Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 16 Loci Locus And Its Constructions

Exercise 16A

Question 1. Given— PQ is perpendicular bisector of side AB of the triangle ABC. Prove— Q is equidistant from A and B.
Answer:
Construction: Join AQ
Proof: In ∆AQP and ∆BQP,
AP = BP (given)
∠QPA = ∠QPB (Each = 90°)
PQ = PQ (Common)
By Side-Angle-Side criterion of congruence, we have
∆AQP ≅ ∆BQP (SAS postulate)
The corresponding parts of the triangle are congruent
∴ AQ = BQ (CPCT)
Hence Q is equidistant from A and B.
In simple words: When a line cuts another line exactly in the middle at 90 degrees, any point on that line is the same distance from both ends. This is a basic rule of geometry.

📝 Teacher's Note: Draw this on the board with a ruler. Show students that P is the exact middle of AB. Then measure AQ and BQ - they will be equal. This helps students see the proof with their eyes.

🎯 Exam Tip: Always write "CPCT" (Corresponding Parts of Congruent Triangles). This is the key phrase examiners look for. Also mention SAS clearly.

Question 2. Given— CP is bisector of angle C of ∆ABC. Prove— P is equidistant from AC and BC.
Answer:
Construction: From P, draw PL ⊥ AB and PM ⊥ CB
Proof: In ∆LPC and ∆MPC,
∠PLC = ∠PMC (Each = 90°)
∠PCL = ∠MCP (Given)
PC = PC (Common)
∴ By Angle-Side-Angle criterion of congruence,
∆LPC ≅ ∆MPC (ASA Postulate)
The corresponding parts of the congruent triangles are congruent.
∴ PL = PM (CPCT)
Hence, P is equidistant from AC and AB
In simple words: When a line divides an angle into two equal parts, any point on that line is the same distance from both sides of the angle. Think of it like standing in the exact middle of a corner.

📝 Teacher's Note: Use a paper corner to show angle bisector. Fold the corner so both sides match perfectly. The fold line is the angle bisector. Any point on it is equally far from both sides.

🎯 Exam Tip: Write "ASA Postulate" clearly. Draw perpendicular lines from P to both sides. This construction is very important for marks.

Question 3. Given— AX bisects angle BAG and PQ is perpendicular bisector of AC which meets AX at point Y. Prove— (i) X is equidistant from AB and AC. (ii) Y is equidistant from A and C.
Answer:
Construction: From X, draw XL ⊥ AC and XM ⊥ AB. Also join YC.
Proof:
(i) In ∆AXL and ∆AXM,
∠XAL = ∠XAM (Given)
AX = AX (Common)
∠XLA = ∠XMA (Each = 90°)
∴ By Angle-Side-Angle criterion of congruence,
∆AXL ≅ ∆AXM (ASA Postulate)
The corresponding parts of the congruent triangles are congruent.
∴ XL = XM (CPCT)
Hence, X is equidistant from AB and AC.
(ii) In ∆YTA and ∆YTC,
AT = CT (∵ PQ is a perpendicular bisector of AC)
∠YTA = ∠YTC (Each = 90°)
YT = YT (Common)
∴ By Side-Angle-Side criterion of congruence,
∆YTA ≅ ∆YTC (SAS Postulate)
The corresponding parts of the congruent triangles are congruent.
∴ YA = YC (CPCT)
Hence, Y is equidistant from A and C.
In simple words: Part (i) uses angle bisector property - any point on it is equally far from both sides. Part (ii) uses perpendicular bisector property - any point on it is equally far from both ends.

📝 Teacher's Note: This combines two important properties. Draw both constructions clearly. Show students that X follows angle bisector rule and Y follows perpendicular bisector rule.

🎯 Exam Tip: Solve both parts separately. Part (i) needs ASA, part (ii) needs SAS. Don't mix up the postulates. Write "Hence proved" for each part.

Question 4. Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.
Answer:
Given: In triangle ABC, AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm
Steps of Construction:
i) Draw a line segment BC = 6.3 cm
ii) With centre B and radius 4.2 cm, draw an arc.
iii) With centre C and radius 5 cm, draw another arc which intersects the first arc at A.
iv) Join AB and AC.
∆ABC is the required triangle.
v) Again with centre B and C and radius greater than \( \frac{1}{2}BC \) draw arcs which intersects each other at L and M.
vi) Join LM intersecting AC at D and BC at E.
vii) Join DB.
Proof: In ∆DBE and ∆DCE
BE = EC (LM is bisector of BC)
∠DEB = ∠DEC (Each = 90°)
DE = DE (Common)
∴ By Side-Angle-Side criterion of congruence, we have
∆DBE ≅ ∆DCE (SAS Postulate)
The corresponding parts of the congruent triangles are congruent.
∴ DB = DC (CPCT)
Hence, D is equidistant from B and C.
In simple words: We first draw the triangle using given measurements. Then we draw the perpendicular bisector of BC. Any point on this line (like D) is equally far from B and C.

📝 Teacher's Note: Do this construction step by step with compass and ruler. Students must see the actual construction. Show them how perpendicular bisector is drawn using arcs from both ends.

🎯 Exam Tip: Write all construction steps clearly. Use exact measurements given. In proof, clearly state which line is the perpendicular bisector. Construction questions need both steps and proof.

Question 5. In each of the given figures: PA = PH and QA = QB.
Answer:
(i) For the first figure:
Since PA = PB (given) and QA = QB (given), we can conclude:
- P lies on the perpendicular bisector of AB (since PA = PB)
- Q lies on the perpendicular bisector of AB (since QA = QB)
Therefore, PQ is the perpendicular bisector of AB.
(ii) For the second figure:
Given PA = PB and QA = QB, we have:
- P is equidistant from A and B
- Q is equidistant from A and B
This means both P and Q lie on the perpendicular bisector of AB.
Therefore, PQ is the perpendicular bisector of AB, and APBQ forms a kite (quadrilateral with two pairs of equal adjacent sides).
In simple words: When two points are both equally far from the ends of a line, then the line joining those two points cuts the original line exactly in the middle at 90 degrees.

📝 Teacher's Note: Use paper folding to show this. Fold a paper so that two corners meet exactly. The fold line will be equidistant from both corners. This is how perpendicular bisector works.

🎯 Exam Tip: When you see equal distances from two points, immediately think "perpendicular bisector." State clearly that both points lie on the perpendicular bisector of the line segment.

[Diagram: Figure (i) shows triangle PAB with point Q inside, where PA = PB and QA = QB are marked with equal signs. Figure (ii) shows a kite shape APBQ where PA = PB and QA = QB are marked with equal signs.]

Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of points equidistant from two given fixed points.
Answer:
Construction: Join PQ which meets AB in D.

Proof:
P is equidistant from A and B.
\( \therefore \) P lies on the perpendicular bisector of AB.

Similarly, Q is equidistant from A and B.
\( \therefore \) Q lies on perpendicular bisector of AB.
\( \therefore \) P and Q both lie on the perpendicular bisector of AB.
\( \therefore \) PQ is perpendicular bisector of AB.

Hence, locus of the points which are equidistant from two fixed points, is a perpendicular bisector of the line joining the fixed points.
In simple words: When two points are same distance from two fixed points, they make a line. This line is the perpendicular bisector. It cuts the line between the fixed points at 90 degrees and divides it into two equal parts.

[Diagram: Two geometric diagrams showing points P and Q equidistant from points A and B, with PQ forming the perpendicular bisector of AB.]

📝 Teacher's Note: Draw two dots on the board as A and B. Ask students to find all points same distance from both dots. They will see it makes a straight line in the middle.

🎯 Exam Tip: Always write "perpendicular bisector" clearly. Show that both points P and Q lie on it. This proves the whole line PQ is the perpendicular bisector.

Question 6. Construct a right angled triangle PQR, in which \( \angle Q = 90° \), hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meets PR at point T. Prove that T is equidistant from PQ and QR.
Answer:
Steps of Construction:
(i) Draw a line segment QR = 4.5 cm
(ii) At Q, draw a ray QX making an angle of 90°
(iii) With centre R and radius 8 cm, draw an arc which intersects QX at P.
(iv) Join RP.
\( \triangle PQR \) is the required triangle.
(v) Draw the bisector of \( \angle PQR \) which meets PR in T.
(vi) From T, draw perpendicular TL and TM respectively on PQ and QR.

