Concise Mathematics Selina Solutions Class 10 ICSE Chapter 16 Loci (Locus and its Contructions)

This page provides the most comprehensive ICSE Class 10 Maths Loci Solutions, covering Chapter 16 from the Class – 10 Concise Mathematics Selina textbook. The chapter, titled ‘Loci (Locus and its Contructions)’, introduces the fascinating concept of a locus as the path traced by a point moving under specific geometrical conditions. You will learn to identify and construct various loci, such as the perpendicular bisector of a line segment (the locus of points equidistant from two fixed points) and the bisector of an angle (the locus of points equidistant from two intersecting lines). Mastering these constructions is crucial as they form the building blocks for more complex problems in coordinate geometry and other advanced topics.

If you are stuck on a specific construction or need to verify your method for a particular problem, you’ve come to the right place. This guide provides clear, step-by-step solutions for all 51 questions found in Exercise 16(A), Exercise 16(B), and the Test Yourself section. Each solution is crafted to follow the exact methodology and presentation style expected by the ICSE board, ensuring you learn the correct way to approach these problems in your exams. Here, you will find accurate and easy-to-understand solutions to help you master the entire chapter.

Exercise 16(A)

Question 1(a)

P is a point on perpendicular bisector of side BC of △ABC, then :

• (a) PA = PB
• (b) PB = PC
• (c) PA = PC
• (d) none of these

In geometry, the locus of a point that is equidistant from two specific points is known as the perpendicular bisector of the line segment connecting those points. Here, since point P is located on the perpendicular bisector of side BC in △ABC, it must be equidistant from points B and C.

∴ PB = PC.

Hence, Option 2 is the correct option.

Question 1(b)

P is a point on bisector of angle A of △ABC. Then:

• (a) P is equidistant from the vertices B and C
• (b) PA = PC
• (c) PA = PB
• (d) P is equidistant from sides AB and AC of the △ABC.

Consider the property of angle bisectors: The locus of a point that maintains an equal distance from two intersecting lines is precisely the angle bisector of those lines.

Here, since P lies on the bisector of angle A in △ABC, it means that P is at the same distance from the sides AB and AC.

∴ P is equidistant from sides AB and AC of the △ABC.

Hence, Option 4 is the correct option.

Question 1(c)

The perpendicular bisector of side AB and bisector of angle A of △ABC meet at point P. Then :

• (a) PA = PB
• (b) PA = PC
• (c) PB = PC
• (d) PB bisects ∠ABC

Consider the concept that the locus of points equidistant from two specific points lies along the perpendicular bisector of the segment connecting those points. Here, the perpendicular bisector of side AB and the bisector of angle A in △ABC intersect at point P. This implies that point P is equidistant from points A and B. ∴ PA = PB.

Hence, Option 1 is the correct option.

Question 1(d)

Using the information in the given diagram, state if :

• (a) AD = DC
• (b) BD = DC
• (c) CD bisects angle ACB
• (d) angle CAD is greater than angle DAB

Observing the diagram, we assume that AD = DB = AC = BC = x.

Consider the triangles △ ADC and △ BDC:

⇒ AD = DB (both are equal to x)

⇒ AC = BC (both are equal to x)

⇒ CD = CD (this side is common to both triangles)

∴ By the S.S.S. (Side-Side-Side) congruence criterion, △ ADC ≅ △ BDC.

Since corresponding parts of congruent triangles are equal, we have:

∴ ∠ACD = ∠BCD

This indicates that CD divides angle ACB into two equal parts.

∴ CD bisects angle ACB.

Hence, Option 3 is the correct option.

Question 1(e)

Bisector of angle B of triangle ABC intersects side AC at point P, then point P is :

• (a) equidistant from vertices A and C
• (b) PA = PB
• (c) PB = PC
• (d) equidistant from sides AB and BC

Consider the property that the locus of a point which maintains equal distance from two intersecting lines is the angle bisector of those lines.

In this scenario, the bisector of \angle B in triangle ABC meets side AC at point P.

∴ Point P is at an equal distance from the sides AB and BC.

Hence, Option 4 is the correct option.

Question 2

Given: PQ is a perpendicular bisector of side AB of the triangle ABC.

Prove: Q is equidistant from A and B.

To demonstrate that point Q is equidistant from points A and B, consider joining A and Q as depicted in the diagram.

In triangles ∆AQP and ∆BQP:

⇒ We have AP = BP, as given.

⇒ Angles ∠QPA and ∠QPB are both 90° since PQ is a perpendicular bisector.

