Loci and its Constructions

Q: Two lines intersect at an angle of 50°. Find the angles that the bisectors of the angles between them make with each given line, and the angle between the two bisectors.

One bisector makes 25° with each line; the other bisects the 130° angles and makes 65° with each line. The bisectors meet at 25° + 65° = 90°.

Q: Find the point on the y-axis that is equidistant from A(4, 0) and B(0, 2).

Let the point be (0, k). Then 16 + k2 = (k-2)2 = k2 - 4k + 4, so 4k = -12 and k = -3. The point is (0, -3); both distances equal 5 units.

ICSE Class 10 · Mathematics · Chapter 16

What this chapter covers

Locus: definition and meaning
Theorems on loci
Locus of a point at a fixed distance from a fixed point
Locus of a point equidistant from two given points
Locus of a point equidistant from two intersecting lines
Points that satisfy two conditions
Circumcentre and incentre of a triangle
Constructions based on loci in triangles and quadrilaterals
Common mistakes with loci

Key definitions

Locus is the set of all points, and only those points, which satisfy a given geometrical condition.

Key points

A locus is the set of all points, and only those points, satisfying a given condition; the plural is loci.
A figure is a locus only if every point on it satisfies the condition and every point satisfying the condition lies on it.
The distance of a point from a line is always the perpendicular distance.
Fixed distance r from a fixed point O: the locus is the circle with centre O and radius r.
Equidistant from two points A and B: the locus is the perpendicular bisector of AB.
Equidistant from two intersecting lines: the locus is the pair of angle bisectors, which are perpendicular to each other; inside a single angle it is that angle's bisector.
Each locus theorem has a converse; proofs are not required, but the statements are used as reasons.
To find a point satisfying two conditions, construct both loci and mark every intersection; there may be two, one or no such points.
The perpendicular bisectors of the sides of a triangle meet at the circumcentre, equidistant from the vertices.
The bisectors of the angles of a triangle meet at the incentre, equidistant from the sides.
Keep all construction arcs and label every locus in words.

Questions answered

Define the locus of a point.

The locus of a point is the set of all points, and only those points, which satisfy a given geometrical condition; equivalently, the path traced by a point moving so that it always satisfies the condition.

State the locus of a point which is always 3.5 cm from a fixed point O.

A circle with centre O and radius 3.5 cm.

State the locus of a point equidistant from two given points A and B.

The perpendicular bisector of the segment AB.

State the locus of a point equidistant from two intersecting lines.

The pair of lines bisecting the angles between the two given lines; these two bisectors are perpendicular to each other.

Why must the compass radius be more than half of AB when constructing the perpendicular bisector of AB?

With a radius of half AB or less, the arcs from A and B do not cut each other at two points, so the two points that fix the bisector cannot be found.

Two lines intersect at an angle of 50^°. Find the angles that the bisectors of the angles between them make with each given line, and the angle between the two bisectors.

One bisector makes 25^° with each line; the other bisects the 130^° angles and makes 65^° with each line. The bisectors meet at 25^° + 65^° = 90^°.

Find the equation of the locus of a point equidistant from A(-2, 4) and B(6, 4).

The mid-point of AB is (2, 4) and AB is horizontal, so the perpendicular bisector is the vertical line x = 2.

Find the point on the y-axis that is equidistant from A(4, 0) and B(0, 2).

Let the point be (0, k). Then 16 + k² = (k-2)² = k² - 4k + 4, so 4k = -12 and k = -3. The point is (0, -3); both distances equal 5 units.

Past board exam questions

ICSE 2027 sample paper · 4 marks

Use ruler and compass for this question: (a) Draw line segment AB of length 6 cm. (b) Construct the locus of points which are equidistant from A and B. Mark a point on the locus which is 4.5 cm from A. Name this point as O. (c) Construct the locus of all points which are 4.5 cm from O. (d) Mark the point of intersection of loci (b) and (c) as P, which is on the same side of AB as O. (e) Construct the locus of points which are equidistant from AP and AB.

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Steps of construction: (a) Draw AB = 6 cm. (b) The locus of points equidistant from A and B is the perpendicular bisector of AB: with A and B as centres and radius more than 3 cm, draw arcs on both sides of AB meeting at two points; join them. With A as centre and radius 4.5 cm, cut this bisector at O (OA = OB = 4.5 cm). (c) The locus of points 4.5 cm from O is the circle with centre O and radius 4.5 cm (it passes through A and B). (d) The perpendicular bisector meets the circle at two points; the one on the same side of AB as O (farther from AB) is P. Join AP. (e) The locus of points equidistant from AP and AB is the bisector of ∠PAB: with A as centre draw an arc cutting AB and AP; …

ICSE 2026 sample paper · 5 marks

Use a ruler and compass to answer this question. (a) Construct a circle of radius 4.5cm and draw a chord AB of length 6.5 cm. (b) At A, construct ∠CAB=75°, where C lies on the circumference of the circle. (c) Construct the locus of all points equidistant from A and B. (d) Construct the locus of all points equidistant from CA and BA. (e) Mark the point of intersection of the two loci as P. Measure and write down the length of CP.