Proof: In \( \triangle TLQ \) and \( \triangle TMQ \)
\( \angle TLQ = \angle TMQ \) (Each = 90°)
\( \angle LQT = \angle TQM \) (QT is angle bisector)
QT = QT (Common)
\( \therefore \) By Angle-Angle-Side criterion of congruence,
\( \triangle TLQ \cong \triangle TMQ \) (AAS Postulate)
The corresponding parts of the congruent triangles are congruent.
\( \therefore \) TL = TM (CPCT)

Hence, T is equidistant from PQ and QR.
In simple words: When we draw the angle bisector, any point on it is same distance from both sides of the angle. We prove this by showing two triangles are exactly the same.

[Diagram: Right triangle PQR with angle bisector QT meeting hypotenuse at T, showing perpendiculars TL and TM to the sides.]

📝 Teacher's Note: Make students understand that angle bisector means the line that cuts an angle into two equal parts. Any point on this line is same distance from both sides.

🎯 Exam Tip: Always write the congruence criteria clearly. Use CPCT (Corresponding Parts of Congruent Triangles) to conclude TL = TM. This shows equal distance.

Question 7. Construct a triangle ABC in which angle ABC = 75°. AB = 5 cm and BC = 6.4 cm. Draw perpendicular bisector of side BC and also the bisector of angle ACB. If these bisectors intersect each other at point P; prove that P is equidistant from B and C; and also from AC and BC.
Answer:
Steps of Construction:
(i) Draw a line segment BC = 6.4 cm
(ii) At B, draw a ray BX making an angle of 75° with BC and cut off BA = 5 cm.
(iii) Join AC.
\( \triangle ABC \) is the required triangle.
(iv) Draw the perpendicular bisector of BC.
(v) Draw the angle bisector of angle ACB which intersects the perpendicular bisector of BC at P.
(vi) Join PB and draw PL \( \perp \) AC.

Proof: In \( \triangle PBQ \) and \( \triangle PCQ \)
PQ = PQ (Common)
\( \angle AQB = \angle PQC \) (Each = 90°)
BQ = QC (PQ is the perpendicular bisector of BC)
\( \therefore \) By Side-Angle-Side criterion of congruence
\( \triangle PBQ \cong \triangle PCQ \) (SAS Postulate)
The corresponding parts of the congruent triangles are congruent.
\( \therefore \) PB = PC (CPCT)

Hence, P is equidistant from B and C.

\( \angle PQC = \angle PLC \) (Each = 90°)
\( \angle PCQ = \angle PCL \) (Given)
PC = PC (common)
Again in \( \triangle PQC \) and \( \triangle PLC \): By Angle-Angle-Side criterion of congruence,
\( \triangle PQC \cong \triangle PLC \) (AAS Postulate)
The corresponding parts of the congruent triangles are congruent.
\( \therefore \) PQ = PL (CPCT)

Hence, P is equidistant from AC and BC.
In simple words: Point P lies on two special lines. One makes it same distance from B and C. The other makes it same distance from the two sides AC and BC.

[Diagram: Triangle ABC with perpendicular bisector of BC and angle bisector of angle C meeting at point P.]

📝 Teacher's Note: Explain that perpendicular bisector makes all points on it same distance from the endpoints. Angle bisector makes all points on it same distance from the two sides.

🎯 Exam Tip: Prove two separate things: P is equidistant from B and C (using perpendicular bisector property), and P is equidistant from AC and BC (using angle bisector property). Use congruence in both proofs.

Question 8. In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC.
Answer:
Construction: From P, draw PL \( \perp \) AB and PM \( \perp \) BC

Proof: In \( \triangle PLB \) and \( \triangle PMB \)
\( \angle PLB = \angle PMB \) (Each = 90°)
\( \angle PBL = \angle PBM \) (Given)
PB = PB (common)
\( \therefore \) By Angle-Angle-Side criterion of congruence,
\( \triangle PLB \cong \triangle PMB \) (AAS Postulate)
The corresponding parts of the congruent triangles are congruent.
\( \therefore \) PL = PM (CPCT)

Hence, P is equidistant from AB and BC.
In simple words: Since PB cuts angle B into two equal parts, point P is same distance from both sides AB and BC. We prove this by showing two right triangles are the same.

[Diagram: Parallelogram ABCD with point P on diagonal AC, showing perpendiculars PL and PM to sides AB and BC respectively.]

📝 Teacher's Note: Remind students that in a parallelogram, opposite sides are parallel and equal. But here we only need the angle bisector property: any point on angle bisector is equidistant from the two sides.

🎯 Exam Tip: Draw perpendiculars from P to both sides AB and BC. Use the angle bisector property with congruent triangles to prove PL = PM.

Question 9. In triangle LMN, bisectors of interior angles at L and N intersect each other at point A. Prove that - (i) point A is equidistant from all the three sides of the triangle. (ii) AM bisects angle LMN.
Answer:
Construction: Join AM
Proof:
\( \because \) A lies on bisector of \( \angle N \)
\( \therefore \) A is equidistant from MN and LN.

Again, A lies on bisector of \( \angle L \)
\( \therefore \) A is equidistant from LN and LM.

Hence, A is equidistant from all sides of the triangle LMN.
\( \therefore \) A lies on the bisector of \( \angle M \)

(i) Point A is equidistant from all the three sides of the triangle.
(ii) AM bisects angle LMN.
In simple words: When two angle bisectors meet at a point, that point is same distance from all three sides. This means the third line from that point also bisects the third angle.

[Diagram: Triangle LMN with angle bisectors from L and N meeting at point A, with AM drawn to complete the third angle bisector.]

📝 Teacher's Note: This is about the incenter of a triangle. All three angle bisectors meet at one point inside the triangle. This point is same distance from all three sides.

🎯 Exam Tip: Use the property that any point on an angle bisector is equidistant from the two sides. Since A is equidistant from all three sides, it must lie on all three angle bisectors.

Question 10. Use ruler and compasses only for this question: (i) construct \( \triangle ABC \), where AB = 3.5 cm, BC = 6 cm and \( \angle ABC = 60° \). (ii) Construct the locus of points inside the triangle which are equidistant from BA and BC. (iii) Construct the locus of points inside the triangle which are equidistant from B and C. (iv) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.
Answer:
Construction Steps:
(i) Draw line segment BC = 6 cm. At B, draw angle of 60° and mark AB = 3.5 cm. Join AC. Triangle ABC is formed.

(ii) Draw the angle bisector of \( \angle ABC \). This is the locus of points equidistant from BA and BC.

(iii) Draw the perpendicular bisector of BC. This is the locus of points equidistant from B and C.

(iv) Mark point P where the angle bisector and perpendicular bisector intersect. Measure PB with ruler.

Note: The actual measurement of PB will depend on accurate construction. Typically PB ≈ 3.0 cm to 3.5 cm for this triangle.
In simple words: We draw two special lines inside the triangle. One makes points same distance from two sides. The other makes points same distance from two vertices. Where they cross is our special point P.

📝 Teacher's Note: Show students how to use compass to bisect an angle and construct perpendicular bisector. The intersection point P has both properties we need.

🎯 Exam Tip: Use only ruler and compass as stated. Draw construction lines lightly and final answer lines darkly. Measure PB carefully and write the measurement clearly.

Solution:
Answer: Steps of construction:
(i) Draw line BC = 6 cm and an angle CBX = 60°. Cut off AB = 3.5. Join AC, triangle ABC is the required triangle.
(ii) Draw perpendicular bisector of BC and bisector of angle B
(iii) Bisector of angle B meets bisector of BC at P.
\( \implies \) BP is the required length, where, PB = 3.5 cm
(iv) P is the point which is equidistant from BA and BC, also equidistant from B and C.
PB=3.6 cm

[Diagram: This diagram shows triangle ABC with BC = 6 cm, angle CBX = 60°, AB = 3.5 cm. Point P is found by drawing perpendicular bisector of BC and angle bisector of B, which meet at P. Distance PB = 3.6 cm is marked.]

📝 Teacher's Note: Show students how to use compass and ruler to draw perpendicular bisectors and angle bisectors. P is special because it is the same distance from three things at once.

🎯 Exam Tip: Always show construction steps clearly. Mark all measurements. Write "equidistant" to get full marks.

Question 11. The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point E. Prove that: (i) F is equidistant from A and B. (ii) F is equidistant from AB and AC.
Answer:

[Diagram: This diagram shows triangle ABC with AD as angle bisector of angle BAC, and EG as perpendicular bisector of AB intersecting AD at point E.]

(i) F is equidistant from A and B.
Construction: Join FB and FC
Proof: In \( \triangle AFE \) and \( \triangle BFE \),
AE = EB (E is the mid-point of AB)
\( \angle FEA = \angle FEB \) (Each = 90°)
FE = FE (Common)
\( \therefore \) By Side-Angle-Side criterion of congruence,
\( \triangle AFE \cong \triangle BFE \) (SAS Postulate)
The corresponding parts of the congruent triangles are congruent.
\( \therefore \) AF = FB (CPCT)
Hence, F is equidistant from A and B.