⇒ The segment PQ is common to both triangles.

∴ By the SAS (Side-Angle-Side) congruence criterion, ∆AQP is congruent to ∆BQP.

∴ As a result of congruence, AQ = BQ by C.P.C.T. (Corresponding Parts of Congruent Triangles).

Hence, proved that Q is equidistant from A and B.

Question 3

Given: CP is the bisector of angle C of ∆ABC.

Prove: P is equidistant from AC and BC.

Construct perpendiculars from point P to the lines AC and BC, naming them PL and PM respectively.

Consider the triangles ∆LPC and ∆MPC.

⇒ Notice that ∠PLC and ∠PMC are both 90°, as they are perpendiculars.

⇒ Since CP bisects angle C, ∠PCL equals ∠MCP.

⇒ The side PC is common to both triangles.

∴ By the Angle-Angle-Side (AAS) congruence criterion, ∆LPC is congruent to ∆MPC.

∴ From this congruence, PL equals PM by the property of Corresponding Parts of Congruent Triangles (C.P.C.T.).

Hence, proved that P is equidistant from AC and BC.

Question 4

Given: AX bisects angle BAC 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.

Start by constructing perpendiculars from point X to lines AC and AB, labeling them as XL and XM respectively. Then, connect point Y to point C.

(i) Consider triangles ∆AXL and ∆AXM:

⇒ Notice that ∠XAL = ∠XAM because AX bisects ∠BAC.

⇒ The segment AX is common to both triangles.

⇒ Both ∠XLA and ∠XMA are right angles, measuring 90° each.

∴ By the Angle-Angle-Side (AAS) criterion, ∆AXL is congruent to ∆AXM.

∴ From the congruency, XL = XM by Corresponding Parts of Congruent Triangles (C.P.C.T.).

Thus, X is equidistant from lines AB and AC.

(ii) Now, examine triangles ∆YTA and ∆YTC:

⇒ AT = CT because PQ is the perpendicular bisector of AC.

⇒ Both ∠YTA and ∠YTC are right angles, each measuring 90°.

⇒ The segment YT is common to both triangles.

∴ By the Side-Angle-Side (SAS) criterion, ∆YTA is congruent to ∆YTC.

∴ From the congruency, YA = YC by C.P.C.T.

Thus, Y is equidistant from points A and C.

Question 5

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.

Construction Steps:

1. Start by drawing the line segment BC with a length of 6.3 \text{ cm}.
2. Using B as the center, draw an arc with a radius of 4.2 \text{ cm}.
3. Next, take C as the center and draw another arc with a radius of 5 \text{ cm}. Ensure this arc intersects the first arc at point A.
4. Connect the points A to B and A to C to form the triangle \triangle ABC.
5. Construct the perpendicular bisector LM of BC.
6. Note where LM meets AC at point D, and BC at point E.
7. Draw a line segment joining D and B.

Proof:

Consider the triangles \triangle DBE and \triangle DCE.

• BE = EC because LM is the perpendicular bisector of BC.
• \angle DEB = \angle DEC = 90^\circ, as LM is perpendicular to BC.
• DE = DE since it is common to both triangles.

By the SAS (Side-Angle-Side) criterion, \triangle DBE \cong \triangle DCE.

Therefore, DB = DC by CPCT (Corresponding Parts of Congruent Triangles).

Hence, proved that D is equidistant from B and C.

Question 6

In each of the given figures; PA = PB and QA = QB.

(i)

(ii)

Prove in each case, that PQ (produce, if required) is perpendicular bisector of AB.

Hence, state the locus of the points equidistant from two given fixed points.

(i) Let’s connect points P and Q, and let their line intersect AB at point D.

We know that PA = PB, which means P is equally distant from A and B. This implies that P must be on the perpendicular bisector of AB.

Similarly, QA = QB indicates that Q is also equally distant from A and B, placing Q on the perpendicular bisector of AB as well.

Since both P and Q lie on this line, PQ itself is the perpendicular bisector of AB.

Hence, PQ is confirmed as the perpendicular bisector of AB.

(ii) Again, connect P and Q, allowing their line to meet AB at D.

Given PA = PB, it follows that P is equidistant from points A and B, placing it on the perpendicular bisector of AB.

Similarly, QA = QB ensures that Q is equidistant from A and B, meaning Q also lies on the perpendicular bisector of AB.

Thus, both P and Q are on the perpendicular bisector of AB, confirming that PQ is indeed the perpendicular bisector of AB.

