Why is PB approximately 3.5 cm in Exercise 16(A) Question 10?

In that construction, P is the intersection of the angle bisector of angle B and the perpendicular bisector of BC. With BC = 6 cm and angle ABC = 60 degrees, P lies 3 cm horizontally from B and at a height of square root 3 cm, so PB = 2 square root 3 cm, which is about 3.46 cm and is measured as about 3.5 cm.

ICSE Class 10 Maths Loci Solutions: Selina Chapter 16

Written by Greya LakshmiPublished on 19 May 2026Updated on 10 June 2026

ICSE Class 10 Maths Loci: direct answer

ICSE Class 10 Maths Loci in Selina Chapter 16 is about finding and constructing the path followed by a moving point when a fixed condition is given. The most used results are: points equidistant from two fixed points lie on the perpendicular bisector of the segment joining them, and points equidistant from the two arms of an angle lie on the angle bisector.

This rewritten page gives a teacher-style study resource for Concise Mathematics Selina Solutions Class 10 ICSE Chapter 16 Loci. It explains the rules, shows the construction logic, gives worked proofs for the exercise types available from the existing page, and adds original practice examples so students can apply the same method in board-style geometry work.

Concise Mathematics Selina Solutions Class 10 ICSE Chapter 16 Loci

In this chapter, a locus is not a single point. It is the set of all points satisfying a given condition. In construction questions, the answer is usually a line, a part of a line, a circle, an arc, or an intersection point obtained from two loci.

The wording of the question is the key. If the question says “equidistant from $A$ and $B$ ,” the fixed objects are two points. If it says “equidistant from $AB$ and $BC$ ,” the fixed objects are two lines or sides. In Maths geometry, distance from a point to a line always means the length of the perpendicular drawn from the point to the line.

Concept snapshot: read a locus like a rule-made track

Think of a moving point $P$ as a pencil tip that is allowed to move only when one rule is obeyed. If the rule is $PA = PB$ , the pencil tip must stay exactly halfway between fixed points $A$ and $B$ . The track it can draw is the perpendicular bisector of $AB$ . If the rule is “equal distance from two arms of an angle,” the track is the angle bisector because both perpendicular distances remain equal there.

Condition in the question	Locus to construct	Reason
Point $P$ is equidistant from two fixed points $A$ and $B$	Perpendicular bisector of $AB$	Every point on it satisfies $PA = PB$
Point $P$ is equidistant from two intersecting lines $l$ and $m$	Angle bisector of the angle formed by $l$ and $m$	Perpendicular distances from $P$ to the two lines are equal
Point $P$ is at a fixed distance $r$ from a fixed point $O$	Circle with centre $O$ and radius $r$	Every point on the circle satisfies $OP = r$
Point $P$ is at a fixed perpendicular distance from a line	A line parallel to the given line	The perpendicular separation remains constant
Two conditions are given together	Intersection of the two loci	The required point must satisfy both conditions at once

For related geometry revision, students can also use Selina Class 10 constructions solutions, Selina Class 10 circles solutions and the wider ICSE Class 10 Maths solutions directory.

Locus rules and construction methods

Before solving Exercise 16(A) and Exercise 16(B), keep these construction rules in mind. They are standard results used in ICSE geometry and in Selina Concise Mathematics.

Rule 1: Perpendicular bisector theorem

Statement: If a point $P$ lies on the perpendicular bisector of $AB$ , then $PA = PB$ .

Converse: If $PA = PB$ , then $P$ lies on the perpendicular bisector of $AB$ .

How to use it: Whenever a question uses the phrase “equidistant from two fixed points,” draw or identify the perpendicular bisector of the segment joining those points.

Rule 2: Angle bisector theorem for distance from two lines

Statement: A point on the bisector of an angle is equidistant from the two arms of the angle.

Converse: A point inside an angle and equidistant from its two arms lies on the angle bisector.

How to use it: Whenever the question says “equidistant from sides $AB$ and $AC$ ,” draw the bisector of $\angle BAC$ . The distances must be perpendicular distances to the two sides.