(ii) F is equidistant from AB and AC.
Construction: Draw LF ⊥ AC
Proof: In \( \triangle AFL \) and \( \triangle AFE \),
\( \angle FEA = \angle FLA \) (Each = 90°)
\( \angle LAF = FAE \) (AD bisects \( \angle BAC \))
AF = AF (Common)
\( \therefore \) By Angle-Angle-Side criterion of congruence,
\( \triangle AFL \cong \triangle AFE \) (AAS Postulate)
The corresponding parts of the congruent triangles are congruent.
\( \therefore \) FE = FL (CPCT)
Hence, F is equidistant from AB and AC.

In simple words: We prove F is the same distance from A and B by showing two triangles are exactly the same shape. Then we prove F is the same distance from two lines by showing another pair of triangles are the same.

📝 Teacher's Note: Draw the triangles separately to help students see which sides and angles are equal. Use different colors for equal parts.

🎯 Exam Tip: Always write "By SAS" or "By AAS" clearly. Write "CPCT" after proving triangles are congruent. This gets you method marks.

Question 12. The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect each other at point P. Show that P is equidistant from the opposite sides AB and CD.
Answer:

[Diagram: This diagram shows quadrilateral ABCD with angle bisectors of angles B and C meeting at point P inside the quadrilateral.]

Since P lies on the bisector of angle B,
therefore, P is equidistant from AB and BC .... (1)
Similarly, P lies on the bisector of angle C,
therefore, P is equidistant from BC and CD .... (2)
From (1) and (2),
Hence, P is equidistant from AB and CD.

In simple words: Point P sits on two angle bisectors. Each bisector makes P the same distance from two sides. When we combine this, P ends up being the same distance from opposite sides AB and CD.

📝 Teacher's Note: Explain that angle bisector property means equal perpendicular distances to the two sides. Use the idea of a person standing exactly between two walls.

🎯 Exam Tip: Write the angle bisector property clearly: "Point on angle bisector is equidistant from the two sides." Number your steps (1) and (2) to show the logic.

Question 13. Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
Answer:

[Diagram: This diagram shows line segment AB = 6 cm with its perpendicular bisector LM drawn vertically through the midpoint. Point P is shown on the bisector with PA = PB marked.]

Steps of construction:
(i) Draw a line segment AB of 6 cm.
(ii) Draw perpendicular bisector LM of AB. LM is the required locus.
(iii) Take any point on LM say P.
(iv) Join PA and PB.
Since, P lies on the right bisector of line AB.
Therefore, P is equidistant from A and B.
i.e. PA = PB
Hence, Perpendicular bisector of AB is the locus of all points which are equidistant from A and B.

In simple words: The locus is a straight line that cuts AB exactly in the middle at 90°. Any point on this line is exactly the same distance from A and B. This line goes up and down forever.

📝 Teacher's Note: Use compass to show students that any point on the perpendicular bisector really is the same distance from A and B. Let them measure with ruler to check.

🎯 Exam Tip: Draw the perpendicular bisector neatly using compass and ruler. Write "perpendicular bisector" as your final answer. Show one point P with equal distances PA = PB.

Question 14. Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
Answer:

[Diagram: This diagram shows angle ABC = 75° with ray BA and ray BC. The angle bisector BP is drawn, and point D is shown on BP with perpendicular distances DE to AB and DF to BC marked equal.]

Steps of Construction:
i) Draw a ray BC.
ii) Construct a ray RA making an angle of 75° with BC. Therefore, ABC= 75°.
iii) Draw the angle bisector BP of ∠ABC.
BP is the required locus.
iv) Take any point D on BP.
v) From D, draw DE ⊥ AB and DF ⊥ BC
Since D lies on the angle bisector BP of ∠ABC
D is equidistant from AB and BC.
Hence, DE = DF
Similarly, any point on BP is equidistant from AB and BC.
Therefore, BP is the locus of all points which are equidistant from AB and BC.

In simple words: The locus is the angle bisector ray. This ray cuts the angle exactly in half. Any point on this ray is exactly the same perpendicular distance from both sides AB and BC.

📝 Teacher's Note: Show students how to construct 75° angle using compass. Then show how angle bisector divides any angle into two equal parts. Use perpendicular distances, not slant distances.

🎯 Exam Tip: Draw the angle bisector accurately using compass method. Mark perpendicular distances from a point to both sides. Write "angle bisector" as your final answer.

Question 15. Draw an angle ABC = 60°, having AB = 4.6 cm and BC = 5 cm. Find a point P equidistant from AB and BC; and also equidistant from A and B.
Answer:

Steps of construction:
(i) Draw ray BC = 5 cm.
(ii) At B, construct angle ABC = 60°.
(iii) Cut off AB = 4.6 cm on the ray BA.
(iv) Draw the angle bisector of ∠ABC.
(v) Draw the perpendicular bisector of AB.
(vi) Let the angle bisector and perpendicular bisector meet at point P.
P is the required point.
P is equidistant from AB and BC (because P lies on angle bisector).
P is equidistant from A and B (because P lies on perpendicular bisector of AB).

In simple words: We need a point that is the same distance from two lines AND the same distance from two points. We draw two locus lines and find where they cross. That crossing point P has both properties we want.

📝 Teacher's Note: Explain that P must satisfy two conditions at the same time. So we draw two different locus lines and find their intersection point.

🎯 Exam Tip: Draw both the angle bisector and perpendicular bisector clearly. Mark their intersection as P. State both properties of P in your answer to get full marks.

Steps of Construction:

(i) Draw a line segment BC = 5 cm

(ii) At B, draw a ray BX making an angle of 60° and cut off BA = 4.6 cm.

(iii) Draw the angle bisector of ∠ABC.

(iv) Draw the perpendicular bisector of AB which intersects the angle bisector at P.

P is the required point which is equidistant from AB and BC, as well as from A and B.

[Diagram: This diagram shows triangle ABC with BC = 5 cm, angle ABC = 60°, and BA = 4.6 cm. Point P is shown where the angle bisector of angle ABC meets the perpendicular bisector of AB.]

Question 16. In the figure given below, find a point P on CD equidistant from points A and B.
Answer:

Steps of Construction:

(i) AB and CD are the two lines given.

(ii) Draw a perpendicular bisector of line AB which intersects CD in P.

P is the required point which is equidistant from A and B.
Since P lies on perpendicular bisector of AB; PA = PB.

[Diagram: This diagram shows two lines AB and CD, with the perpendicular bisector of AB intersecting CD at point P.]

In simple words: When we want to find a point on line CD that is the same distance from A and B, we draw the perpendicular bisector of AB. Where it cuts CD is our answer point P.

📝 Teacher's Note: Show students that any point on the perpendicular bisector of a line segment is always the same distance from both ends. Use a ruler to measure and prove this.

🎯 Exam Tip: Always write "perpendicular bisector of AB" clearly. Draw it with a compass to show equal distances. This gets full marks.

Question 17. In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
Answer:

Steps of Construction:

(i) In the given triangle, draw the angle bisector of ∠BAC.

(ii) Draw the perpendicular bisector of BC which intersects the angle bisector at P.

P is the required point which is equidistant from AB and AC as well as from B and C.
Since P lies on angle bisector of ∠BAC,
It is equidistant from AB and AC.
Again, P lies on perpendicular bisector of BC,
Therefore, it is equidistant from B and C.

[Diagram: This diagram shows triangle ABC with the angle bisector of angle BAC and the perpendicular bisector of BC intersecting at point P.]

In simple words: We need a point that is equally far from two sides (AB and AC) and also equally far from two corners (B and C). The angle bisector helps with sides, the perpendicular bisector helps with corners.

📝 Teacher's Note: Explain that angle bisectors make equal distances from sides, while perpendicular bisectors make equal distances from endpoints. The meeting point satisfies both conditions.

🎯 Exam Tip: Draw both bisectors clearly and mark their intersection as P. Write "equidistant from AB and AC" and "equidistant from B and C" separately for full marks.

Question 18. Construct a triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ABC = 60°. Locate by construction the point P such that:
(i) P is equidistant from B and C.
(ii) P is equidistant from AB and BC.
(iii) Measure and record the length of PB.
Answer:

Steps of Construction:

1) Draw a line segment AB = 7 cm.

2) Draw angle ∠ABC = 60° with the help of compass.

3) Cut off BC = 8 cm.

4) Join A and C.

5) The triangle ABC so formed is the required triangle.

(i) Draw the perpendicular bisector of BC. The point situated on this line will be equidistant from B and C.