Hence, PQ is confirmed as the perpendicular bisector of AB.

In conclusion, the locus of points equidistant from two fixed points is the perpendicular bisector of the segment joining those points.

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.

Let’s go through the steps of construction:

1. Begin by drawing line segment BC measuring 6.4 cm.
2. At point B, construct a ray BX that forms a 75° angle with BC, and mark point A such that AB = 5 cm.
3. Connect points A and C to form the triangle \triangle ABC.
4. Construct the perpendicular bisector of BC, which will intersect BC at point Q.
5. Draw the angle bisector of \angle ACB, denoted as CX. This bisector will intersect the perpendicular bisector of BC at point P.
6. Connect P to B and draw PL such that it is perpendicular to AC.

Now, let’s analyze the congruence of triangles:

In \triangle PBQ and \triangle PCQ:

⇒ PQ = PQ (This is the common side)

⇒ \angle PQB = \angle PQC (Both are 90°)

⇒ BQ = QC (Since PQ is the perpendicular bisector of BC)

∴ \triangle PBQ \cong \triangle PCQ by the SAS congruence axiom.

∴ PB = PC by C.P.C.T. (Corresponding Parts of Congruent Triangles).

This confirms that P is equidistant from B and C.

Additionally, consider \triangle PQC and \triangle PLC:

⇒ \angle PQC = \angle PLC (Both are 90°)

⇒ \angle PCQ = \angle PCL (Since CX is the angle bisector of \angle ACB)

⇒ PC = PC (This is the common side)

∴ \triangle PQC \cong \triangle PLC by the AAS congruence axiom.

∴ PQ = PL by C.P.C.T.

Thus, P is equidistant from AC and BC.

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

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.

Steps to construct and prove:

1. Begin by sketching a parallelogram ABCD where side AB is longer than side BC.
2. Connect diagonal AC.
3. Construct BX, the angle bisector of ∠ABC, and let it meet AC at point P.
4. From P, draw perpendiculars PL to AB and PM to BC.

Now, consider triangles ∆PLB and ∆PMB:

⇒ Both ∠PLB and ∠PMB are right angles (90° each).

⇒ Since BX is the angle bisector of ∠ABC, ∠PBL equals ∠PBM.

⇒ The side PB is common to both triangles.

∴ By the Angle-Angle-Side (AAS) criterion, ∆PLB is congruent to ∆PMB.

∴ This implies PL equals PM.

Thus, point P is equidistant from lines AB and BC.

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.

Steps for constructing the solution:

1. Begin by drawing triangle LMN.
2. Construct the angle bisectors for angles at L and N. These bisectors will intersect at point A.
3. Connect point A to point M.

(i) Since point A is located on the bisector of ∠N,

∴ A is equidistant from the sides MN and LN.

Additionally, as A is on the bisector of ∠L,

∴ A is equidistant from the sides LN and LM.

Thus, point A is equidistant from all three sides of triangle LMN.

(ii) From the earlier part, we established that:

A is equidistant from MN and LN, and also from LN and LM.

This implies,

A is equidistant from MN and LM.

∴ A must be on the angle bisector of ∠LMN.

Therefore, AM bisects ∠LMN.

Question 10

Use ruler and compasses only for this question.

(i) Construct ∆ABC, where AB = 3.5 cm, BC = 6 cm and ∠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.

Steps to follow for the construction:

(i) Begin by drawing a line segment BC with a length of 6 cm. At point B, construct an angle CBX measuring 60°. From point B, measure and mark a point A on BX such that AB equals 3.5 cm. Connect point A to point C to form the triangle ∆ABC.

(ii) To find the locus of points equidistant from the sides BA and BC, draw the angle bisector of ∠ABC. This bisector is the line BX, which represents the locus inside the triangle.

(iii) For the locus of points equidistant from points B and C, construct the perpendicular bisector of the line segment BC. This line, labeled YZ, lies within the triangle and represents the desired locus.

(iv) Identify point P where the angle bisector BX intersects the perpendicular bisector YZ. This point P is equidistant from the sides AB, BC, and also from points B and C. Measure the distance from P to B.

Hence, PB = 3.5 cm

Exercise 16(B)

Question 1(a)

The locus of the centers of all circles, which are tangents to the arms AB and BC of angle ABC is :

• (a) perpendicular bisector of arm AB
• (b) perpendicular bisector of arm BC
• (c) bisector of angle ABC
• (d) none of these

Consider the diagram provided.