Rule 3: Locus inside a bounded figure

Step 1: Draw the full locus line first, such as a perpendicular bisector or an angle bisector.

Step 2: Keep only the part of the locus that lies inside the triangle, circle, or region specified in the question.

Step 3: If another condition is given, mark the intersection of the two loci and name it clearly.

Exercise 16(A) solutions

Exercise 16(A) mainly checks whether students can recognise the correct locus theorem and support it with a short geometric proof. The solutions below use the question statements visible in the existing page and rewrite the working in original school format.

Question 1(a): $P$ is on the perpendicular bisector of $BC$

Given: $P$ lies on the perpendicular bisector of side $BC$ of $\triangle ABC$ .

Step 1: A point on the perpendicular bisector of a segment is equidistant from the two end-points of that segment.

Step 2: Here the segment is $BC$ , so the two fixed points are $B$ and $C$ .

$PB = PC$

Final answer: Option $b$ , $PB = PC$ .

Question 1(b): $P$ is on the bisector of $\angle A$

Given: $P$ lies on the bisector of $\angle A$ of $\triangle ABC$ .

Step 1: A point on an angle bisector is equidistant from the two arms of the angle.

Step 2: The arms of $\angle A$ are $AB$ and $AC$ .

Final answer: Option $d$ , $P$ is equidistant from sides $AB$ and $AC$ .

Question 1(c): Perpendicular bisector of $AB$ and bisector of $\angle A$ meet at $P$

Given: $P$ lies on the perpendicular bisector of $AB$ .

Step 1: The perpendicular bisector of $AB$ is the locus of points equidistant from $A$ and $B$ .

$PA = PB$

Step 2: The fact that $P$ also lies on the bisector of $\angle A$ gives another distance condition from sides, but the option asked here follows directly from the perpendicular bisector.

Final answer: Option $a$ , $PA = PB$ .

Question 1(d): Decide the correct statement from the figure

Assumption (edition may vary): The equal marks in the figure show $AD = DB$ and $AC = BC$ , with $CD$ joined.

Step 1: Compare $\triangle ACD$ and $\triangle BCD$ .

$AD = DB,\quad AC = BC,\quad CD = CD$

Step 2: By the SSS congruence criterion,

$\triangle ACD \cong \triangle BCD$

Step 3: Corresponding angles are equal.

$\angle ACD = \angle BCD$

Step 4: Therefore $CD$ divides $\angle ACB$ into two equal parts.

Final answer: Option $c$ , $CD$ bisects $\angle ACB$ .

Question 1(e): Bisector of $\angle B$ meets $AC$ at $P$

Given: $BP$ is the bisector of $\angle B$ in $\triangle ABC$ .

Step 1: The arms of $\angle B$ are $BA$ and $BC$ .

Step 2: A point on an angle bisector is equidistant from the two arms of the angle.

Final answer: Option $d$ , $P$ is equidistant from sides $AB$ and $BC$ .

Question 2: $PQ$ is the perpendicular bisector of $AB$

Given: $PQ$ is the perpendicular bisector of side $AB$ of $\triangle ABC$ . Let $P$ be the midpoint of $AB$ , so $AP = PB$ , and $PQ \perp AB$ .

To prove: $Q$ is equidistant from $A$ and $B$ , that is $QA = QB$ .

Step 1: Join $QA$ and $QB$ .

Step 2: In $\triangle AQP$ and $\triangle BQP$ ,

$AP = PB \quad \text{and} \quad \angle QPA = \angle QPB = 90^\circ \quad \text{and} \quad QP = QP$

Step 3: Therefore, by SAS congruence,

$\triangle AQP \cong \triangle BQP$

Step 4: Corresponding sides are equal.

$QA = QB$

Hence proved: $Q$ is equidistant from $A$ and $B$ .

Question 3: $CP$ bisects $\angle C$ of $\triangle ABC$

Given: $CP$ is the bisector of $\angle C$ .

To prove: $P$ is equidistant from $AC$ and $BC$ .