(ii) Draw the angle bisector of ∠ABC. Any point situated on this angular bisector is equidistant from lines AB and BC.

The point which fulfills the condition required in (i) and (ii) is the intersection point of bisector of line BC and angular bisector of ∠ABC.

P is the required point which is equidistant from AB and AC as well as from B and C.
On measuring the length of line segment PB, it is equal to 4.5 cm.

[Diagram: This diagram shows triangle ABC with AB = 7 cm, BC = 8 cm, angle ABC = 60°, with the perpendicular bisector of BC and angle bisector of angle ABC meeting at point P.]

In simple words: We first make the triangle with given measurements. Then we draw two special lines - one for equal distance from B and C, another for equal distance from sides AB and BC. Where they meet is point P.

📝 Teacher's Note: Show students how to use a compass to draw the 60° angle accurately. Emphasize that P must satisfy both conditions at the same time.

🎯 Exam Tip: Construct the triangle first, then draw both bisectors. Measure PB carefully with a ruler. Write the measurement clearly - this gets marks too.

Question 19. On a graph paper, draw lines x = 3 and y = -5. Now, on the same graph paper, draw the locus of the point which is equidistant from the given lines.
Answer:

On the graph, draw axis XOX' and YOY'
Draw a line l, x = 3 which is parallel to y-axis
And draw another line m, y = -5, which is parallel to x-axis
These two lines intersect each other at P.
Now draw the angle bisector p of angle P.
Since p is the angle bisector of P, any point on P is equidistant from l and m.
Therefore, this line p is equidistant from l and m.

[Diagram: This diagram shows a coordinate plane with vertical line x = 3 and horizontal line y = -5 intersecting at point (3, -5), with their angle bisector drawn as a diagonal line through this point.]

In simple words: We draw two straight lines - one going up-down (x = 3) and one going left-right (y = -5). The locus is a diagonal line that cuts the angle between them exactly in half.

📝 Teacher's Note: Use graph paper to show how x = 3 is a vertical line and y = -5 is a horizontal line. The angle bisector will be at 45° to both lines.

🎯 Exam Tip: Draw the coordinate axes first, then the two given lines clearly. The locus is the angle bisector - draw it as a straight diagonal line through their intersection point.

Question 20. On a graph paper, draw the line x = 6. Now, on the same graph paper, draw the locus of the point which moves in such a way that its distance from the given line is always equal to 3 units.
Answer:

On the graph, draw axis XOX' and YOY'
Draw a line l, x = 6 which is parallel to y-axis
Take points P and Q which are at a distance of 3 units from the line l.
Draw lines m and n from P and Q parallel to l
With locus = 3, two lines can be drawn x = 3 and x = 9.

[Diagram: This diagram shows a coordinate plane with a vertical line x = 6, and two parallel vertical lines at x = 3 and x = 9, each 3 units away from the original line.]

In simple words: We have one vertical line at x = 6. The locus is two more vertical lines - one 3 steps to the left (x = 3) and one 3 steps to the right (x = 9).

📝 Teacher's Note: Show students that distance from a vertical line means counting horizontal steps. 3 units left gives x = 3, and 3 units right gives x = 9.

🎯 Exam Tip: Draw x = 6 first. Then draw two parallel lines at x = 3 and x = 9. Write the equations clearly - both lines are part of the locus.

Exercise 16B

Question 1. Describe the locus of a point at a distance of 3 cm from a fixed point.
Answer:

The locus of a point at a distance of 3 cm from a fixed point is a circle with the fixed point as center and radius = 3 cm.

[Diagram: This diagram shows a circle with center O and radius 3 cm, with point P on the circumference.]

In simple words: When a point stays exactly 3 cm away from a fixed center point, it traces out a perfect circle. Think of it like a dog walking around a pole with a 3 cm rope.

📝 Teacher's Note: Use a compass to show students how all points on a circle are the same distance from the center. This is the most basic locus example.

🎯 Exam Tip: Always write "circle with center at the fixed point and radius 3 cm." Draw a neat circle with center marked clearly. This gets full marks.

Question 2. Describe the locus of a point at a distance of 2 cm from a fixed line.
Answer: The locus of a point at a distance of 2 cm from a fixed line AB is a pair of straight lines l and m which are parallel to the given line at a distance of 2 cm.

[Diagram: Shows a horizontal line AB with two parallel lines above and below it, each 2 cm away from the original line]

In simple words: When you want all points that are exactly 2 cm away from a line, you get two new lines. One is 2 cm above the original line and one is 2 cm below it.

📝 Teacher's Note: Draw a straight line on the board. Then draw two more lines parallel to it - one above and one below. Show students that every point on these new lines is exactly the same distance from the middle line.

🎯 Exam Tip: Always write "pair of straight lines" and "parallel to the given line". Don't forget to mention the distance. These are key words for full marks.

Question 3. Describe the locus of the centre of a wheel of a bicycle going straight along a level road.
Answer: The locus of the centre of a wheel, which is going straight along a level road will be a straight line parallel to the road at a distance equal to the radius of the wheel.

[Diagram: Shows three circles representing a wheel at different positions, with their centers forming a straight line parallel to the road surface]

In simple words: As the bicycle wheel rolls on the road, the center of the wheel moves in a straight line. This line is always the same height above the road - exactly one radius distance.

📝 Teacher's Note: Show students a coin rolling on the table. The center of the coin traces a straight path parallel to the table surface. This makes the concept very clear.

🎯 Exam Tip: Write "straight line parallel to the road" and mention "at a distance equal to radius". Both parts are needed for full marks.

Question 4. Describe the locus of the moving end of the minute hand of a clock.
Answer: The locus of the moving end of the minute hand of the clock will be a circle where radius will be the length of the minute hand.

[Diagram: Shows a clock face with the minute hand and a circle traced by the tip of the minute hand]

In simple words: The tip of the minute hand goes around in a perfect circle. The size of this circle depends on how long the minute hand is.

📝 Teacher's Note: Use a real clock or draw one on the board. Point to the minute hand and trace its path with your finger. Students can see the circular motion clearly.

🎯 Exam Tip: Always write "circle" and "radius equals length of minute hand". These are the exact words examiners look for.

Question 5. Describe the locus of a stone dropped from the top of a tower.
Answer: The locus of a stone which is dropped from the top of a tower will be a vertical line through the point from which the stone is dropped.

[Diagram: Shows a vertical line from the top of a tower straight down to ground level]

In simple words: When you drop a stone, it falls straight down. The path it takes is a straight vertical line from where you dropped it to the ground.

📝 Teacher's Note: Drop a chalk or eraser in class. Students can see it falls in a straight line downward. This is the simplest example of locus.

🎯 Exam Tip: Write "vertical line" clearly. Don't write "straight line" - it must be "vertical" to get full marks.

Question 6. Describe the locus of a runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge.
Answer: The locus of the runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge will be the circumference of a circle whose radius is equal to the radius of the inner circular track plus 1.5 m.

[Diagram: Shows two concentric circles - the inner track and the outer path where the runner moves, with the gap between them marked as 1.5 m]

In simple words: The runner moves in a bigger circle around the track. This bigger circle is 1.5 meters away from the inner edge all around.

📝 Teacher's Note: Draw two circles on the board - one inside the other. Show that the outer circle is always the same distance from the inner one. Like lanes on a race track.

🎯 Exam Tip: Write "circumference of a circle" and "radius = inner radius + 1.5 m". The addition part is very important for marks.

Question 7. Describe the locus of the door handle as the door opens.
Answer: The locus of the door handle will be the circumference of a circle with centre at the axis of rotation of the door and radius equal to the distance between the door handle and the axis of rotation of the door.

[Diagram: Shows a door handle tracing a curved path as the door opens, with the hinge as the center point]

In simple words: As the door opens and closes, the handle moves in a curved path. This curved path is part of a circle centered at the door hinge.

📝 Teacher's Note: Open and close the classroom door slowly. Ask students to watch the door handle's movement. They can see it makes a curved path around the hinges.

🎯 Exam Tip: Write "circumference of a circle" and mention "centre at axis of rotation". Also state the radius clearly for full marks.

Question 8. Describe the locus of a point inside a circle and equidistant from two fixed points on the circumference of the circle.
Answer: The locus of the points inside the circle which are equidistant from the fixed points on the circumference of a circle will be the diameter which is perpendicular bisector of the line joining the two fixed points on the circle.

[Diagram: Shows a circle with two fixed points on the circumference and a diameter line that bisects the line segment between these two points at right angles]

In simple words: All points that are the same distance from two fixed points lie on a straight line. This line cuts the line between the two fixed points exactly in half at a right angle.

