Loci Class 10 ICSE — Bansal Chapter 16
Loci Class 10 ICSE Maths, Chapter 16 in Bansal’s book, is the study of the path traced by a moving point that satisfies a given condition. The chapter teaches the definition of a locus, the two symmetry theorems, the four centres of a triangle, and the construction methods used in exam questions. It builds directly on your ruler-and-compass skills.
Download Loci Class 10 ICSE Mathematics Chapter 16 PDF
Download the Loci Class 10 ICSE Mathematics chapter PDF from the Bansal textbook to revise offline or print for practice. It is the full chapter as printed in the book — 11 pages, 5 sections, 53 questions.
What is inside Chapter 16
11 pages · 5 sections · 6 worked examples · 53 exercise questions
| Section | Title |
|---|---|
| 16.2 | Definition |
| 16.3 | Theorems Based on Symmetry |
| 16.4 | Applications |
| 16.5 | Summary |
| 16.6 | Important |
- EXERCISE 16(A): 20 questions
- EXERCISE 16(B): 33 questions
16.2 Definition
Locus is the path traced by a moving point that satisfies a given condition. The plural of locus is loci, pronounced “losai”. The method is condition-first: state the moving condition, mark points that satisfy it, then join them to reveal the figure.
Worked pattern 1 (parallel lines). Take two parallel lines 8 cm apart. A point moves so it stays equidistant from both lines. The separation is 8 cm, so each marked point sits at \frac{8}{2} = 4 cm from each line. Mark several such points and join them. The result is a straight line parallel to both given lines and midway between them.
Worked pattern 2 (fixed point). Take a fixed point O and a moving point P that stays at a constant 5 cm from O. Mark points A, B, C, D, each 5 cm from O. Draw a free-hand curve through them. The final figure is a circle with centre O and radius 5 cm.
This condition-first method is the backbone of the whole chapter. Every locus question in this ICSE chapter starts the same way: write the condition, place points, join them. For more on the constructions that build on this idea, see the Class 10 ICSE Maths hub.
16.3 Theorems Based on Symmetry
The chapter gives two core theorems based on symmetry.
Theorem 3. The locus of a point equidistant from two intersecting lines is the bisector of the angles between the lines.
Theorem 4. The locus of a point equidistant from two given points is the perpendicular bisector of the line joining the two points.
Each proof follows the structure Given, To Prove, Construction. Each also has two parts. Part (i) shows that every point satisfying the condition lies on the locus. Part (ii), the converse, shows that every point on the locus satisfies the condition.
For Theorem 3, draw perpendiculars PL and PM from P to the two lines. In triangles POL and POM, PL = PM (given), both angles are 90°, and PO is common. So \Delta \mathrm{POL} \cong \Delta \mathrm{POM} by R.H.S. Congruent angles follow, so P lies on the bisector. The converse uses A.A.S. and C.P.C.T.C. The same argument works for any of the four angles the intersecting lines form.
For Theorem 4, triangles AOP and BOP have PA = PB (given), AO = BO (O is the midpoint), and PO common. So \Delta \mathrm{AOP} \cong \Delta \mathrm{BOP} by S.S.S., giving \angle AOP = \angle BOP = 90^\circ. The converse uses S.A.S.
Why does the converse matter? A locus must be exact. Checking both directions proves the figure is the locus, not merely part of it. If you only prove one direction, a point outside the figure could also satisfy the condition.
16.4 Applications
Section 16.4 turns the two theorems into proof and construction tools.
Proof patterns. To show a point on an angle bisector is equidistant from the sides, drop perpendiculars and use A.A.S. congruence, then C.P.C.T.C. To show a point on a perpendicular bisector is equidistant from the endpoints, join to the endpoints and use S.A.S., then C.P.C.T.C.
Construction pattern. To find a point equidistant from B and C, and also equidistant from AB and BC: draw the perpendicular bisector of BC and the bisector of angle ABC. Their intersection is the required point.
The scalene triangle pattern uses one locus only. A point on base BC equidistant from sides AB and AC lies on the bisector of angle BAC. The point where that bisector meets BC is the answer, because every point on an angle bisector is equidistant from the arms.
The quadrilateral diagonal pattern uses a perpendicular bisector. In a quadrilateral ABCD, a point P on diagonal AC equidistant from B and C lies where the perpendicular bisector of BC meets AC.
Original example. Construct a triangle with AB = 6 cm, BC = 7 cm, CA = 6.5 cm. Draw the bisector of angle ABC and the perpendicular bisector of BC. They meet at a point P equidistant from B and C, and from AB and BC.