Step 1: Draw $PL \perp AC$ and $PM \perp BC$ . The distances from $P$ to $AC$ and $BC$ are $PL$ and $PM$ .

Step 2: In $\triangle LPC$ and $\triangle MPC$ ,

$\angle PLC = \angle PMC = 90^\circ$

$\angle LCP = \angle PCM \quad \text{because } CP \text{ bisects } \angle C$

$PC = PC \quad \text{common side}$

Step 3: Hence, by AAS congruence,

$\triangle LPC \cong \triangle MPC$

Step 4: Therefore,

$PL = PM$

Hence proved: $P$ is equidistant from $AC$ and $BC$ .

Question 4: $AX$ bisects $\angle BAC$ and $PQ$ is the perpendicular bisector of $AC$

Given: $AX$ bisects $\angle BAC$ , and $PQ$ is the perpendicular bisector of $AC$ , meeting $AX$ at $Y$ .

Part (i) to prove: $X$ is equidistant from $AB$ and $AC$ .

Step 1: Draw $XM \perp AB$ and $XL \perp AC$ . Then $XM$ and $XL$ are the perpendicular distances from $X$ to the two sides.

Step 2: In $\triangle AXM$ and $\triangle AXL$ ,

$\angle XMA = \angle XLA = 90^\circ$

$\angle MAX = \angle XAL \quad \text{because } AX \text{ bisects } \angle BAC$

$AX = AX \quad \text{common side}$

Step 3: By AAS congruence, $\triangle AXM \cong \triangle AXL$ , so

$XM = XL$

Part (i) proved: $X$ is equidistant from $AB$ and $AC$ .

Part (ii) to prove: $Y$ is equidistant from $A$ and $C$ .

Step 4: Since $Y$ lies on the perpendicular bisector of $AC$ , every point on this locus is equidistant from $A$ and $C$ .

$YA = YC$

Part (ii) proved: $Y$ is equidistant from $A$ and $C$ .

Question 5: Construct $\triangle ABC$ with $AB = 4.2\text{ cm}$ , $BC = 6.3\text{ cm}$ , $AC = 5\text{ cm}$

Given: $AB = 4.2\text{ cm}$ , $BC = 6.3\text{ cm}$ , and $AC = 5\text{ cm}$ . The perpendicular bisector of $BC$ meets $AC$ at $D$ .

Construction step 1: Draw $BC = 6.3\text{ cm}$ .

Construction step 2: With centre $B$ and radius $4.2\text{ cm}$ , draw an arc.

Construction step 3: With centre $C$ and radius $5\text{ cm}$ , draw another arc cutting the first arc at $A$ .

Construction step 4: Join $AB$ and $AC$ . Construct the perpendicular bisector of $BC$ . Let it meet $BC$ at $E$ and $AC$ at $D$ . Join $DB$ and $DC$ .

Proof step 1: In $\triangle DBE$ and $\triangle DCE$ ,

$BE = EC \quad \text{because } E \text{ is the midpoint of } BC$

$\angle DEB = \angle DEC = 90^\circ$

$DE = DE \quad \text{common side}$

Proof step 2: Therefore, by SAS congruence,

$\triangle DBE \cong \triangle DCE$

Proof step 3: Corresponding sides are equal.

$DB = DC$

Hence proved: $D$ is equidistant from $B$ and $C$ .

Question 6: $PA = PB$ and $QA = QB$

Given: $PA = PB$ and $QA = QB$ .

To prove: $PQ$ , produced if required, is the perpendicular bisector of $AB$ .

Step 1: Since $PA = PB$ , point $P$ is equidistant from $A$ and $B$ . Hence $P$ lies on the perpendicular bisector of $AB$ .

Step 2: Since $QA = QB$ , point $Q$ is also equidistant from $A$ and $B$ . Hence $Q$ lies on the same perpendicular bisector of $AB$ .

Step 3: Only one straight line can pass through two distinct points $P$ and $Q$ . Therefore the line $PQ$ is the perpendicular bisector of $AB$ . If the drawn segment is short, produce it to meet $AB$ .