📝 Teacher's Note: Take two points on a circle. Show students that the perpendicular bisector is the line where any point is equidistant from both fixed points. Use a compass to demonstrate.

🎯 Exam Tip: Write "diameter" and "perpendicular bisector" clearly. Both words must be there. Also mention it bisects the line joining the two fixed points.

Question 9. Describe the locus of the centers of all circles passing through two fixed points.
Answer: The locus of the centre of all the circles which pass through two fixed points will be the perpendicular bisector of the line segment joining the two given fixed points.

[Diagram: Shows multiple circles of different sizes all passing through two fixed points, with their centers lying on the perpendicular bisector of the line segment connecting the two points]

In simple words: If you draw many circles that all pass through the same two points, the centers of all these circles will lie on one straight line. This line cuts the line between the two points in half at a right angle.

📝 Teacher's Note: Draw two fixed points. Then draw several circles passing through both points with different radii. Show students that all centers lie on the perpendicular bisector.

🎯 Exam Tip: Write "perpendicular bisector" and "line segment joining the two fixed points". Don't forget to mention both fixed points for complete answer.

Question 10. Describe the locus of vertices of all isosceles triangles having a common base.
Answer: The locus of vertices of all isosceles triangles having a common base will be the perpendicular bisector of the common base of the triangles.

[Diagram: Shows multiple isosceles triangles with the same base but different heights, with all their vertices lying on the perpendicular bisector of the base]

In simple words: All isosceles triangles with the same base have their top vertex somewhere on a straight line. This line stands straight up from the middle of the base.

📝 Teacher's Note: Draw a base line. Then draw several isosceles triangles with this same base but different heights. Show that all top vertices lie on the perpendicular bisector.

🎯 Exam Tip: Write "perpendicular bisector of the common base". The word "common" is important because all triangles share the same base.

Question 11. Describe the locus of a point in space which is always at a distance of 4 cm from a fixed point.
Answer: The locus of a point in space is the surface of the sphere whose centre is the fixed point and radius equal to 4 cm.
In simple words: In 3D space, all points that are exactly 4 cm away from one fixed point form the surface of a ball (sphere) with radius 4 cm.

📝 Teacher's Note: Hold a ball and explain that every point on its surface is the same distance from the center. This is different from a circle which is flat - a sphere is 3D.

🎯 Exam Tip: Write "surface of sphere" not just "sphere". Also mention "radius = 4 cm" and "centre is the fixed point" for full marks.

Question 12. Describe the locus of a point P so that: AB² = AP² + BP², where A and B are two fixed points.
Answer: The locus of the point P is the circumference of a circle with AB as diameter and satisfies the condition AB² = AP² + BP².

[Diagram: Shows a circle with AB as diameter and point P on the circumference, forming a right triangle]

In simple words: This condition means angle APB is a right angle (90 degrees). All points P that make a right angle with A and B lie on a circle with AB as diameter.

📝 Teacher's Note: This is Pythagoras theorem in reverse. The condition AB² = AP² + BP² means triangle APB has a right angle at P. Show students this creates a semicircle.

🎯 Exam Tip: Write "circumference of circle" and "AB as diameter". Also mention the condition is satisfied. This shows you understand the geometry.

Question 13. Describe the locus of a point in rhombus ABCD, so that it is equidistant from
(i) AB and BC
(ii) B and D.
Answer:
(i) The locus of points equidistant from sides AB and BC is the angle bisector of angle ABC.
(ii) The locus of points equidistant from vertices B and D is the perpendicular bisector of diagonal BD.
In simple words: (i) Points the same distance from two sides of an angle lie on the line that cuts the angle in half. (ii) Points the same distance from two vertices lie on the line that cuts the diagonal between them in half.

📝 Teacher's Note: For part (i), show that angle bisector divides the angle equally. For part (ii), show that perpendicular bisector cuts BD in half at right angles.

🎯 Exam Tip: For (i) write "angle bisector of angle ABC". For (ii) write "perpendicular bisector of diagonal BD". These are the exact geometric terms needed.

Question 13. The locus of the point in a rhombus ABCD which is equidistant from AB and BC will be the diagonal BD.

[Diagram: This diagram shows a rhombus ABCD with its diagonals BD drawn as a dashed line.]

Answer: The locus of the point in a rhombus ABCD which is equidistant from AB and BC will be the diagonal BD.

ii)
The locus of the point in a rhombus ABCD which is equidistant from B and D will be the diagonal AC.
In simple words: In a rhombus, if you want points that are the same distance from two sides, you get the diagonal. If you want points that are the same distance from two opposite corners, you get the other diagonal.

📝 Teacher's Note: Draw a rhombus on the board and show students how the diagonal cuts the angle in half. This makes all points on the diagonal equally far from the two sides.

🎯 Exam Tip: Always write "diagonal BD" or "diagonal AC" clearly. Mention which diagonal for which condition. This shows you understand the concept.

Question 14. The speed of sound is 332 meters per second. A gun is fired. Describe the locus of all the people on the Earth's surface, who hear the sound exactly one second later.
Answer: The locus of all the people on Earth's surface is the circumference of a circle whose radius is 332 m and centre is the point where the gun is fired.
In simple words: Sound travels at 332 m/s. In 1 second, it covers 332 meters. So all people who hear it after 1 second are exactly 332 meters away from the gun. This makes a circle.

📝 Teacher's Note: Drop a stone in water and show the circular ripples. Sound spreads the same way in all directions. After 1 second, it reaches all points that are 332 meters away.

🎯 Exam Tip: Write "circumference of a circle" and give the radius (332 m) and center (gun position). These three parts get you full marks.

Question 15. Describe:
(i) The locus of points at distances less than 3 cm from a given point.
(ii) The locus of points at distances greater than 4 cm from a given point.
(iii) The locus of points at distances less than or equal to 2.5 cm from a given point.
(iv) The locus of points at distances greater than or equal to 35 mm from a given point.
(v) The locus of the centre of a given circle which rolls around the outside of a second circle and is always touching it.
(vi) The locus of the centers of all circles that are tangent to both the arms of a given angle.
(vii) The locus of the mid-points of all chords parallel to a given chord of a circle.
(viii) The locus of points within a circle that are equidistant from the end points of a given chord.
Answer:
(i) The locus is the space inside of the circle whose radius is 3 cm and the centre is the fixed point which is given.
(ii) The locus is the space outside of the circle whose radius is 4 cm and centre is the fixed point which is given.
(iii) The locus is the space inside and circumference of the circle with a radius of 2.5 cm and the centre is the given fixed point.
(iv) The locus is the space outside and circumference of the circle with a radius of 35 mm and the centre is the given fixed point.
(v) The locus is the circumference of the circle concentric with the second circle whose radius is equal to the sum of the radii of the two given circles.
(vi) The locus of the centre of all circles whose tangents are the arms of a given angle is the bisector of that angle.
(vii) The locus of the mid-points of the chords which are parallel to a given chords is the diameter perpendicular to the given chords.
(viii) The locus of the points within a circle which are equidistant from the end points of a given chord is the diameter which is perpendicular bisector of the given chord.
In simple words: For distances less than a number, you get inside a circle. For distances more than a number, you get outside a circle. For rolling circles, the center traces a bigger circle.

[Diagram: This diagram shows a large circle with a smaller circle rolling around its outside, illustrating the locus concept for part (v).]

📝 Teacher's Note: Use a coin rolling around a plate to show part (v). The center of the coin traces a circle bigger than the plate. Students see this easily.

🎯 Exam Tip: For "less than" write "inside circle". For "greater than" write "outside circle". For "equal to" add "and circumference". Be very clear about these words.

Question 16. Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude.
Answer: Draw a line XY parallel to the base BC from the vertex A. This line is the locus of vertex A of all the triangles which have the base BC and length of altitude equal to AD.
In simple words: If the base BC is fixed and the height is fixed, then the top point A can be anywhere on a line parallel to BC. This line is at the same height above BC.

[Diagram: This diagram shows a triangle ABC with the vertex A and multiple positions A1, A2 for the vertex, all lying on a line XY parallel to base BC.]

📝 Teacher's Note: Draw several triangles with the same base on the board. Show students that all the top vertices lie on the same horizontal line parallel to the base.

🎯 Exam Tip: Always draw the line parallel to the base. Write "line parallel to base BC at distance equal to altitude". This gets full marks.

Question 17. In the given figure, obtain all the points equidistant from lines m and n ; and 2.5 cm from O.
Answer: Draw an angle bisector PQ and XY of angles formed by the lines m and n. From O, draw arcs with radius 2.5 cm, which intersect the angle bisectors at a, b, c and d respectively. Hence, a, b, c and d are the required four points.
In simple words: Points equidistant from two lines lie on the angle bisector. Points 2.5 cm from O lie on a circle around O. Where these meet, we get our answer points.

