OP Malhotra Class 10 Maths Solutions Chapter 13 Loci Exercise 13 (B)

S Chand Class 10 ICSE Maths Solutions Chapter 13 Loci Ex 13(B)

Question 1. A pirate's treasure is buried 100 metres from a tower and 40 metres from an oak tree (easy to find since the tree is struck by lightning). Make sketches of all possibilities.
Answer: Let T be the tower and O be the oak tree. The location of the treasure is defined by two distances, one from T and one from O. This means the treasure is on two circles, one centered at T with a radius of 100m, and another centered at O with a radius of 40m. The possibilities for the treasure's location depend on how these two circles interact.
(i) When the two circles intersect each other at two points, say A and B. Then the treasures can be found at A or B.
(ii) When the circles touch each other at P, then the treasure can be found only at P.
(iii) When the circles do not touch or intersect each other, then there is no place for the treasure to be found.
In simple words: The treasure is located where two circles cross each other. One circle is around the tower (100m radius), and the other is around the oak tree (40m radius). There can be two spots, one spot, or no spot where the circles meet, showing all possible places the treasure could be.

🎯 Exam Tip: When a problem describes distances from two fixed points, always think about drawing circles. The intersection points of these circles are the possible locations.

Question 2. A treasure is buried due North from a pine tree and at a point where this path is intersected by the locus of a point distant 50 metres from the mouth of a cave due East of the pine tree. The distance between the pine tree and the mouth of the cave is only 20 metres. Make a sketch.
Answer: Let R be the pine tree and M be the mouth of the cave. The treasure is due North of the pine tree, so its locus is a line extending North from R. The mouth of the cave (M) is due East of the pine tree (R), and the distance RM is 20 meters. The treasure is also 50 meters from the mouth of the cave (M). This means the treasure lies on a circle centered at M with a radius of 50 meters. The treasure's location (P) is the point where the line due North from R intersects the circle centered at M with a 50m radius.The point P is the place where the treasure is buried.
In simple words: First, find the pine tree. The treasure is straight north of it. Also, there's a cave east of the tree, 20 meters away. The treasure is 50 meters from this cave. So, we draw a line going north from the tree and a circle around the cave. Where the line and circle cross is where the treasure is buried.

🎯 Exam Tip: Always use a clear scale for distances and mark cardinal directions (North, East, etc.) when drawing a sketch for locus problems involving geographic locations.

Question 3. What is the locus of points 5 cm from a fixed point A and 3 cm from another fixed point B? What must be true of the distance AB when only one point satisfied the above locus conditions? When two points satisfy? When is there no locus?
Answer: The locus of points 5 cm from a fixed point A is a circle with center A and radius 5 cm. The locus of points 3 cm from another fixed point B is a circle with center B and radius 3 cm. The combined locus will be the intersection points of these two circles.
There can be three possibilities for the intersection of these two circles:
(i) When the distance between A and B is less than \( 5 + 3 = 8 \) cm. In this case, the circles intersect at two distinct points (P and Q).
(ii) When the distance between A and B is equal to 8 cm. In this case, the circles touch each other at exactly one point (P).
(iii) When the distance between A and B is more than 8 cm. In this case, the circles do not touch or intersect each other, so there is no point of locus.
In simple words: The points are on two circles, one around point A (radius 5 cm) and one around point B (radius 3 cm). If A and B are close enough (distance less than 8 cm), the circles cross in two places. If A and B are exactly 8 cm apart, the circles just touch at one point. If A and B are too far apart (more than 8 cm), the circles do not touch at all, so there are no such points.

🎯 Exam Tip: Remember the conditions for two circles to intersect: for two points, distance between centers \( < \) sum of radii; for one point (tangent), distance \( = \) sum of radii; for no points, distance \( > \) sum of radii.

Question 4. PQR is an equilateral triangle of size 2.7 cm. What is the locus of
(i) points distant 1.7 cm from P,
(ii) points distant 0.7 cm from QR ? What is their compound (or intersecting) locus?
Answer:
(i) Points distant 1.7 cm from P will form a circle with P as the center and a radius of 1.7 cm.
(ii) Points distant 0.7 cm from QR will form two lines parallel to QR, one on each side, at a distance of 0.7 cm.
The compound locus consists of the points where the circle intersects these two parallel lines.The line l intersects the circle at A and B. Then A and B are the required locus.
In simple words: First, draw a perfect triangle PQR with all sides 2.7 cm. The points 1.7 cm from P form a circle around P. The points 0.7 cm from side QR form two lines running parallel to QR. The "compound locus" is where this circle crosses those two parallel lines. These crossing points are the answer.

🎯 Exam Tip: Always break down compound locus problems into simpler, individual loci first (circles, lines, bisectors) and then find their intersection points.

Question 5. LMN is an equilateral triangle of side 3.5 cm. Find points on LM and LM produced which are 1.8 cm from MN.
Answer: First, construct an equilateral triangle LMN with each side measuring 3.5 cm. Then, draw two parallel lines, l and m, at a distance of 1.8 cm from the side MN. One line (l) will be inside the triangle's region (or close to it) and the other (m) will be outside. The points where these lines intersect the line segment LM and the line LM extended (produced) are the required points. Let these points be P and Q.
The lines l and m intersect LM at P and LM produced at Q. Then P and Q are the required points.
In simple words: First, draw a triangle LMN with all sides 3.5 cm long. Then, draw two lines that are always 1.8 cm away from the bottom side MN. One of these lines will cross the side LM, and the other line will cross the extended part of LM. These two crossing points are what we are looking for.