Final statement: The locus of points equidistant from two fixed points is the perpendicular bisector of the segment joining those fixed points.

Question 7: Construct triangle with $\angle ABC = 75^\circ$ , $AB = 5\text{ cm}$ , $BC = 6.4\text{ cm}$

Given: $AB = 5\text{ cm}$ , $BC = 6.4\text{ cm}$ , and $\angle ABC = 75^\circ$ . The perpendicular bisector of $BC$ and the bisector of $\angle ACB$ meet at $P$ .

Construction step 1: Draw $BC = 6.4\text{ cm}$ .

Construction step 2: At $B$ , construct $\angle CBX = 75^\circ$ . A ruler-and-compass method is to construct $60^\circ$ and $90^\circ$ at $B$ , then bisect the angle between them.

Construction step 3: On ray $BX$ , mark $A$ such that $AB = 5\text{ cm}$ . Join $AC$ .

Construction step 4: Draw the perpendicular bisector of $BC$ . Draw the bisector of $\angle ACB$ . Let the two loci meet at $P$ .

Proof step 1: Since $P$ lies on the perpendicular bisector of $BC$ ,

$PB = PC$

Proof step 2: Draw perpendiculars $PL \perp AC$ and $PM \perp BC$ . Since $P$ lies on the bisector of $\angle ACB$ ,

$PL = PM$

Hence proved: $P$ is equidistant from $B$ and $C$ , and also equidistant from $AC$ and $BC$ .

Question 8: Locus point in a parallelogram

Given: In parallelogram $ABCD$ , $AB > BC$ , and $P$ is a point on $AC$ such that $PB$ bisects $\angle ABC$ .

To prove: $P$ is equidistant from $AB$ and $BC$ .

Step 1: Draw $PL \perp AB$ and $PM \perp BC$ . Then $PL$ and $PM$ are the perpendicular distances from $P$ to the sides.

Step 2: In $\triangle PLB$ and $\triangle PMB$ ,

$\angle PLB = \angle PMB = 90^\circ$

$\angle PBL = \angle PBM \quad \text{because } PB \text{ bisects } \angle ABC$

$PB = PB \quad \text{common side}$

Step 3: Hence, by AAS congruence,

$\triangle PLB \cong \triangle PMB$

Step 4: Therefore,

$PL = PM$

Hence proved: $P$ is equidistant from $AB$ and $BC$ .

Question 9: Two angle bisectors meet at $A$ in $\triangle LMN$

Given: The bisectors of the interior angles at $L$ and $N$ meet at point $A$ .

Part (i) to prove: $A$ is equidistant from all three sides of $\triangle LMN$ .

Step 1: Since $A$ lies on the bisector of $\angle L$ , $A$ is equidistant from $LM$ and $LN$ .

Step 2: Since $A$ lies on the bisector of $\angle N$ , $A$ is equidistant from $LN$ and $MN$ .

Step 3: Therefore the perpendicular distances from $A$ to $LM$ , $LN$ , and $MN$ are all equal.

Part (i) proved: $A$ is equidistant from all three sides of the triangle.

Part (ii) to prove: $AM$ bisects $\angle LMN$ .

Step 4: From part (i), $A$ is equidistant from the sides $ML$ and $MN$ .

Step 5: A point inside an angle and equidistant from the two arms of that angle lies on the angle bisector.

Part (ii) proved: $AM$ bisects $\angle LMN$ .

Question 10: Ruler and compass construction with two loci

Given: Construct $\triangle ABC$ , where $AB = 3.5\text{ cm}$ , $BC = 6\text{ cm}$ , and $\angle ABC = 60^\circ$ .

Part (i): Draw $BC = 6\text{ cm}$ . At $B$ , construct $\angle CBX = 60^\circ$ . On ray $BX$ , mark $A$ such that $AB = 3.5\text{ cm}$ . Join $AC$ .

Part (ii): The locus of points inside the triangle equidistant from $BA$ and $BC$ is the internal bisector of $\angle ABC$ .