[Diagram: This diagram shows two intersecting lines m and n with their angle bisectors PQ and XY, point O at the intersection, and a circle of radius 2.5 cm centered at O, with intersection points a, b, c, and d marked.]

📝 Teacher's Note: Show students that angle bisectors are like the center line between two walls. Any point on this line is equally far from both walls.

🎯 Exam Tip: Draw both angle bisectors (there are two). Draw the circle around O. Mark all four intersection points clearly. Label them a, b, c, d.

Question PQ. By actual drawing obtain the points equidistant from lines m and n and 6 cm from the point P, where P is 2 cm above m, m is parallel to n and m is 6 cm above n.
Answer:
Steps of construction:
(i) Draw a line n.
(ii) Take a point L on n and draw a perpendicular to n.
(iii) Cut off LM = 6 cm and draw a line q, the perpendicular bisector of LM.
(iv) At M, draw a line m making an angle of 90°.
(v) Produce LM and mark a point P such that PM = 2 cm.
(vi) From P, draw an arc with 6 cm radius which intersects the line q, the perpendicular bisector of LM, at A and B.

A and B are the required points which are equidistant from m and n and are at a distance of 6 cm from P.
In simple words: We draw two parallel lines 6 cm apart. The middle line between them has all points equally far from both lines. Then we find points on this middle line that are exactly 6 cm from P.

[Diagram: This diagram shows the construction with parallel lines m and n, point P above line m, the perpendicular bisector q, and the intersection points A and B marked.]

📝 Teacher's Note: Explain that parallel lines have a middle line (perpendicular bisector) where all points are equally far from both lines. This is like standing exactly in the middle of a road.

🎯 Exam Tip: Follow the construction steps exactly. Draw neatly. Mark points A and B clearly where the arc meets the perpendicular bisector.

Question 18. A straight line AB is 8 cm long. Draw and describe the locus of a point which is:
(i) always 4 cm from the line AB
(ii) equidistant from A and B.
Mark the two points X and Y, which are 4 cm from AB and equidistant from A and B.
Answer:
(i) The locus of points always 4 cm from line AB is two parallel lines, one on each side of AB, at a distance of 4 cm from AB.
(ii) The locus of points equidistant from A and B is the perpendicular bisector of AB.

Points X and Y are where the perpendicular bisector of AB intersects the two parallel lines that are 4 cm away from AB.
In simple words: Points 4 cm from a line make two parallel lines (one above, one below). Points equally far from A and B make a line through the middle of AB. Where these meet, we get points X and Y.

📝 Teacher's Note: Draw line AB on the board. Use a ruler to show all points 4 cm above AB make a line, and all points 4 cm below AB make another line. The middle line cuts both.

🎯 Exam Tip: Draw two parallel lines 4 cm above and below AB. Draw the perpendicular bisector through the middle of AB. Mark intersection points as X and Y clearly.

Describe the figure AXBY.
Answer:
Construction Steps:
(i) Draw a line segment AB = 8 cm.
(ii) Draw two parallel lines l and m to AB at a distance of 4 cm.
(iii) Draw the perpendicular bisector of AB which intersects the parallel lines l and m at X and Y respectively then, X and Y are the required points.
(iv) Join AX, AY, BX and BY.

The figure AXBY is a square as its diagonals are equal and intersect at 90°.
In simple words: We make a square by placing points X and Y exactly above and below the middle of line AB. All sides are equal and corners are 90°.

[Diagram: This diagram shows a square AXBY with diagonal AB horizontal and diagonal XY vertical, intersecting at the center O. Lines l and m are drawn parallel to AB at 4 cm distance above and below.]

📝 Teacher's Note: Show students how the perpendicular bisector finds the exact middle. When we place points at equal distance from this middle, we get a perfect square.

🎯 Exam Tip: Always mention that diagonals are equal and meet at 90° to prove it is a square. Draw neat parallel lines for full marks.

Question 19. Angle ABC = 60° and BA = BC = 8 cm. The mid-points of BA and BC are M and N respectively. Draw and describe the locus of a point which is:
(i) equidistant from BA and BC.
(ii) 4 cm from M.
(iii) 4 cm from N.
Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP, and describe the figure BMPN.
Answer:
Construction Steps:
(i) Draw an angle of 60° with AB = BC = 8 cm
(ii) Draw the angle bisector BX of ∠ABC
(iii) With centre M and N, draw circles of radius equal to 4 cm, which intersects each other at P. P is the required point.
(iv) Join MP, NP

BMPN is a rhombus since MP = BM = NB = NP = 4 cm
In simple words: We find point P by drawing two circles from M and N. Where they meet is our point P. The shape BMPN has all sides equal, so it is a rhombus.

[Diagram: This diagram shows angle ABC = 60° with equal sides BA = BC = 8 cm, angle bisector BX, and circles from midpoints M and N intersecting at point P.]

📝 Teacher's Note: Remind students that angle bisector gives equidistant points. Two circles of same radius from two points will meet at two places - choose the right one.

🎯 Exam Tip: Write "rhombus" clearly and explain all sides are equal (4 cm each). Show the angle bisector construction step clearly.

Question 20. Draw a triangle ABC in which AB = 6 cm, BC = 4.5 cm and AC = 5 cm. Draw and label:
(i) the locus of the centers of all circles which touch AB and AC.
(ii) the locus of the centers of all circles of radius 2 cm which touch AB.
Hence, construct the circle of radius 2 cm which touches AB and AC.
Answer:
Construction Steps:
(i) Draw a line segment BC = 4.5 cm
(ii) With B as centre and radius 6 cm and C as centre and radius 5 cm, draw arcs which intersect each other at A.
(iii) Join AB and AC.
ABC is the required triangle.
(iv) Draw the angle bisector of ∠BAC
(v) Draw lines parallel to AB and AC at a distance of 2 cm, which intersect each other and AD at O.
(vi) With centre O and radius 2 cm, draw a circle which touches AB and AC.
In simple words: First we make the triangle using given measurements. Then we draw the angle bisector and parallel lines to find where our circle center should be.

[Diagram: This diagram shows triangle ABC with the angle bisector of angle A, parallel lines at 2 cm distance from AB and AC, and the inscribed circle of radius 2 cm touching both sides.]

📝 Teacher's Note: Show students that angle bisector contains all centers of circles touching both sides. Parallel lines at fixed distance contain centers of circles with fixed radius.

🎯 Exam Tip: Draw neat angle bisector and clearly mark the 2 cm distance for parallel lines. Circle must touch both lines exactly.

Question 21. Construct a triangle ABC, having given AB = 4.8 cm. AC = 4 cm and ∠A = 75°. Find a point P.
(i) inside the triangle ABC.
(ii) outside the triangle ABC.
Equidistant from B and C; and at a distance of 1.2 cm from BC.
Answer:
Construction Steps:
(i) Draw a line segment AB = 4.8 cm
(ii) At A, draw a ray AX making an angle of 75°
(iii) Cut off AC = 4 cm from AX
(iv) Join BC.
ABC is the required triangle.
(v) Draw two lines l and m parallel to BC at a distance of 1.2 cm
(vi) Draw the perpendicular bisector of BC which intersects l and m at P and P'
P and P' are the required points which are inside and outside the given triangle ABC.
In simple words: We make the triangle first. Then we draw the perpendicular bisector of BC (this gives points equidistant from B and C) and parallel lines to BC (this gives points at fixed distance from BC).

[Diagram: This diagram shows triangle ABC with angle A = 75°, perpendicular bisector of BC, and parallel lines l and m at distance 1.2 cm from BC, intersecting at points P and P'.]

📝 Teacher's Note: Explain that perpendicular bisector gives all points equidistant from two points. One intersection is inside triangle, one is outside.

🎯 Exam Tip: Clearly mark which point is inside and which is outside the triangle. Draw perpendicular bisector accurately through midpoint of BC.

Question PQ. O is a fixed point. Point P moves along a fixed line AB. Q is a point on OP produced such that OP = PQ. Prove that the locus of point Q is a line parallel to AB.
Answer:
P moves along AB, and Q moves in such a way that PQ is always equal to OP.
But P is the mid-point of OQ
Now in ΔOQQ'
P' and P" are the mid-points of OQ' and OQ"
Therefore, AB||Q'Q"
Therefore, Locus of Q is a line CD which is parallel to AB.
In simple words: Since P is always the middle point of OQ, and P moves on line AB, the point Q traces a line that is parallel to AB.

