Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 16 Loci

Exercise 16.1

Answer 2.
Answer: The locus of path traced by point P equidistant from A and B is the perpendicular bisector of the line segment joining the two points.
In simple words: When a point moves so that it stays equal distance from two fixed points, it traces a straight line. This line cuts the line joining the two points at the middle and at right angles.

📝 Teacher's Note: Draw two points A and B on the board. Ask students to find points that are same distance from both A and B. They will see these points form a straight line in the middle.

🎯 Exam Tip: Always write "perpendicular bisector" clearly. This is the key term examiners look for. Draw a neat diagram to show the concept.

Answer 3.
Answer: The locus of a point which moves so that its distance from a fixed straight line is same is a pair of straight lines parallel to the given line, one on each side of it and at the given distance from it.
In simple words: If you want all points that are 3 cm away from a line, you get two new lines. One line is 3 cm above the original line, and one line is 3 cm below it.

[Diagram: Shows a horizontal line with two parallel lines above and below it at equal distances]

📝 Teacher's Note: Use a ruler to show a straight line on the board. Mark points 2 cm away from it on both sides. Students will see two parallel lines form.

🎯 Exam Tip: Write "pair of parallel lines" and mention "on each side". Draw both parallel lines clearly in your diagram.

Answer 4.
Answer: The locus of a point so that its perpendicular distance from two given lines is always equal is a line AB parallel to given lines L and M.
In simple words: When a point stays equal distance from two parallel lines, it moves along a straight line that runs exactly in the middle of the two given lines.

[Diagram: Shows three parallel horizontal lines with perpendicular distance markers between them]

📝 Teacher's Note: Draw two parallel lines on the board. Ask students where points would be if they are same distance from both lines. They will see it forms a line in the middle.

🎯 Exam Tip: Write "parallel to both given lines" and "equidistant from both lines". Show equal distance marks in your diagram.

Answer 5.
Answer: The locus of point P is a circle with AB as diameter.
In simple words: When a point moves so that it always makes a right angle with two fixed points, it traces a circle. The line joining the two fixed points becomes the diameter of this circle.

[Diagram: Shows a circle with diameter AB and several points P on the circle, each forming right angles with A and B]

📝 Teacher's Note: This is a very important theorem. Any point on a semicircle makes a right angle with the diameter. Use a compass to draw this and show students.

🎯 Exam Tip: Write "circle with AB as diameter" exactly. This is a standard result. Always mention that angle APB is 90 degrees.

Answer 6.
Answer:
(a) The locus of points at a distance of 4 cm from a fixed line.
The locus of points at a distance of 4 cm from fixed line AB are lines L and M which are parallel to AB.
In simple words: All points that are exactly 4 cm away from a line form two new parallel lines — one 4 cm above and one 4 cm below the original line.

[Diagram: Shows three parallel lines with 4 cm distances marked between them]

(b) The locus of points inside a circle and equidistant from two fixed points on the circle.
The locus of the points inside the circle which are equidistant from the fixed points on the circle will be the diameter which is perpendicular bisector of the line joining the two fixed points on the circle.
In simple words: Inside a circle, if you want points that are same distance from two fixed points on the circle edge, you get a straight line that cuts the circle in half through its center.

[Diagram: Shows a circle with two points A and B on the circumference and a diameter CD that is the perpendicular bisector of chord AB]

(c) The locus of the mid-points of all parallel chords of a circle.
The locus of the mid-points of all parallel chords of a circle is the diameter perpendicular to those chords.
In simple words: If you draw many parallel lines across a circle and mark the middle point of each line, all these middle points form a straight line through the center of the circle.

📝 Teacher's Note: For part (c), draw a circle and several parallel chords. Mark midpoints and show they lie on a diameter. This helps students visualize the concept.

🎯 Exam Tip: For each part, clearly state what type of locus it is — parallel lines, diameter, or circle. Use proper geometric terms like "perpendicular bisector".

Answer 7.
Answer:
(a) Midpoint of radii of a circle.
The locus of mid-point of radii of a circle is a concentric circle of radius equal to half the radius of given circle.
In simple words: If you mark the middle point of every radius of a circle, all these points form a smaller circle inside the original circle. The small circle has half the radius of the big circle.