Part (iii): The locus of points inside the triangle equidistant from $B$ and $C$ is the perpendicular bisector of $BC$ .

Part (iv): Let the angle bisector of $\angle ABC$ meet the perpendicular bisector of $BC$ at $P$ . Then $P$ is equidistant from $AB$ and $BC$ , and also equidistant from $B$ and $C$ .

Measurement check: Since $BC = 6\text{ cm}$ , the perpendicular bisector of $BC$ is $3\text{ cm}$ from $B$ along $BC$ . The angle bisector of $60^\circ$ makes $30^\circ$ with $BC$ . Thus the height of $P$ above $BC$ is $3\tan 30^\circ = \sqrt{3}\text{ cm}$ .

$PB = \sqrt{3^2 + \left(\sqrt{3}\right)^2} = \sqrt{12} = 2\sqrt{3}\text{ cm}$

$2\sqrt{3}\text{ cm} \approx 3.46\text{ cm}$

Final answer: $PB \approx 3.5\text{ cm}$ when measured from the construction.

Exercise 16(B) solutions

Exercise 16(B) applies the same locus ideas to circles, chords and moving points. A common pattern is that a moving point is not random; its motion is restricted by an equality condition.

Question 1(a): Centres of circles tangent to arms $AB$ and $BC$

Given: Circles are tangent to the arms $AB$ and $BC$ of $\angle ABC$ .

Step 1: The centre of a circle tangent to a line is at a perpendicular distance equal to the radius from that line.

Step 2: If a circle is tangent to both arms $AB$ and $BC$ , its centre is equidistant from the two arms.

Step 3: The locus of points equidistant from two intersecting lines is the angle bisector.

Final answer: Option $c$ , the bisector of $\angle ABC$ .

Question 1(b): Moving point inside a circle equidistant from chord endpoints $A$ and $B$

Given: $AB$ is a chord of a circle, and $P$ moves inside the circle such that $PA = PB$ .

Step 1: The condition $PA = PB$ means that $P$ lies on the perpendicular bisector of $AB$ .

Step 2: In a circle, the perpendicular bisector of a chord passes through the centre of the circle.

Step 3: The part of this line inside the circle is a diameter perpendicular to the chord $AB$ .

Final answer: Option $d$ , the diameter of the circle which is perpendicular to chord $AB$ .

Question 1(c): $\triangle APB$ is always isosceles with base $AB$

Given: $AB$ is a line segment, and $P$ moves so that $\triangle APB$ is always isosceles with base $AB$ .

Step 1: If $AB$ is the base of the isosceles triangle, the equal sides are $PA$ and $PB$ .

$PA = PB$

Step 2: The locus of points equidistant from $A$ and $B$ is the perpendicular bisector of $AB$ .

Final answer: Option $c$ , the line which is the perpendicular bisector of $AB$ .

Original worked examples for practice

The following examples are not copied from the exercise. They are included to help students transfer the same ideas to new loci questions in ICSE Class 10 Maths.

Worked Example 1: Point equidistant from two fixed points and lying on a given line

Question: Points $A$ and $B$ are fixed. A point $P$ is equidistant from $A$ and $B$ , and $P$ also lies on a given line $l$ . Describe the construction of $P$ .

Step 1: Since $PA = PB$ , construct the perpendicular bisector of $AB$ .

Step 2: The point $P$ must also lie on line $l$ . Therefore $P$ is at the intersection of line $l$ and the perpendicular bisector of $AB$ .

Step 3: If the line $l$ does not meet the perpendicular bisector, there is no point $P$ satisfying both conditions in the drawn plane region.

Final answer: $P = l \cap \text{perpendicular bisector of } AB$ , if the two lines intersect.

Worked Example 2: Centre of a circle touching two sides of an angle

Question: A circle touches both arms of $\angle XYZ$ . Where can its centre lie?

Step 1: Let the centre be $O$ . The radius to a tangent line is perpendicular to the tangent at the point of contact.