Here, we see that PD = PE represents the radius of a circle with center P, and P’F = P’G represents the radius of another circle with center P’. This indicates that both centers, P and P’, are equidistant from the lines AB and BC.

Recall that the locus of a point that is equidistant from two intersecting lines is the angle bisector of those lines. Therefore, the path traced by the centers of all such circles, tangent to the arms AB and BC of angle ABC, must be the bisector of angle ABC.

Hence, Option 3 is the correct option.

Question 1(b)

P is a moving point and AB is a chord of a circle. If P moves, within this circle, in such a way that it is at equal distances from points A and B. The locus of P is :

• (a) a chord perpendicular to chord AB
• (b) a chord that bisects the chord AB
• (c) a diameter of the circle
• (d) the diameter of the circle which is perpendicular to chord AB

Consider a circle with center O and a chord AB. Imagine drawing a line XY that is the perpendicular bisector of AB.

Notice that the locus of a point that is equidistant from two fixed points, A and B, is the perpendicular bisector of the segment joining these points.

∴ The line XY serves as the locus of point P.

Remember that any line drawn perpendicular to a chord and passing through the circle’s center is a diameter.

Thus, the locus of point P is indeed the diameter of the circle which stands perpendicular to chord AB.

Hence, Option 4 is the correct option.

Question 1(c)

AB is a line segment. A point P moves in such a way that the triangle APB is always an isosceles triangle with base AB. The locus of point P is the line which :

• (a) is parallel to AB
• (b) is perpendicular to AB
• (c) is perpendicular bisector of AB
• (d) passes through the mid-point of AB.

To find the locus of point P, follow these steps:

1. Begin by sketching the line segment AB.
2. Next, construct the line XY, which is the perpendicular bisector of AB.

Recall that the locus of a point, which is equidistant from two fixed points, lies along the perpendicular bisector of the segment connecting those points.

∴ Any point P located on line XY will satisfy the condition PA = PB, forming an isosceles triangle with base AB.

Hence, Option 3 is the correct option.

Question 1(d)

AB is a line segment and P is a moving point that moves in such a way that it is always equidistant from AB. The locus of point P is the line which :

• (a) is parallel to AB and through point P.
• (b) is perpendicular to AB through point P.
• (c) is perpendicular bisector of AB.
• (d) passes through the mid-point of AB.

To construct the locus of a point equidistant from a line segment AB, follow these steps:

1. Begin by sketching the line segment AB.
2. Construct two lines, labeled as l and m, that are parallel to AB and situated on opposite sides of it.

Recall that the locus of a point maintaining a constant distance from a given line is itself a line that runs parallel to the original line.

Hence, Option 1 is the correct option.

Question 1(e)

A point P moves in such a way that it is at a distance less than or equal to 5 cm from a fixed point O. The locus of point P is :

• (a) a circle with radius 5 cm.
• (b) a circle with OP as radius.
• (c) a circle with diameter of 10 cm.
• (d) a circle of radius 5 cm and the fixed point O as its center.

Steps for constructing the locus:

1. Begin by drawing a circle with a center at point O and a radius of 5 cm.

Observe the diagram:

Every point located on the circle’s boundary maintains a distance of exactly 5 cm from O, while any point situated within the circle is at a distance less than 5 cm from O.

Hence, Option 4 is the correct option.

Question 2

Describe the locus of a point at a distant 3 cm from a fixed point.

Steps for the construction:

Consider a fixed point labeled as A. Let B be any point that lies on the circle.

1. With A as the center, construct a circle with a radius of 3 cm, which is the distance from A to B.

The locus of a point that is 3 cm from a fixed point is the circumference of a circle with a radius of 3 cm, where the fixed point serves as the center of the circle.

Question 3

Describe the locus of points at a distance 2 cm from a fixed line.

Construction steps:

1. Consider a fixed line, denoted as AB.
2. Construct two lines, labeled l and m, each parallel to AB but situated on opposite sides. Ensure these lines are exactly 2 cm away from AB.

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) parallel to the given fixed line and at a distance of 2 cm from it.

Question 4

Describe the locus of the center of a wheel of a bicycle going straight along a level road.

Consider a wheel with a radius of r units.

As the wheel moves in a straight path on a flat road, the path traced by its center forms a straight line. This line runs parallel to the road and is positioned at a height equal to the wheel’s radius.

Question 5

Describe the locus of the moving end of the minute hand of a clock.

Consider the length of the minute hand to be r units.

The path traced by the moving end of the minute hand forms the circumference of a circle, where the radius is equal to the length of the minute hand.