[Diagram: Shows two concentric circles with radii marked and midpoints highlighted]

(b) Centre of a ball, rolling along a straight line on a level floor.
The locus of the centre of a ball, rolling along a straight line on a level floor will be a straight line parallel to the floor at a distance equal to the radius of the ball.
In simple words: When a ball rolls on a flat floor, its center moves in a straight line above the floor. This line is at a height equal to the ball's radius.

[Diagram: Shows a ball rolling on a flat surface with its center path marked as a parallel line above the surface]

(c) Point in a plane equidistant from a given line.
The locus of all points in a plane equidistant from a fixed line is represented by two parallel lines either side of it at a distance d away.
In simple words: All points that are the same distance away from a line form two new parallel lines — one on each side of the original line.

[Diagram: Shows three parallel horizontal lines with equal distances marked between them]

(d) Centre of a circle of varying radius and touching the two arms of ∠ABC.
In simple words: When circles of different sizes all touch both arms of an angle, their centers lie on the angle bisector — the line that divides the angle into two equal parts.

📝 Teacher's Note: For part (d), draw an angle and show circles of different sizes touching both arms. Mark their centers and show they form the angle bisector.

🎯 Exam Tip: Remember key phrases: "concentric circle" for (a), "parallel line" for (b) and (c), and "angle bisector" for (d). Always mention distances clearly.

Answer 8.
Answer:
Steps of construction:
(i) Draw a line segment BC = 3.5 cm.
(ii) With B as centre and radius 5 cm draw an arc.
(iii) With C as centre and radius 4 cm draw another arc which intersects the first arc at A.
(iv) Join AB and AC.
(v) Draw perpendicular bisector of BC.
(vi) Draw the angle bisector of angle ABC which intersects the perpendicular bisector of BC at P.

P is the required point which is equidistant from AB, BC, B and C.
The length of PB = 2.5 cm
In simple words: We first make a triangle with given side lengths. Then we find a special point P that is the same distance from all three sides of the triangle and also from two vertices.

[Diagram: Shows the constructed triangle ABC with point P marked as the intersection of perpendicular bisector and angle bisector]

📝 Teacher's Note: This combines two loci concepts — perpendicular bisector (equidistant from points) and angle bisector (equidistant from lines). Show each step clearly with compass and ruler.

🎯 Exam Tip: Write each construction step clearly with proper geometric terms. Mention "perpendicular bisector" and "angle bisector" exactly as shown.

Answer 9.
Answer:
Steps of construction:
(i) Draw a line segment BC = 7.3 cm.
(ii) With B as centre and radius 6 cm draw an arc.
(iii) With C as centre and radius 5.2 cm draw another arc which intersects the first arc at A.
(iv) Join AB and AC.
(v) Draw perpendicular bisector of BC, AB and AC.

In triangle ABC, P is the point of intersection of AB, AC and BC.
Therefore, PA = PB, PB = PC, PC = PA.
Thus, circum-centre of a triangle is the point which is equidistant from all its vertices.
In simple words: The circumcenter is the center of a circle that passes through all three corners of a triangle. It is equidistant from all three vertices of the triangle.

[Diagram: Shows triangle ABC with point P as the circumcenter and perpendicular bisectors of all three sides intersecting at P]

📝 Teacher's Note: The circumcenter is where all three perpendicular bisectors meet. From this point, you can draw a circle through all three vertices of the triangle.

🎯 Exam Tip: Write "circumcenter" clearly and state that it is "equidistant from all vertices". Show that PA = PB = PC in your answer.

Answer 10.
Answer:

[Diagram: This diagram shows two roads AB and CD crossing at angle 75°, with point P on one road at distance 800m from intersection, and construction lines showing how to find point O equidistant from P, X and both roads.]

Steps of construction:

1. Draw two lines AB and CD crossing at an angle of 75°
2. Draw an angle bisector for ∠BPD
3. Draw perpendicular from X on angle bisector meeting at O.
4. From point Y, PX=PY, draw a perpendicular on angle bisector meeting at O.
5. O is the point which is equidistant from P, X and both the roads.