Step 2: Since the same circle touches both arms, the perpendicular distance from $O$ to $YX$ equals the perpendicular distance from $O$ to $YZ$ .

Step 3: A point equidistant from the two arms of an angle lies on the angle bisector.

Final answer: The centre lies on the bisector of $\angle XYZ$ .

Worked Example 3: Two conditions inside a triangle

Question: In $\triangle ABC$ , locate a point $P$ inside the triangle such that $P$ is equidistant from $B$ and $C$ , and also equidistant from sides $AB$ and $AC$ .

Step 1: The condition “equidistant from $B$ and $C$ ” gives the perpendicular bisector of $BC$ .

Step 2: The condition “equidistant from $AB$ and $AC$ ” gives the internal angle bisector of $\angle BAC$ .

Step 3: The required point must satisfy both conditions, so mark the intersection of these two loci inside the triangle.

Final answer: $P$ is the intersection of the perpendicular bisector of $BC$ and the internal bisector of $\angle BAC$ .

Examiner’s mindset for loci constructions

In loci construction answers, the examiner looks for both the drawing and the reason. A neat construction alone may not be enough if the question asks to prove a distance property. Students should name the locus, show the required construction marks, label the intersection point, and then write one or two lines explaining why the point satisfies the given condition.

For perpendicular bisector problems: mention the equality $PA = PB$ or $PB = PC$ , depending on the fixed points.
For angle bisector problems: draw or mention perpendicular distances to the two sides, because distance from a point to a line is measured perpendicularly.
For measurement questions: write the measured value with $\text{cm}$ . If the value is obtained from construction, a small drawing tolerance is expected; do not present a rough measurement as an exact theorem unless you have proved it.
For proof questions: use a named reason such as SAS, AAS, SSS, CPCT, perpendicular bisector theorem or angle bisector property.

Common mistakes students make

Mistake 1: Confusing points and lines. If the condition is equal distance from $A$ and $B$ , use a perpendicular bisector. If the condition is equal distance from $AB$ and $AC$ , use an angle bisector.
Mistake 2: Measuring distance to a side along a slant line. The distance from a point to a line is always the perpendicular distance. In proofs, draw $PL \perp AB$ and $PM \perp AC$ before comparing $PL$ and $PM$ .
Mistake 3: Drawing the full locus when the question says inside the triangle. First draw the full locus lightly, then keep the required part inside the given region.
Mistake 4: Forgetting to prove the constructed point satisfies both conditions. When two loci meet at $P$ , write both conclusions, such as $PB = PC$ and equal perpendicular distances from $P$ to the two sides.
Mistake 5: Writing exact values for measured constructions without proof. In Question 10, $PB$ is measured as about $3.5\text{ cm}$ . If using coordinate checking, the exact value is $2\sqrt{3}\text{ cm}$ , which rounds to $3.46\text{ cm}$ .

Quick answer index

Exercise	Question	Answer / status
16(A)	1(a)	$PB = PC$
16(A)	1(b)	$P$ is equidistant from $AB$ and $AC$
16(A)	1(c)	$PA = PB$
16(A)	1(d)	$CD$ bisects $\angle ACB$ , under the equal-marked-side figure assumption
16(A)	1(e)	$P$ is equidistant from $AB$ and $BC$
16(A)	2	$QA = QB$
16(A)	3	Perpendicular distances from $P$ to $AC$ and $BC$ are equal
16(A)	4(i)	$X$ is equidistant from $AB$ and $AC$
16(A)	4(ii)	$YA = YC$
16(A)	5	$DB = DC$
16(A)	6	$PQ$ , produced if required, is the perpendicular bisector of $AB$
16(A)	7	$PB = PC$ and the perpendicular distances from $P$ to $AC$ and $BC$ are equal
16(A)	8	$P$ is equidistant from $AB$ and $BC$
16(A)	9(i)	$A$ is equidistant from all three sides of $\triangle LMN$
16(A)	9(ii)	$AM$ bisects $\angle LMN$
16(A)	10	$PB \approx 3.5\text{ cm}$
16(B)	1(a)	Bisector of $\angle ABC$
16(B)	1(b)	Diameter perpendicular to chord $AB$
16(B)	1(c)	Perpendicular bisector of $AB$

Frequently Asked Questions

What is the locus of points equidistant from two fixed points in ICSE Class 10 Maths?