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Steps of construction: (a) Draw a circle with centre O and radius 4.5 cm. With A on the circle as centre and radius 6.5 cm, cut the circle at B; join AB. (b) At A construct ∠BAX = 75° (60° + half of the angle between 60° and 90°), on the side of AB containing the centre; let AX meet the circle at C. (c) The locus of points equidistant from A and B is the perpendicular bisector of AB (it passes through O). (d) The locus of points equidistant from CA and BA is the bisector of ∠CAB. (e) The perpendicular bisector of AB and the bisector of ∠CAB meet at P. On measuring, CP ≈ 5.1 cm (calculation gives 5.1 cm; the board's draft marking scheme gives CP = 4.9 cm, within construction tolerance).

ICSE 2025 board paper · 1 mark

Statement 1: The point which is equidistant from three non-collinear points D, E and F is the circumcentre of the ΔDEF. Statement 2: The incentre of a triangle is the point where the bisector of the angles intersects.

Both the statements are true.
Both the statements are false.
Statement 1 is true, and Statement 2 is false.
Statement 1 is false, and Statement 2 is true.

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Both the statements are true.

ICSE 2025 compartment paper · 1 mark

The locus of a toy bird fixed at the tip of one of the blades of a rotating ceiling fan is a:

straight line
circle
semi-circular arc
diameter of the circle so formed

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circle

ICSE 2025 sample paper · 4 marks

Use ruler and compasses for the following question taking a scale of 10 m=1 cm. A park in a city is bounded by straight fences AB, BC, CD and DA. Given that AB = 50 m, BC = 63 m, ∠ABC = 75°. D is a point equidistant from the fences AB and BC. If ∠BAD = 90°, construct the outline of the park ABCD. Also locate a point P on the line BD for the flag post which is equidistant from the corners of the park A and B.

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Scale 10 m = 1 cm, so AB = 5 cm and BC = 6.3 cm. Steps of construction: 1. Draw AB = 5 cm. 2. At B construct ∠ABC = 75° and cut BC = 6.3 cm. 3. Construct the bisector of ∠ABC (the locus of points equidistant from the fences AB and BC). 4. At A construct ∠BAD = 90°; this perpendicular meets the bisector of ∠ABC at D. 5. Join CD. ABCD is the outline of the park. 6. Construct the perpendicular bisector of AB (the locus of points equidistant from A and B). It meets BD at P, the position of the flag post. (Check: ∠ABD = 37.5°, so AD = 5 tan 37.5° ≈ 3.8 cm, i.e. about 38 m; PA = PB = 2.5 ÷ cos 37.5° ≈ 3.2 cm, i.e. about 32 m.)

ICSE 2025 compartment paper · 4 marks

Use ruler and compass for the following constructions: Construct: (a) an isosceles ΔABC in which AB = AC = 7 cm and BC = 6 cm. (b) the locus of points which moves such that it is 2.5 cm from the point A. (c) the locus of points equidistant from B and C. Mark point P which satisfies both the conditions mentioned in (b) and (c). (d) a circle passing through P, B and C.

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(a) Draw BC = 6 cm. With B and C as centres and radius 7 cm draw arcs meeting at A. Join AB and AC. (b) With centre A and radius 2.5 cm draw a circle; this is the locus of points 2.5 cm from A. (c) Draw the perpendicular bisector of BC; it is the locus of points equidistant from B and C and passes through A. Mark P where it cuts the circle of (b) (the point inside the triangle, 2.5 cm from A towards BC; the bisector also cuts the circle at a second point above A). (d) Draw the perpendicular bisector of PB; it meets the perpendicular bisector of BC at O. With centre O and radius OB draw the circle; it passes through P, B and C. (For P inside the triangle: altitude from A $= \sqrt{49 - 9} …

ICSE 2024 sample paper · 1 mark

The three vertices of a scalene triangle are always equidistant from a fixed point. The point is:

Orthocentre of the triangle.
Incentre of the triangle.
Circumcentre of the triangle.
Centroid of the triangle.

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Circumcentre of the triangle.

ICSE 2024 board paper · 1 mark

The circumcentre of a triangle is the point which is:

at equal distance from the three sides of the triangle.
at equal distance from the three vertices of the triangle.
the point of intersection of the three medians.
the point of intersection of the three altitudes of the triangle.

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at equal distance from the three vertices of the triangle.