Calculation:
\( \cos \theta = \frac{\text{hypotenuse}}{\text{base}} \)
\( \cos \frac{75}{2} = \frac{PO}{PX} \)
\( \cos(37.5) = \frac{PO}{800} \)
\( 0.980243 = \frac{PO}{800} \)
PO = 784.19m

In simple words: We need to find a point that is same distance from P, X and both roads. We use angle bisector method and then calculate using cosine formula with the given angle.

📝 Teacher's Note: Show students that angle bisectors help find points equidistant from two lines. The perpendicular distance concept is key here.

🎯 Exam Tip: Always draw the angle bisector first. Then use trigonometry to find exact distances. Show all calculation steps clearly.

Answer 11.
Answer:

[Diagram: This diagram shows two intersecting lines m and n with angle bisectors PQ and RS, and a circle of radius 2 cm drawn from center that intersects the angle bisectors at points a, b, c and d.]

Draw an angle bisector PQ and RS of angles formed by the lines m and n. From centre draw a circle with radius 2 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: We draw angle bisectors of the angles made by two crossing lines. Then we draw a circle from the center. Where the circle cuts the angle bisectors, we get our four points.

📝 Teacher's Note: Explain that angle bisectors of intersecting lines give us four rays. A circle centered at intersection will cut these rays at equal distances.

🎯 Exam Tip: Draw both angle bisectors clearly. Mark all four intersection points with the circle. Label them as required.

Answer 12.
Answer:

[Diagram: This diagram shows angle bisectors of AB and CD drawn as perpendiculars, with point P marked at distance 1.8 cm from line EF, showing the construction method.]

Draw angle bisector of AB and CD. Draw perpendiculars from AB and CD on angle bisector, say P. P is the required point which is equidistant from AB and CD and at a distance of 1.8 cm from EF.

In simple words: We find the angle bisector first. Then we mark a point on it that is exactly 1.8 cm away from line EF. This point is same distance from both lines AB and CD.

📝 Teacher's Note: Show students that any point on an angle bisector is equidistant from the two lines forming the angle. Use ruler to measure the 1.8 cm distance.

🎯 Exam Tip: Draw the angle bisector first. Then measure exactly 1.8 cm from EF to mark point P. Show that P is equidistant from AB and CD.

Answer 13.
Answer:

[Diagram: This diagram shows the construction of a quadrilateral ABCD with given measurements - AB=6cm, AD=6cm, BC=3.6cm, CD=5cm, and point P marked as the intersection of perpendicular bisectors.]

Steps of construction:

1. Draw a line AB = 6 cm.
2. Draw a ray making an angle of 45° with AB.
3. With a as centre, draw AD = 6 cm on the ray.
4. Draw an angle bisector of angle BAD.
5. With B as centre cut an arc BC = 3.6 cm on the angle bisector.
6. With D as centre cut an arc CD = 5 cm on the angle bisector. ABCD is the required quadrilateral.
7. Join BD.
8. Draw perpendicular bisectors of CD and BC which meet BD on P. P is the required point.

In simple words: We construct a quadrilateral with the given side lengths. Then we find the point P by drawing perpendicular bisectors of two sides that meet on the diagonal BD.

📝 Teacher's Note: Emphasize that perpendicular bisectors help find points equidistant from endpoints of a line segment. Show the construction step by step.

🎯 Exam Tip: Follow the construction steps in order. Make sure all measurements are accurate. Mark point P clearly where the perpendicular bisectors intersect.

Answer 14.
Answer:

[Diagram: This diagram shows a rhombus ABCD construction with AC=6cm and all sides=5cm, with perpendicular bisector of BC meeting AD at point R, where RB=RC=1.2cm.]

Steps of Construction:

1. Draw AC = 6 cm.
2. With A as centre, draw two arcs of 5 cm on both sides of line AC.
3. With C as centre, draw two arcs of 5 cm on both sides of line AC.
4. All the arcs meet at B and D. Join AB, AD, BC and BD. ABCD is the required rhombus.
5. On measuring, ∠ABC = 78°.
6. Draw perpendicular bisector of BC meeting AD at R. R is the point equidistant from B and C, hence RB = RC.
7. On measuring, R = 1.2 cm

In simple words: We construct a rhombus using the diagonal and side length. Then we draw the perpendicular bisector of one side to find point R that is equidistant from the endpoints of that side.