In ICSE Class 10 Maths, the locus of points equidistant from two fixed points is the perpendicular bisector of the line segment joining those two points. For fixed points $A$ and $B$ , any point $P$ on this locus satisfies $PA = PB$ .

How do I decide between a perpendicular bisector and an angle bisector in Selina Chapter 16 Loci?

Use a perpendicular bisector when the condition says equal distance from two fixed points. Use an angle bisector when the condition says equal distance from two intersecting lines or the two arms of an angle.

In loci construction questions, should I write the proof after drawing the figure?

Yes, when the question asks to prove or justify the construction. A correct Selina Chapter 16 Loci answer should name the locus, draw it accurately, mark the required point and then give a short reason using the locus theorem.

Why is

PB

approximately

3.5\text{ cm}

in Exercise 16(A) Question 10?

In that construction, $P$ is the intersection of the angle bisector of $\angle B$ and the perpendicular bisector of $BC$ . With $BC = 6\text{ cm}$ and $\angle ABC = 60^\circ$ , $P$ lies $3\text{ cm}$ horizontally from $B$ and at a height of $\sqrt{3}\text{ cm}$ , so $PB = 2\sqrt{3}\text{ cm} \approx 3.46\text{ cm}$ . On a construction drawing, this is recorded as about $3.5\text{ cm}$ .

Are Loci and Constructions tested as separate ideas in Class 10 Maths?

They are closely linked in Class 10 Maths. A loci question usually starts with a condition such as equal distance from two points or equal distance from two lines, and the construction is the geometric drawing that represents that condition.

Sources referenced

CISCE official Regulations and Syllabuses publications page
Council for the Indian School Certificate Examinations official website
Selina Publishers official website
Selina Concise Mathematics Class 10 textbook, Chapter 16: Loci (Locus and its Constructions)
Standard Euclidean geometry results on perpendicular bisectors, angle bisectors, congruence and circle chords

ICSE Class 10 Maths Loci Solutions: Selina Chapter 16

ICSE Class 10 Maths Loci: direct answer

Concise Mathematics Selina Solutions Class 10 ICSE Chapter 16 Loci

Concept snapshot: read a locus like a rule-made track

Locus rules and construction methods

Rule 1: Perpendicular bisector theorem

Rule 2: Angle bisector theorem for distance from two lines

Rule 3: Locus inside a bounded figure

Exercise 16(A) solutions

Question 1(a): $P$ is on the perpendicular bisector of $BC$

Question 1(b): $P$ is on the bisector of $\angle A$

Question 1(c): Perpendicular bisector of $AB$ and bisector of $\angle A$ meet at $P$

Question 1(d): Decide the correct statement from the figure

Question 1(e): Bisector of $\angle B$ meets $AC$ at $P$

Question 2: $PQ$ is the perpendicular bisector of $AB$

Question 3: $CP$ bisects $\angle C$ of $\triangle ABC$

Question 4: $AX$ bisects $\angle BAC$ and $PQ$ is the perpendicular bisector of $AC$

Question 5: Construct $\triangle ABC$ with $AB = 4.2\text{ cm}$ , $BC = 6.3\text{ cm}$ , $AC = 5\text{ cm}$

Question 6: $PA = PB$ and $QA = QB$

Question 7: Construct triangle with $\angle ABC = 75^\circ$ , $AB = 5\text{ cm}$ , $BC = 6.4\text{ cm}$

Question 8: Locus point in a parallelogram

Question 9: Two angle bisectors meet at $A$ in $\triangle LMN$

Question 10: Ruler and compass construction with two loci

Exercise 16(B) solutions

Question 1(a): Centres of circles tangent to arms $AB$ and $BC$

Question 1(b): Moving point inside a circle equidistant from chord endpoints $A$ and $B$