Question 6

Describe the locus of a stone dropped from the top of a tower.

When a stone is released from the top of a tower, it follows a path that is a straight line. This line is vertical and perpendicular to the ground, originating from the point at the top of the tower from where the stone begins its descent.

Question 7

Describe the locus of a runner, running round a circular track and always keeping a distance of 1.5 m from the inner edge.

Consider the inner edge of the track as a circle with radius r meters.

The locus we are interested in is a circle that shares the same center as the track but has a radius of (r + 1.5) meters. This makes it concentric with the original track, maintaining a consistent distance of 1.5 meters from the inner edge.

Question 8

Describe the locus of the door handle, as the door opens.

When the door swings open, the path traced by the door handle forms a circle. This circle’s center is located at the axis around which the door rotates. The radius of this circle is the length from the axis of rotation to the door handle. Therefore, the locus of the door handle is the circumference of this circle.

Question 9

Describe the locus of points inside a circle and equidistant from two fixed points on the circumference of the circle.

Steps for the construction:

1. Begin by drawing a circle with center O.
2. Identify and mark two points, A and B, on the circle. Connect these two points with a line segment AB.
3. Construct the perpendicular bisector of the line segment AB. This bisector will intersect the circle at its center, O.

Notice that the center O lies on this perpendicular bisector, which means that OA = OB.

Hence, the locus of points inside the circle which are equidistant from the two fixed points on the circumference of a circle will be the diameter which is the perpendicular bisector of the chord joining the two fixed points on the circle.

Question 10

Describe the locus of the centers of all circles passing through two fixed points.

Steps for construction:

1. Assume two fixed points, A and B.
2. Consider circles that include both A and B on their circumferences.
3. Construct the perpendicular bisector of the line segment AB.

From the diagram,

The centers of these circles must be located on the perpendicular bisector of AB.

Hence, the locus of centers 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.

Question 11

Describe the locus of vertices of all isosceles triangles having a common base.

Consider that the locus of a point maintaining equal distance from two fixed points is represented by the perpendicular bisector of the segment joining these points.

Steps for Construction:

1. Construct a line segment labeled as BC, which will act as the common base.
2. Draw line XY, ensuring it is the perpendicular bisector of BC.
3. Select a point P on line XY.

Here, since point P is on the perpendicular bisector, it ensures that PB equals PC.

Hence, the locus of vertices of all isosceles triangles having a common base will be the perpendicular bisector of the common base of the triangles.

Question 12

Describe the locus of a point P, so that :

AB^2 = AP^2 + BP^2,

where A and B are two fixed points.

Consider the property that a diameter subtends a right angle to any point on the circle. This means that if point P lies on the circle with diameter AB, then the angle ∠APB will always be 90°.

From the diagram, we apply the Pythagorean theorem:

⇒ Hypotenuse^2 = Perpendicular^2 + Base^2

⇒ AB^2 = AP^2 + BP^2

This equation is the hallmark of a right-angled triangle, confirming that triangle APB is right-angled at P.

∴ AP ⊥ BP.

Hence, the locus of the point P is the circumference of a circle with AB as diameter.

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.

(i) Consider the property that the locus of a point equidistant from two intersecting lines is the angle bisector of those lines. In the context of a rhombus, the diagonals serve this purpose by bisecting the angles at the vertices.

∴ The locus of a point in rhombus ABCD that maintains equal distance from sides AB and BC is the diagonal BD.

(ii) Recall that the locus of a point equidistant from two given points is the perpendicular bisector of the segment connecting those points. In a rhombus, the diagonals intersect each other at right angles, effectively bisecting each other.

∴ The locus of a point in rhombus ABCD equidistant from points B and D is diagonal AC.

Question 14

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 center 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.

(i) Consider a circle with the given point as its center. The region where all points lie within a distance of 3 cm from this center forms the interior of this circle.

(ii) Now think of a circle with the given point as its center and a radius of 4 cm. The area where all points are located beyond this circle is the exterior region.

(iii) Imagine a circle centered at the given point with a radius of 2.5 cm. The locus includes all points either inside this circle or exactly on its boundary.

(iv) Picture a circle centered at the given point with a radius of 35 mm. The locus consists of all points either outside this circle or precisely on its circumference.

(v) When a circle rolls around another circle while always touching it, the path traced by the center of the rolling circle is another circle. This new circle is concentric with the fixed circle, and its radius is the sum of the radii of the two circles.