📝 Teacher's Note: Show that in a rhombus, all sides are equal. The perpendicular bisector of any side will pass through points equidistant from that side's endpoints.

🎯 Exam Tip: Use compass to draw equal arcs for rhombus construction. Measure angles and distances accurately. Mark point R where perpendicular bisector intersects the opposite side.

Answer 16.
Answer:

[Diagram: This diagram shows triangle PQR construction with PQ=5.5cm, and arcs drawn from P (radius 4.8cm) and Q (radius 3.2cm) meeting at R, with perpendicular bisector of PR and arc from Q (radius 2.5cm) intersecting at point O.]

Steps of construction:

1. Draw PQ = 5.5 cm
2. With P as centre and radius 4.8 cm draw an arc.
3. With Q as centre and radius 3.2 cm cut another arc which meets the first arc at R. Join PR and QR. PQR is the required triangle.
4. Draw perpendicular bisector of PR.
5. Q as centre and radius as 2.5 cm, draw an arc which intersects the perpendicular bisector of PR at O.

O is the required point which is at a distance of 2.5 cm from Q.

In simple words: We first construct triangle PQR with given side lengths. Then we find point O using perpendicular bisector method so that O is exactly 2.5 cm away from point Q.

📝 Teacher's Note: Show students how to use compass to construct triangles with given side lengths. Perpendicular bisectors help find equidistant points.

🎯 Exam Tip: Draw the triangle accurately first. Then use perpendicular bisector and arc intersection to find the required point O. Mark all distances clearly.

Answer 17.
Answer:
Steps of Construction:
(i) Draw YZ = 5 cm
(ii) Draw an arc with angle Y = 120° and radius 4 cm.
(iii) Join XZ.
(iv) Draw perpendicular bisector of YZ.
(v) With X as centre and angle X as 90°, join X to the perpendicular bisector at T. T is the required point.
(vi) Measure TY. TY = 6.8 cm = TZ as T lies on perpendicular bisector of YZ.

In simple words: We first draw the base line. Then we make angles and join points. The perpendicular bisector helps us find the special point T that is equal distance from Y and Z.

[Diagram: This diagram shows a triangle construction with point T on the perpendicular bisector of side YZ, with measurements and construction lines marked.]

📝 Teacher's Note: Show students how to use a compass for the arc and a protractor for the 120° angle. The perpendicular bisector is the key - any point on it is equal distance from both ends.

🎯 Exam Tip: Always write "perpendicular bisector" clearly. Mark all points with letters. Show your measurement of TY = TZ to prove T is equidistant.

Answer 18.
Answer:
Steps of Construction:
(i) Draw line segment PQ.
(ii) With P and Q as centre draw intersecting arcs at R.
(iii) Join PR and RQ.
(iv) Draw angle bisector of angle Q.
(v) Draw perpendicular bisectors of PQ and RQ which meet the angle bisector at S. S is the required point.
(vi) In △QSY and △QSX
SQ = SQ
∠SQY = ∠SQX
∠SYQ = ∠SXQ = 90 degrees.
Therefore, △QSY and △QSX are congruent.
Hence, SY = SX and therefore S is equidistant from PQ and RQ.

In simple words: We make a triangle first. Then we find the special point S that is equal distance from two sides of the triangle. This uses angle bisectors and perpendicular bisectors together.

[Diagram: This diagram shows triangle PQR with point S inside, where S is equidistant from sides PQ and RQ, with construction lines and perpendicular marks shown.]

📝 Teacher's Note: Explain that S lies on the angle bisector, so it is equal distance from the two sides. Use the congruent triangles to prove this property step by step.

🎯 Exam Tip: Write the congruence proof clearly with all three conditions - SAS or ASA. Always state "S is equidistant from PQ and RQ" as your final conclusion.

Answer 20.
Answer:
A is equidistant from B and D. Therefore, A lies on perpendicular bisector of BD.
C is equidistant from B and D. Therefore, C lies on perpendicular bisector of BD.
A and C both lie on perpendicular bisector of BD.
Hence, AC is perpendicular bisector of BD.

In simple words: When two points are equal distance from the ends of a line, those two points must lie on the perpendicular bisector. So line AC is the perpendicular bisector of line BD.