Question 1(c): $\triangle APB$ is always isosceles with base $AB$

Original worked examples for practice

Worked Example 1: Point equidistant from two fixed points and lying on a given line

Worked Example 2: Centre of a circle touching two sides of an angle

Worked Example 3: Two conditions inside a triangle

Examiner’s mindset for loci constructions

Common mistakes students make

Quick answer index

Frequently Asked Questions

Sources referenced

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ICSE Class 10 Maths Loci: direct answer

Concise Mathematics Selina Solutions Class 10 ICSE Chapter 16 Loci

Concept snapshot: read a locus like a rule-made track

Locus rules and construction methods

Rule 1: Perpendicular bisector theorem

Rule 2: Angle bisector theorem for distance from two lines

Rule 3: Locus inside a bounded figure

Exercise 16(A) solutions

Question 1(a): P is on the perpendicular bisector of BC

Question 1(b): P is on the bisector of \angle A

Question 1(c): Perpendicular bisector of AB and bisector of \angle A meet at P

Question 1(d): Decide the correct statement from the figure

Question 1(e): Bisector of \angle B meets AC at P

Question 2: PQ is the perpendicular bisector of AB

Question 3: CP bisects \angle C of \triangle ABC

Question 4: AX bisects \angle BAC and PQ is the perpendicular bisector of AC

Question 5: Construct \triangle ABC with AB = 4.2\text{ cm}, BC = 6.3\text{ cm}, AC = 5\text{ cm}

Question 6: PA = PB and QA = QB

Question 7: Construct triangle with \angle ABC = 75^\circ, AB = 5\text{ cm}, BC = 6.4\text{ cm}

Question 8: Locus point in a parallelogram

Question 9: Two angle bisectors meet at A in \triangle LMN

Question 10: Ruler and compass construction with two loci

Exercise 16(B) solutions

Question 1(a): Centres of circles tangent to arms AB and BC

Question 1(b): Moving point inside a circle equidistant from chord endpoints A and B

Question 1(c): \triangle APB is always isosceles with base AB

Original worked examples for practice

Worked Example 1: Point equidistant from two fixed points and lying on a given line

Worked Example 2: Centre of a circle touching two sides of an angle

Worked Example 3: Two conditions inside a triangle

Examiner’s mindset for loci constructions

Common mistakes students make

Quick answer index

Frequently Asked Questions

Sources referenced

Continue Studying: Related Chapters & Resources

More from this section

Exploring Robotics & AI Solutions Class 10 ICSE Solved

Exploring Robotics & AI Solutions Class 10 ICSE: Ch 1

Exploring Robotics & AI Solutions Class 10: Chapter 2

Question 1(a): $P$ is on the perpendicular bisector of $BC$

Question 1(b): $P$ is on the bisector of $\angle A$

Question 1(c): Perpendicular bisector of $AB$ and bisector of $\angle A$ meet at $P$

Question 1(e): Bisector of $\angle B$ meets $AC$ at $P$

Question 2: $PQ$ is the perpendicular bisector of $AB$

Question 3: $CP$ bisects $\angle C$ of $\triangle ABC$

Question 4: $AX$ bisects $\angle BAC$ and $PQ$ is the perpendicular bisector of $AC$

Question 5: Construct $\triangle ABC$ with $AB = 4.2\text{ cm}$ , $BC = 6.3\text{ cm}$ , $AC = 5\text{ cm}$

Question 6: $PA = PB$ and $QA = QB$

Question 7: Construct triangle with $\angle ABC = 75^\circ$ , $AB = 5\text{ cm}$ , $BC = 6.4\text{ cm}$

Question 9: Two angle bisectors meet at $A$ in $\triangle LMN$

Question 1(a): Centres of circles tangent to arms $AB$ and $BC$

Question 1(b): Moving point inside a circle equidistant from chord endpoints $A$ and $B$

Question 1(c): $\triangle APB$ is always isosceles with base $AB$