(vi) The path formed by the centers of all circles that touch both arms of a given angle is the angle’s bisector.

(vii) For chords of a circle parallel to a given chord, the mid-points of these chords trace a path along the diameter that is perpendicular to the given chord.

(viii) Within a circle, the points that are equidistant from the endpoints of a given chord lie along the diameter that acts as the perpendicular bisector of that chord.

Question 15

In the given figure, obtain all the points equidistant from lines m and n; and 2.5 cm from O.

Construction process:

1. Begin by constructing the angle bisectors of the angles formed by lines m and n. Label these bisectors as PQ and XY.
2. Next, using a compass, draw arcs centered at O with a radius of 2.5 cm. These arcs should intersect the angle bisectors at points a, b, c, and d respectively.

Hence, a, b, c, and d are the required four points.

Question 16

Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude.

Steps for constructing the locus:

1. Begin by drawing the line segment BC.
2. Construct XY, which is the perpendicular bisector of BC, ensuring it intersects BC at point D. From D, measure and mark the point A such that DA equals the given altitude.
3. Next, draw the line segment EF parallel to BC at a distance equal to the altitude, ensuring it passes through point A.

Hence, the line EF is the locus of the vertices of all triangles with a given base and a given altitude.

Question 17

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. Describe the figure AXBY.

Steps for constructing the required loci:

1. Begin by sketching a line segment AB with a length of 8\,\text{cm}.
2. Construct two lines, labeled l and m, parallel to AB. These lines should be at a distance of 4\,\text{cm} from AB on either side.
3. Identify and draw CD, the line that acts as the perpendicular bisector of AB. This line will cross the parallel lines l and m at points X and Y.
4. Connect A to X and Y, and B to X and Y to form the quadrilateral AXBY.

Notice that the diagonals of AXBY are equal in length and intersect each other at right angles.

Hence, AXBY is a square.

(i) Thus, the locus of a point that is 4\,\text{cm} away from AB consists of two lines, each parallel to AB.

(ii) Thus, the locus of a point that is equidistant from A and B is the perpendicular bisector of AB.

Question 18

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 the circles of radius 2 cm which touch AB.

Hence, construct the circle of radius 2 cm which touches AB and AC.

Steps for Construction:

1. Begin by drawing the line segment BC with a length of 4.5 \text{ cm}.
2. Use B as the center to draw an arc with a radius of 6 \text{ cm}, and C as the center for an arc with a radius of 5 \text{ cm}. These arcs will intersect at point A.
3. Connect points A to B and A to C to form the triangle ABC.
4. Construct AD, the angle bisector of \angle BAC.
5. Draw lines l and n parallel to AB, and line m parallel to AC, each at a distance of 2 \text{ cm}. These lines will intersect one another and the angle bisector AD at point O.
6. With O as the center and using a radius of 2 \text{ cm}, draw a circle that is tangent to both AB and AC.

(i) The locus of the centers of all circles that touch AB and AC is AD, the angle bisector of \angle A.

(ii) The locus of the centers of all the circles with a radius of 2 \text{ cm} that touch AB is given by the lines l and n.

Question 19

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.

Steps of construction:

1. Start by drawing ray BC.
2. From point B, construct ray BA such that it forms a 75° angle with ray BC.
3. Construct a line, denoted as line l, which runs parallel to AB and is 2 cm away from it.
4. Next, draw line m parallel to BC, maintaining a distance of 1.5 cm from it. This line m will intersect line l at point P.

Hence, P is the required point.

Question 20

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.

To construct triangle ABC and find the specified points, follow these steps:

1. Begin by drawing a line segment AB with a length of 5.6 cm.
2. Using a compass, take A and B as centers and draw arcs with a radius of 9.2 cm. These arcs will intersect at point C.
3. Connect points C to A and C to B to form triangle ABC.
4. Next, draw two lines parallel to BC, each 2 cm away from BC on either side. Label these lines as n and m.
5. Construct the angle bisector of ∠BAC. This bisector will intersect the parallel lines n and m at points P and Q, respectively.

Upon measuring, the distance between points P and Q is 4.3 cm.

Thus, P and Q are the desired points equidistant from AB and AC, and also 2 cm from BC, with PQ measuring 4.3 cm.

Question 21

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.

Steps for constructing the triangle and finding the required point:

1. Begin by drawing a line segment AB with a length of 6 cm.
2. Using A and B as centers, draw two arcs with a radius of 9 cm each. These arcs should intersect at point C.
3. Connect points C to A and C to B to form triangle ABC.
4. Construct the perpendicular bisector of segment BC.
5. With point A as the center, draw an arc with a radius of 4 cm. This arc should intersect the perpendicular bisector of BC at point P.