📝 Teacher's Note: Use a simple example - if two students stand equal distance from two corners of the classroom, they must be standing on an invisible line that cuts the classroom wall at right angles in the middle.

🎯 Exam Tip: Write "equidistant" clearly and state the perpendicular bisector property. Always conclude with "Hence, AC is perpendicular bisector of BD."

Answer 21.
Answer:
A is equidistant from B and D. Therefore, A lies on perpendicular bisector of BD.
C is equidistant from B and D. Therefore, C lies on perpendicular bisector of BD.
A and C both lie on perpendicular bisector of BD.
Hence, AC is perpendicular bisector of BD.
Since AC is perpendicular bisector of BD so ∠AMB = ∠AMD = right angle.

In simple words: Same as Answer 20, but here we also prove that the angles formed are 90 degrees. This makes AC cut BD at right angles.

📝 Teacher's Note: Show that when a line is a perpendicular bisector, it always makes 90° angles. Draw this clearly on the board with a right angle symbol.

🎯 Exam Tip: Write the right angle conclusion clearly. Use the symbol ∠ and write "= 90°" or "= right angle". This gets you the extra marks.

Answer 22.
Answer:
Since I lies on bisector of ∠R, I is equidistant from PR and QR.
Again I lies on the bisector of ∠Q, I is equidistant from PQ and QR.
Hence, I is equidistant from all sides of the triangle.
Therefore, I lies on the bisector of ∠P i.e ∠QPR

In simple words: Point I is special - it is equal distance from all three sides of the triangle. This means it must lie on all three angle bisectors. The three angle bisectors meet at one point.

[Diagram: This diagram shows triangle PQR with point I inside, where all three angle bisectors meet at I, making it equidistant from all three sides.]

📝 Teacher's Note: Explain that I is the incenter of the triangle. If you put a coin at I, it touches all three sides equally. The three angle bisectors always meet at one point.

🎯 Exam Tip: Write "equidistant from all sides" clearly. State that I lies on the bisector of the third angle. This proves all three angle bisectors meet at one point.

Answer 23.
Answer:
Since O lies on the perpendicular bisector of AB, O is equidistant from A and B.
OA = OB........(i)
Again, O lies on the perpendicular bisector of AC, O is equidistant from A and C.
OA = OC........(ii)
From (i) and (ii)
OB = OC
Now in △OBM and △OCM,
OB = OC (proved)
OM = OM
BM = CM (M is mid-point of BC)
Therefore, △OBM and △OCM are congruent
∠OMB = ∠OMC
But BMC is a straight line, so
∠OMB = ∠OMC = 90°
Thus, OM meets BC at right angles.

In simple words: Point O is equal distance from all three corners A, B, and C. This means O is the center of a circle that passes through all three points. The line from O to any side makes a right angle.

[Diagram: This diagram shows triangle ABC with point O inside, where O is equidistant from all three vertices, and OM is perpendicular to side BC.]

📝 Teacher's Note: Point O is called the circumcenter. It is where you can draw a circle that passes through all three corners. Show students how to find this by drawing perpendicular bisectors.

🎯 Exam Tip: Prove the triangles are congruent using SSS. Write "∠OMB = ∠OMC = 90°" clearly. State that OM is perpendicular to BC as your final answer.

Answer 26.
Answer:
Steps of Construction:
(i) ABC is the required triangle.
(ii) Draw perpendicular bisector of BC which intersects BA in M, then any point on LM is equidistant from B and C.
(iii) Through A, draw a line m || BC.
(iv) The perpendicular bisector of BC and the parallel line m intersect each other at Q.
(v) 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.

In simple words: We want to find all points that make triangles with the same area as triangle ABC. These points lie on a line parallel to the base BC and passing through vertex A.

[Diagram: This diagram shows triangle ABC with a parallel line through A, and the construction to find the locus of points that create triangles with equal area to triangle ABC.]

📝 Teacher's Note: Explain that triangles with the same base and height have equal areas. The parallel line keeps the height constant, so any point on it gives the same area.

🎯 Exam Tip: Write "locus of all points" clearly in your answer. State that the line is parallel to BC and passes through A. Mention that all triangles formed have equal area.