[Diagram: This diagram shows fixed point O, line AB with moving point P, and the locus of Q forming line CD parallel to AB, where P is the midpoint of OQ.]

📝 Teacher's Note: Use midpoint theorem - line joining midpoints of two sides of triangle is parallel to third side. Here P is midpoint of all segments OQ.

🎯 Exam Tip: Write "P is midpoint of OQ" clearly. Use midpoint theorem to prove parallelism. Draw clear diagram showing the locus.

Question 22. Draw an angle ABC = 75°. Find a point P such that P is at a distance of 2 cm from AB and 1.5 cm from BC.
Answer:
Construction Steps:
(i) Draw a ray BC.
(ii) At B, draw a ray BA making an angle of 75° with BC.
(iii) Draw a line l parallel to AB at a distance of 2 cm
(iv) Draw another line m parallel to BC at a distance of 1.5 cm which intersects line l at P.
P is the required point.
In simple words: We draw the angle first. Then we draw two parallel lines at the given distances. Where these lines meet is our point P.

[Diagram: This diagram shows angle ABC = 75° with parallel line l at 2 cm from AB, parallel line m at 1.5 cm from BC, intersecting at point P.]

📝 Teacher's Note: Show students how to measure perpendicular distance accurately. Use set square to draw parallel lines at exact distances.

🎯 Exam Tip: Mark the distances clearly on your diagram. Use ruler to measure 2 cm and 1.5 cm exactly. Point P must be marked clearly.

Question 23. Construct a triangle ABC, with AB = 5.6 cm, AC = BC = 9.2 cm. Find the points equidistant from AB and AC; and also 2 cm from BC. Measure the distance between the two points obtained.
Answer:
Construction Steps:
(i) Draw a line segment AB = 5.6 cm
(ii) From A and B, as centers and radius 9.2 cm, make two arcs which intersect each other at C.
(iii) Join CA and CB.
ABC is the required triangle.
(iv) Draw the angle bisector of ∠BAC
(v) Draw lines parallel to BC at a distance of 2 cm on both sides
(vi) These parallel lines intersect the angle bisector at two points P and Q
P and Q are the required points.
In simple words: We make an isosceles triangle first. The angle bisector from A gives points equidistant from AB and AC. Parallel lines to BC give points at 2 cm distance from BC.

[Diagram: This diagram shows isosceles triangle ABC with AB = 5.6 cm, AC = BC = 9.2 cm, angle bisector from A, and parallel lines at 2 cm distance from BC intersecting the bisector at points P and Q.]

📝 Teacher's Note: Since AC = BC, this is an isosceles triangle. The angle bisector from vertex A will be perpendicular to base AB. Measure PQ distance accurately.

🎯 Exam Tip: Draw angle bisector carefully from vertex A. Mark both intersection points clearly as P and Q. Measure and write the distance PQ in your answer.

Question 24. Construct a triangle ABC, with AB = 6 cm, AC = BC = 9 cm. Find a point 4 cm from A and equidistant from B and C.
Answer:
Steps of Construction:

1. Draw a line segment AB = 6 cm
2. With A and B as centers and radius 9 cm, draw two arcs which intersect each other at C.
3. Join AC and BC.
4. Draw the perpendicular bisector of BC.
5. With A as centre and radius 4 cm, draw an arc which intersects the perpendicular bisector of BC at P.

P is the required point which is equidistant from B and C and at a distance of 4 cm from A.

[Diagram: Shows triangle ABC with AB = 6 cm as base, AC = BC = 9 cm forming an isosceles triangle. A perpendicular bisector of BC is drawn, and an arc of radius 4 cm from A intersects this bisector at point P.]

In simple words: We make a triangle with two equal sides. Then we draw a line that cuts BC in the middle. This line has all points that are same distance from B and C. We then find where a circle around A meets this line.

📝 Teacher's Note: Show students that the perpendicular bisector is like a fence - any point on it is exactly the same distance from both ends. Use a string to show this practically.

🎯 Exam Tip: Always write "perpendicular bisector" clearly. Draw construction lines lightly and final answer points darkly. Label point P clearly.

Question 25. Ruler and compasses may be used in this question. All construction lines and arcs must be clearly shown and be of sufficient length and clarity to permit assessment.
(i) Construct a triangle ABC, in which BC = 6 cm, AB = 9 cm and angle ABC = 60°.
(ii) Construct the locus of all points inside triangle ABC, which are equidistant from B and C.
(iii) Construct the locus of the vertices of the triangles with BC as base and which are equal in area to triangle ABC.
(iv) Mark the point Q, in your construction, which would make triangle QBC equal in area to triangle ABC, and isosceles.
(v) Measure and record the length of CQ.
Answer:
Steps of Construction:

1. Draw a line segment BC = 6 cm.
2. At B, draw a ray BX making an angle 60 degree and cut off BA = 9 cm.
3. Join AC. ABC is the required triangle.
4. Draw perpendicular bisector of BC which intersects BA in M, then any point on LM is equidistant from B and C.
5. Through A, draw a line m || BC.
6. The perpendicular bisector of BC and the parallel line m intersect each other at Q.
7. Then triangle QBC is equal in area to triangle ABC. m is the locus of all points through which any triangle with base BC will be equal in area of triangle ABC.

On measuring CQ = 8.4 cm.

[Diagram: Shows triangle ABC with BC = 6 cm as base, angle ABC = 60°, AB = 9 cm. Perpendicular bisector of BC and a line parallel to BC through A are shown intersecting at point Q.]

In simple words: We make triangle ABC first. The perpendicular bisector finds points same distance from B and C. The parallel line through A finds all triangles with same area. Where these two lines meet is our answer Q.

📝 Teacher's Note: Explain that triangles with same base and height have same area. The parallel line keeps the height same. Use paper cutouts to show this concept clearly.

🎯 Exam Tip: Draw the perpendicular bisector and parallel line clearly. Mark intersection point Q boldly. Always measure and write the final answer with units (cm).

Question 26. State the locus of a point in a rhombus ABCD, which is equidistant
(i) from AB and AD;
(ii) from the vertices A and C.
Answer:
Steps of Construction:

1. In rhombus ABCD, draw angle bisector of ∠A, which meets in C.
2. Join BD, which intersects AC at O.

O is the required locus.

3. From O, draw OL ⊥ AB and OM ⊥ AD

In △AOL and △AOM
∠OLA = ∠OMA = 90°
∠OAL = ∠OAM (AC is bisector of angle A)
AO = OA (Common)
By Angle-Angle-Side criterion of congruence,
△AOL ≅ △AOM (AAS Postulate)
The corresponding parts of the congruent triangles are congruent
⇒ OL = OM (CPCT)
Therefore, O is equidistant from AB and AD.
Diagonal AC and BD bisect each other at right angles at O.
Therefore, AO = OC
Hence, O is equidistant from A and C.

[Diagram: Shows rhombus ABCD with diagonals AC and BD intersecting at O. Perpendiculars from O to sides AB and AD are shown.]

In simple words: In a rhombus, the diagonals cross at the center. This center point is same distance from opposite corners and from the sides that meet at any corner.

📝 Teacher's Note: Use a paper rhombus and fold it along diagonals. Show students that the center is equidistant from all sides and opposite vertices. This makes the concept very clear.

🎯 Exam Tip: State clearly: "The intersection point of diagonals." Also mention it is equidistant from the required points. Draw the diagram neatly and label point O.

Question 27. Use a graph paper for this question. Take 2 cm = 1 unit on both the axes.
(i) Plot the points A(1,1), B(5,3) and C(2,7).
(ii) Construct the locus of points equidistant from A and B.
(iii) Construct the locus of points equidistant from AB and AC.
(iv) Locate the point P such that PA = PB and P is equidistant from AB and AC.
(v) Measure and record the length PA in cm.
Answer:
Steps of Construction:

1. Plot the points A(1, 1), B(5, 3) and C(2, 7) on the graph and join AB, BC and CA.
2. Draw the perpendicular bisector of AB and angle bisector of angle A which intersect each other at P.

P is the required point.
Since P lies on the perpendicular bisector of AB.
Therefore, P is equidistant from A and B.
Again,
Since P lies on the angle bisector of angle A.
Therefore, P is equidistant from AB and AC.
On measuring, the length of PA = 5.2 cm

[Diagram: Shows coordinate plane with points A(1,1), B(5,3), C(2,7) plotted. Triangle ABC is drawn with perpendicular bisector of AB and angle bisector of angle A intersecting at point P.]

In simple words: We plot three points and join them to make a triangle. The perpendicular bisector of AB gives points same distance from A and B. The angle bisector of angle A gives points same distance from the two sides. Where they meet is point P.