Thus, point P is the desired location, being equidistant from B and C and exactly 4 cm away from A.

Test Yourself

Question 1(a)

The locus of point which is equidistant from two non-parallel lines AB and CD is :

• (a) perpendicular to AB.
• (b) perpendicular to CD.
• (c) bisector of angle between AB and CD.
• (d) perpendicular bisector of CD.

Consider two non-parallel lines, AB and CD. These lines will meet at a certain point since they are not parallel.

Recall that when two lines intersect, the locus of a point that is equidistant from both lines is the angle bisector of the angles formed between the lines.

Hence, Option 3 is the correct option.

Question 1(b)

Three isosceles triangles PBC, QBC and RBC are on the same base, then :

• (a) P, Q and R are collinear.
• (b) △PQR is isosceles triangle.
• (c) Q lies on the circumference of a circle with BC as diameter.
• (d) Q is mid-point of line segment PR.

To construct the triangles, follow these steps:

1. Start by drawing the line segment BC.
2. Next, construct the line XY, which is the perpendicular bisector of BC.

It’s important to remember that the perpendicular bisector of a line segment is the locus of all points that are equidistant from the segment’s endpoints. Therefore, any point located on line XY will be equidistant from points B and C.

∴ The points P, Q, and R will all lie on line XY.

Hence, Option 1 is the correct option.

Question 1(c)

Locus of the centers of the circles passing through two fixed points A and B is :

• (a) a line parallel to line segment AB.
• (b) the bisector of the line segment AB.
• (c) perpendicular to line segment AB.
• (d) perpendicular bisector of line segment AB.

Consider the centers of circles, denoted by O and O’, which pass through the fixed points A and B.

Construction steps:

1. Connect the points A and B with a line segment.
2. Construct a perpendicular line from O to the line segment AB, and similarly, from O’ to AB.

Recall the geometric principle that a perpendicular line drawn from the center of a circle to a chord will bisect that chord.

Given that both perpendiculars bisect the segment AB, it follows that the line OO’ is both perpendicular to AB and bisects it.

Hence, Option 4 is the correct option.

Question 1(d)

A point is equidistant from the sides of an obtuse angle triangle. The point is called :

• (a) circumcenter of the triangle.
• (b) incenter of the triangle.
• (c) centroid of the triangle.
• (d) orthocenter of the triangle.

Consider an obtuse triangle labeled as ABC.

Construction steps:

1. Draw the angle bisectors AX, BY, and CZ for the angles at vertices A, B, and C respectively.
2. These angle bisectors will intersect at a point, which we will label as I.

Recall that the locus of a point that is equidistant from two intersecting lines is along the angle bisector of those lines.

∴ BY is equidistant from the sides AB and BC.

∴ AX is equidistant from the sides AB and AC.

∴ CZ is equidistant from the sides AC and BC.

∴ The intersection point I, where AX, BY, and CZ meet, is equidistant from all three sides of triangle ABC. This point is known as the incenter.

Hence, Option 2 is the correct option.

Question 1(e)

ABC is a triangle. Point P moves with vertex B as center and radius 2.8 cm. The locus of point P is :

• (a) bisector of angle ABC.
• (b) a line parallel to BC and at a distance of 2.8 cm from it.
• (c) circle with center at point B and radius = 2.8 cm.
• (d) perpendicular bisector of BC.

When a point moves such that it remains at a constant distance from a fixed point, the path it traces is the circumference of a circle. Here, the fixed point is vertex B, and the constant distance is 2.8 cm.

∴ The path traced by point P is a circle centered at point B with a radius of 2.8 cm.

Hence, Option 3 is the correct option.

Question 2

Draw an ∠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.

Construction steps:

1. Begin by drawing the line segment BC with a length of 5 cm.
2. From point B, construct a ray BX making an angle of 60° with BC. Measure 4.6 cm along this ray to locate point A, so BA = 4.6 cm.
3. Construct the angle bisector BY of ∠ABC.
4. Next, draw MN, the perpendicular bisector of the segment AB.
5. The intersection of MN and BY will give you point P.

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

Question 3

On a graph paper, draw the 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.

Steps for constructing the locus:

1. Begin by sketching the line l with the equation x = 3 and the line m with the equation y = -5 on graph paper.
2. Identify the point P where these two lines intersect.
3. Construct the line n, which serves as the angle bisector of the angle at point P.