16.5 Summary
The summary collects eight locus rules into one reference.
- Equidistant from two fixed points → perpendicular bisector of the joining segment.
- Equidistant from two intersecting lines → pair of angle bisectors.
- Fixed distance from a fixed point → circumference of a circle.
- Equidistant from two parallel lines → line midway between them.
- Given distance from a line → pair of parallel lines.
- Mid-points of equal chords → concentric circle.
- Mid-points of parallel chords → perpendicular diameter.
- Equidistant from two concentric circles → concentric circle midway between them.
The chapter adds three memory notes. State the kind of figure and its position. Every point satisfying the condition lies on the locus, and every point on the locus satisfies the condition. The locus may be straight or curved.
The angle-in-a-semicircle pattern uses a circle. Take AB = 10 cm. Draw a circle with AB as diameter. The locus of point P with \angle APB = 90^\circ is the circumference of this circle, since the angle in a semicircle is always 90°.
The chord pattern is similar. For a chord AB, the perpendicular bisector of the chord is a diameter of the circle, and that diameter is the locus of points equidistant from A and B.
The chapter notes that circle-based locus problems should be attempted after completing the circles chapter, which appears later in this book.
16.6 Important
Every triangle has four special points of concurrency. This ICSE chapter calls them the centres of the triangle.
| Centre | How it is found | Equidistant from | Special property |
|---|---|---|---|
| Centroid | Three medians | — | Divides each median 2:1 |
| Incentre | Angle bisectors | Sides | \mathrm{ID} = \mathrm{IE} = \mathrm{IF}, the inradius |
| Circumcentre | Perpendicular bisectors | Vertices | \mathrm{OA} = \mathrm{OB} = \mathrm{OC}, the circumradius |
| Orthocentre | Altitudes | — | Point of concurrency of altitudes |
The centroid is where the three medians meet. It divides each median in the ratio 2:1, so \mathrm{AG} : \mathrm{GP} = 2 : 1. The incentre is where the angle bisectors meet, and it is equidistant from the sides.
The circumcentre is where the perpendicular bisectors meet, and it is equidistant from the vertices. The orthocentre is where the three altitudes meet.
For an isosceles triangle, the median from the apex is also the angle bisector, the perpendicular bisector of the base, and the altitude. In an equilateral triangle, all four centres coincide at a single point.
What to practise from Loci Class 10 ICSE
EXERCISE 16(A) has 20 questions on proofs and constructions. EXERCISE 16(B) has 33 questions that ask you to describe or sketch loci.
Common mistakes to avoid:
- Forgetting the converse part of a locus — a locus must be exact in both directions.
- Treating the angle-bisector locus as a single line instead of a pair of bisectors.
- Assuming a locus is always straight — it can be a circle.
- Placing the midway line between parallel lines at the wrong distance — it sits at half the separation.
Several questions carry past-paper tags: 2000, 2001, 2007, 2008, 2010. These are mostly ruler-and-compass constructions. The official CISCE site keeps the Class 10 Mathematics syllabus and exam format for this ICSE chapter.
Loci Class 10 ICSE frequently asked questions
What is the locus of a point equidistant from two fixed points in loci class 10 ICSE?
The locus is the perpendicular bisector of the line segment joining the two fixed points. Every point on that bisector is equidistant from both points, and every equidistant point lies on the bisector. Theorem 4 in this ICSE chapter proves both directions.
How do I remember the four centres of a triangle for the loci chapter?
Pair each centre with how it is built. The centroid comes from medians, the incentre from angle bisectors, the circumcentre from perpendicular bisectors, and the orthocentre from altitudes. The incentre is equidistant from the sides, while the circumcentre is equidistant from the vertices.
What is the plural of locus and how is it pronounced?
The plural of locus is loci, pronounced “losai”. One condition traces one locus; several conditions trace several loci.
Why does the locus of a point equidistant from two parallel lines lie midway between them?
Because the point must stay exactly halfway between them. For two parallel lines 8 cm apart, the midpoint distance is \frac{8}{2} = 4 cm from each line. Symmetry forces the locus to be a straight line parallel to both given lines and midway between them.
Which exercises from the Loci chapter should I practise?
Work through EXERCISE 16(A), which has 20 proof and construction questions, then EXERCISE 16(B) with 33 description questions. The past-paper tagged questions from 2000, 2001, 2007, 2008, and 2010 deserve extra attention.
Reference: ICSE Class 10 Mathematics textbook material, chapter 16: Loci — CISCE-aligned.
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