{13^2-5^2}=12$ cm from $AB$, so 24 cm apart.

6. Q6 · 4 marks · Short answerHarder board level

   Construct triangle $ABC$ with $BC=6$ cm, $AB=4.5$ cm and $\angle ABC=90^{\circ }$. Construct the locus of points equidistant from $BA$ and $BC$, meeting $AC$ at $D$. Measure $BD$.

   Hint

   D is the same distance t from both legs; t = ab ÷ (a + b).

   Answer

   $D$ is $t=\frac{4.5\times 6}{10.5}\approx 2.57$ cm from each leg, so $BD=t\sqrt{2}\approx 3.6$ cm.

7. Q7 · 3 marks · Long answerBoard level

   State the locus of: (a) points 2 cm from a fixed point $O$; (b) centres of circles that pass through two fixed points $A$ and $B$; (c) points inside $\angle ABC$ equidistant from $BA$ and $BC$.

   Hint

   One for each of the three loci.

   Answer

   (a) Circle, centre $O$, radius 2 cm. (b) Perpendicular bisector of $AB$. (c) Bisector of $\angle ABC$.

8. Q8 · 4 marks · Long answerHarder board level

   Construct triangle $PQR$ with $PQ=PR=5$ cm and $QR=6$ cm. Construct its incircle and measure its radius. Check by calculation using area $=r\times s$, where $s$ is the semi-perimeter.

   Hint

   The height from P to QR is 4 cm.

   Answer

   Area $=\frac{1}{2}\times 6\times 4=12$ cm², $s=8$ cm, so $r=12÷8=1.5$ cm.

9. Q9 · 4 marks · Long answerHOTS / competency

   Using loci, explain why the perpendicular bisectors of the three sides of any triangle meet at a single point. Where is this point for (a) a right-angled triangle, (b) an obtuse-angled triangle?

   Hint

   Two bisectors meet at O; what does that tell you about OA, OB, OC?

   Answer

   The bisectors of $AB$ and $BC$ meet at $O$ with $OA=OB$ and $OB=OC$. So $OA=OC$, which means $O$ is also on the bisector of $AC$: all three meet at $O$. (a) At the mid-point of the hypotenuse. (b) Outside the triangle, beyond the longest side.

10. Q10 · 4 marks · Long answerHarder board level

    $AB=8$ cm. $P$ is equidistant from $A$ and $B$ and is 3 cm from the line $AB$. How many such points are there? Find $PA$.

    Hint

    Measure 3 cm along the perpendicular bisector on each side.

    Answer

    Two points (one each side of $AB$). $PA=\sqrt{4^2+3^2}=5$ cm.

11. Q11 · 1 mark · MCQHOTS / competency

    The locus of the centres of all circles of radius 2 cm that pass through a fixed point $A$ is:
    (a) the point $A$
    (b) a circle with centre $A$ and radius 2 cm
    (c) a circle with centre $A$ and radius 4 cm
    (d) a line through $A$

    Hint

    Each centre is 2 cm from A.

    Answer

    (b)

12. Q12 · 5 marks · Long answerHOTS / competency

    Construct triangle $ABC$ with $AB=BC=6$ cm and $\angle ABC=120^{\circ }$. (a) Construct its circumcentre $O$. (b) Measure $OA$. (c) Show that $O$ lies outside the triangle and that triangle $ABO$ is equilateral.

    Hint

    The bisector of AC passes through B. In △ABO, OA = OB.

    Answer

    (b) $OA\approx 6$ cm (the circumradius is $\frac{AC}{2\sin 120^{\circ }}=\frac{6\sqrt{3}}{\sqrt{3}}=6$ cm). (c) The triangle is obtuse, so $O$ lies outside, beyond $AC$. $O$ is on the bisector of $AC$, which is also the bisector of $\angle B$, so $\angle ABO=60^{\circ }$; with $OA=OB$, triangle $ABO$ has all angles $60^{\circ }$: equilateral, so $OA=AB=6$ 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. 1.Locus of points at a fixed distance from a fixed point?Show answer

   A circle with that point as centre and that distance as radius.

2. 2.Locus of points equidistant from two points A and B?Show answer

   The perpendicular bisector of AB.

3. 3.Locus of points equidistant from two intersecting lines?Show answer

   The bisectors of the angles between them.

4. 4.Point equidistant from the three vertices of a triangle?Show answer

   The circumcentre.

5. 5.Point equidistant from the three sides of a triangle?Show answer

   The incentre.

6. 6.Where is the circumcentre of a right triangle?Show answer

   At the mid-point of the hypotenuse.

7. 7.Are proofs of the locus theorems required?Show answer

   No.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you name the locus for each of the three conditions?
• ☐Can you construct a perpendicular bisector and an angle bisector cleanly?
• ☐Can you construct 90° and 60° without a protractor?
• ☐Do you show every arc?
• ☐Can you locate the circumcentre and incentre?
• ☐Do you measure and state lengths to the nearest mm?

Loci: frequently asked questions