Notice that line n is the angle bisector of ∠P, which implies that any point situated on line n maintains equal distance from both lines l and m.

Hence, the locus of the point which is equidistant from the given lines is line n.

Question 4

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.

Steps for constructing the required locus:

1. Begin by sketching line l, which is defined by x = 6. This line is parallel to the y-axis.
2. Identify points P and Q such that they are situated 3 units away on either side of the line x = 6.
3. Draw lines m and n that are parallel to line l, passing through points P and Q, respectively.

From the diagram:

The equation for line m is x = 3, and for line n, it is x = 9.

Hence, locus of the point which moves in such a way that its distance from the given line is always equal to 3 units are lines x = 3 and x = 9.

Question 5

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 △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 △QBC equal in area to △ABC, and isosceles.

(v) Measure and record the length of CQ

(i) To construct △ABC:

1. Begin by drawing a line segment BC with a length of 6 cm.
2. At point B, construct a ray BX such that it forms a 60° angle with BC. Measure and mark a point A on this ray such that BA is 9 cm.
3. Connect points A and C with a straight line.

Thus, △ABC is constructed as required.

(ii) For the locus of points equidistant from B and C:

1. Construct the perpendicular bisector of the segment BC. This bisector will intersect line BA at point M and BC at point L.
2. Any point located on line LM will be equidistant from points B and C.

Therefore, LM is the required locus.

(iii) To find the locus of vertices of triangles with BC as base and equal area to △ABC:

1. Draw a line m through point A that is parallel to BC.
2. Identify the intersection point Q where the perpendicular bisector of BC meets the parallel line m.
3. Connect Q to both B and C.

Consequently, line m serves as the locus for the vertices of triangles with BC as base and equal area to △ABC.

(iv) The triangle △QBC has an area equal to △ABC.

Since point Q lies on the perpendicular bisector of BC, it follows that QB = QC.

Hence, △QBC is an isosceles triangle.

(v) When measuring, the length of CQ is found to be 8.4 cm.

Question 6

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

Steps for constructing the required figure:

1. Begin by drawing the line segment AB with a length of 6 cm.
2. Using a compass, take A and B as centers and draw arcs with a radius of 4 cm each. These arcs will intersect at point C.
3. Connect points C to A and C to B to form triangle ABC.
4. Construct the internal bisector of ∠C and mark a point P on this bisector such that CP measures 5 cm.
5. Draw a line m that runs parallel to AB and is located 5 cm away from it.
6. With P as the center, draw arcs with a radius of 5 cm. These arcs will intersect the line m at points Q and R.
7. Finally, connect points P to Q and Q to R to complete the construction.

Question 7

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.

(i) Steps for constructing the triangle and rectangle:

1. Begin by drawing the line segment BC with a length of 5 cm.
2. Using B as the center, draw an arc BX at a 45° angle.
3. From the arc BX, mark off a segment BP measuring 4 cm.
4. Connect the points P and C. ∴, triangle BPC is formed as required.
5. Draw perpendicular lines from points B and C.
6. With P as the center and a radius equal to PC, draw an arc intersecting the perpendicular from C at a point D.
7. Using D as the center, draw a line parallel to BC that meets the perpendicular from B at point A.

Thus, rectangle ABCD is constructed as needed.

(ii) Upon measuring, we find that:

AB = 5.7 cm.

Question 8

Use ruler and compasses only for the following question. All construction lines and arcs must be clearly shown.

(i) Construct a △ABC in which BC = 6.5 cm, ∠ABC = 60° and 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 a distance of 3.5 cm from A and also equidistant from AC and BC. Measure XY.

(i) Steps for construction:

1. Begin by drawing the line segment BC = 6.5 \text{ cm}.
2. Place the compass at point B and draw an arc such that it forms an angle of 60^\circ with BC.
3. On this arc, measure and cut off a segment BA = 5 \text{ cm}.
4. Connect point A to point C to complete triangle \triangle ABC.
5. With A as the center, draw a circle that has a radius of 3.5 \text{ cm}.
6. Construct CZ, the angle bisector of \angle C.
7. Identify the intersection points of CZ with the circle centered at A, labeling these points as X and Y.

(ii) The locus of points at a distance of 3.5 \text{ cm} from A is a circle with radius 3.5 \text{ cm} centered at A.

(iii) The locus of points equidistant from AC and BC is the angle bisector CZ of \angle C.

(iv) Upon measuring, the distance XY = 4.8 \text{ cm}.