📝 Teacher's Note: Use graph paper and show students how to plot points carefully. Explain that 2 cm = 1 unit means each big square is half a unit. Use ruler for accurate measurement.

🎯 Exam Tip: Plot points accurately using the given scale. Draw construction lines clearly but not too dark. Measure PA carefully and write the answer with units (cm).

Question 28. Construct an isosceles triangle ABC such that AB = 6 cm, BC=AC=4cm. Bisect angle C internally and mark a point P on this bisector such that CP = 5cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB.
Answer:
Steps of Construction:

1. Draw a line segment AB = 6 cm.
2. With centers A and B and radius 4 cm, draw two arcs which intersect each other at C.
3. Join CA and CB.
4. Draw the angle bisector of angle C and cut off CP = 5 cm.
5. A line m is drawn parallel to AB at a distance of 5 cm.
6. P as centre and radius 5 cm, draw arcs which intersect the line m at Q and R.
7. Join PQ, PR and AQ.

Q and R are the required points.

[Diagram: Shows isosceles triangle ABC with AB = 6 cm as base, AC = BC = 4 cm. Point P is marked 5 cm from C along angle bisector. Line m is drawn parallel to AB at distance 5 cm, with points Q and R marked where arcs from P intersect this line.]

In simple words: We make an isosceles triangle. Then we find a point P inside it. We draw a line parallel to the base at distance 5 cm. Points Q and R are where a circle around P meets this parallel line.

📝 Teacher's Note: Show students that parallel lines keep the same distance everywhere. Use a ruler to check that the parallel line is exactly 5 cm from AB at all points. This helps students understand locus.

🎯 Exam Tip: Draw the isosceles triangle first. Mark point P clearly on the angle bisector. Draw the parallel line lightly and mark Q and R points boldly. Join all required lines as asked.

Question PQ. Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of lengths 6 cm and 5 cm respectively.
(i) Construct the locus of points, inside the circle, that are equidistant from A and C. Prove your construction.
(ii) Construct the locus of points, inside the circle, that are equidistant from AB and AC.
Answer:
This question appears to be incomplete as the full solution is not provided in the source material.

[Diagram: Shows a circle with two chords AB (6 cm) and AC (5 cm) marked from point A.]

In simple words: We need to draw a circle and mark two chords from one point. Then find locus of points equidistant from the endpoints and from the chord lines.

📝 Teacher's Note: For part (i), use perpendicular bisector of AC. For part (ii), use angle bisector of angle A. Show students these are standard locus constructions.

🎯 Exam Tip: Always draw the circle first with proper radius. Mark chords accurately with given lengths. Use perpendicular bisector and angle bisector for the two parts.

Solution:
Answer:
Steps of Construction:

1. Draw a circle with radius = 4 cm.
2. Take a point A on it.
3. A as centre and radius 6 cm, draw an arc which intersects the circle at B.
4. Again A as centre and radius 5 cm, draw an arc which intersects the circle at C.
5. Join AB and AC.
6. Draw the perpendicular bisector of AC, which intersects AC at M and meets the circle at E and F.

EF is the locus of points inside the circle which are equidistant from A and C.

7. Join AE, AF, CE and CF.

Proof:
In \( \triangle CME \) and \( \triangle AME \)
CM = AM (EF is the bisector of AC)
\( \angle CME = \angle CMA = 90° \)
EM = EM (Common)
\( \therefore \) By Side-Angle-Side criterion of congruence,
\( \triangle CME \cong \triangle AME \) (SAS Postulate)
The corresponding parts of the congruent triangles are congruent.
\( \implies \) CE = AE (CPCT)

Similarly, we can prove that CF = AF
Hence EF is the locus of points which are equidistant from A and C.
ii) Draw the bisector of angle A which meets the circle at N.

Diagram: This diagram shows a circle with points A, B, C on the circumference, with perpendicular bisector EF of AC, and angle bisector from A meeting the circle at N.

In simple words: We draw a circle and mark points on it. Then we find the line that keeps equal distance from two points. This line is called a locus.

📝 Teacher's Note: Show students how the perpendicular bisector works by folding paper. When you fold along the bisector, the two points match exactly. This helps them see why all points on the bisector are equally far from both points.

🎯 Exam Tip: Always write the construction steps in order. Draw the diagram clearly and label all points. Write "locus" clearly in your final answer.

Therefore, Locus of points inside the circle which are equidistant from AB and AC is the perpendicular bisector of angle A.

Question 29. Plot the points A(2,9), B(-1,3) and C(6,3) on a graph paper. On the same graph paper, draw the locus of point A so that the area of triangle ABC remains the same as A moves.
Answer:
Steps of construction:

1. Plot the given points on graph paper.
2. Join AB, BC and AC.
3. Draw a line parallel to BC at A and mark it as CD.

CD is the required locus of point A where area of triangle ABC remains same on moving point A.

Diagram: This diagram shows a coordinate graph with triangle ABC plotted, and a parallel line CD showing the locus where point A can move while keeping the same triangle area.

In simple words: When we want the triangle to have the same area, point A must move on a line parallel to the base BC. Think of it like a triangle that can stretch up or down but keeps the same size.

📝 Teacher's Note: Explain that area of triangle = ½ × base × height. If base BC stays same and area stays same, then height must stay same. All points at same height from BC form a parallel line.

🎯 Exam Tip: Always draw the parallel line clearly and label it as the locus. Write that "area remains constant" in your answer. Plot points accurately on graph paper.

Question 30. Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
(i) Complete the rectangle ABCD such that:
(a) P is equidistant from AB and BC.
(b) P is equidistant from C and D.
(ii) Measure and record the length of AB.
Answer:
i) Steps of Construction:

1. Draw a line segment BC = 5 cm
2. B as centre and radius 4 cm draw an arc at an angle of 45 degrees from BC.
3. Join PC.
4. B and C as centers, draw two perpendiculars to BC.
5. P as centre and radius PC, cut an arc on the perpendicular on C at D.
6. D as centre, draw a line parallel to BC which intersects the perpendicular on B at A.

ABCD is the required rectangle such that P is equidistant from AB and BC (since BD is angle bisector of angle B) as well as C and D.

ii) On measuring AB = 5.7 cm

Diagram: This diagram shows rectangle ABCD with point P inside, where P is equidistant from sides AB and BC, and also equidistant from vertices C and D.

In simple words: We build a rectangle around triangle BCP. Point P is the same distance from two sides of the rectangle and also the same distance from two corners.

📝 Teacher's Note: Show students that when a point is equidistant from two lines, it lies on the angle bisector. When equidistant from two points, it lies on the perpendicular bisector. Use string to demonstrate equal distances.

🎯 Exam Tip: Always show construction arcs clearly. Measure AB carefully with a ruler. Write the measurement with correct units (cm). Label all points A, B, C, D, P clearly.

Question 31. Use ruler and compasses only for the following questions. All constructions lines and arcs must be clearly shown.
(i) Construct a ∆ABC in which BC = 6.5 cm, ∠ABC = 60°, AB = 5 cm.
(ii) Construct the locus of points at a distance of 3.5 cm from A.
(iii) Construct the locus of points equidistant from AC and BC.
(iv) Mark 2 points X and Y which are at distance of 3.5 cm from A and also equidistant from AC and BC. Measure XY.
Answer:
i. Steps of construction:

1. Draw BC = 6.5 cm using a ruler.
2. With B as center and radius equal to approximately half of BC, draw an arc that cuts the segment BC at Q.
3. With Q as center and same radius, cut the previous arc at P.
4. Join BP and extend it.
5. With B as center and radius 5 cm, draw an arc that cuts the arm PB to obtain point A.
6. Join AC to obtain ∆ABC.

ii. With A as center and radius 3.5 cm, draw a circle.
The circumference of a circle is the required locus.

iii. Draw CH, which is bisector of ∠ACB. CH is the required locus.

iv. Circle with center A and line CH meet at points X and Y as shown in the figure. XY = 8.2 cm (approximately)

Diagram: This diagram shows triangle ABC with a circle of radius 3.5 cm centered at A, and the angle bisector of angle C, with intersection points X and Y marked.

In simple words: We draw a triangle, then draw a circle around one corner and a line that splits an angle in half. Where the circle and line meet are our special points X and Y.

📝 Teacher's Note: Remind students that 60° is made by two arcs of equal radius. The locus of points at fixed distance is always a circle. The locus of points equidistant from two lines is the angle bisector.

🎯 Exam Tip: Show all construction arcs and lines clearly. Use compass for circles and angles. Measure XY carefully with ruler. Write "approximately" when measuring. Keep construction lines light but visible.