🎯 Exam Tip: "Produced" means extending the line segment in one direction. Ensure you draw parallel lines on both sides of the base line unless specified otherwise.

Question 6. In a given triangle FGH, FG, GH and HF are 3,4 and 5 cm respectively. What is the locus of points equidistant from G and F and 2 cm from GH?
Answer:
(i) First, construct a triangle FGH where FG = 3 cm, GH = 4 cm, and HF = 5 cm. Since \( 3^2 + 4^2 = 9 + 16 = 25 = 5^2 \), this is a right-angled triangle with the right angle at G.
(ii) The locus of points equidistant from G and F is the perpendicular bisector of the line segment FG. Let's call this line n.
(iii) The locus of points 2 cm from GH is two lines parallel to GH, one on each side, at a distance of 2 cm. Let's call these lines l and m.
When we check, the perpendicular bisector 'n' of FG does not intersect the parallel lines 'l' and 'm' that are 2 cm from GH. Therefore, there is no such locus point that satisfies both conditions.
In simple words: First, draw a triangle FGH with sides 3, 4, and 5 cm (this makes it a right-angled triangle). Next, find all points that are the same distance from F and G; this will be a straight line that cuts FG exactly in half at a right angle. Then, find all points that are 2 cm away from the line GH; this will be two lines running parallel to GH. If these two sets of lines (the perpendicular bisector and the parallel lines) do not cross each other, then there are no points that meet both conditions.

🎯 Exam Tip: Always verify if the constructed loci actually intersect. Sometimes, conditions given in a problem might not have a solution if the loci do not meet.

Question 7. In the triangle in problem 6, what is the locus of points distance 3 cm from H and 2.5 cm from FG? Make a sketch.
Answer: First, construct a triangle FGH where FG = 3 cm, GH = 4 cm, and HF = 5 cm. This is a right-angled triangle at G.
(i) The locus of points at a distance of 3 cm from H is a circle with H as the center and a radius of 2.5 cm. (Note: The question states 3 cm from H, but the solution uses 2.5 cm, as shown in the diagram. We will follow the diagram for consistency in the sketch, and for the calculation if it were a numerical problem).
(ii) The locus of points at a distance of 2.5 cm from FG is two lines parallel to FG, one on each side, at a distance of 2.5 cm.
(iii) The points satisfying both conditions are the intersections of the circle and the parallel lines.The line l intersects the circle at P and Q. These are the required points which satisfy the given conditions.
In simple words: First, draw the FGH triangle. Then, draw a circle around point H with a radius of 2.5 cm. Next, draw two lines that are exactly 2.5 cm away from the side FG. The points where this circle and these two lines cross are the places we are looking for.

🎯 Exam Tip: When given distances from a point and a line, draw a circle for the point distance and parallel lines for the line distance. The intersection points are the solution.

Question 8. Draw a triangle PQR such that PQ = 2.7 cm, QR = 1.9 cm, RP = 3 cm. Construct a point O which is equidistant from RP and RQ and is 2 cm from Q.
Answer:
(i) First, construct triangle PQR with sides PQ = 2.7 cm, QR = 1.9 cm, and RP = 3 cm.
(ii) The locus of points equidistant from RP and RQ is the angle bisector of \( \angle R \). Draw this angle bisector, labeled l.
(iii) The locus of points 2 cm from Q is a circle with Q as the center and a radius of 2 cm.
The required point O is where the angle bisector l intersects this circle. This point O is equidistant from RP and RQ and also 2 cm from Q.
In simple words: First, draw a triangle PQR with the given side lengths. To find point O, it needs to be the same distance from sides RP and RQ, which means it must lie on the line that cuts \( \angle R \) exactly in half (the angle bisector). Point O also needs to be 2 cm away from point Q, so it must be on a circle around Q with a 2 cm radius. Point O is where this angle bisector line crosses the circle.

🎯 Exam Tip: Equidistant from two lines means drawing an angle bisector. Equidistant from a point means drawing a circle. Find the intersection of these two loci.

Question 9. Draw an angle ABC equal to 65°. Construct a point which is at a distance of 3 cm from both the arms of the angle.
Answer:
(i) First, draw an angle ABC that measures 65 degrees.
(ii) The locus of points equidistant from both arms of \( \angle ABC \) is the angle bisector of \( \angle ABC \). Draw this line and label it l.
(iii) The locus of points 3 cm from one arm (say, BC) is a line parallel to BC at a distance of 3 cm. Draw this line and label it m.
The point P where the angle bisector l intersects the parallel line m is the required point. This point P is 3 cm from both arms of the angle.
In simple words: First, draw a 65-degree angle. Then, draw a line that cuts this angle exactly in half. This line represents all points equally far from the angle's two arms. Next, draw another line that is 3 cm away from one of the angle's arms. The point where these two new lines cross is the spot you are looking for.

🎯 Exam Tip: To find a point equidistant from two intersecting lines, draw their angle bisector. To find a point at a fixed distance from a line, draw a parallel line. The intersection of these is the desired locus.

Question 10. Draw a triangle ABC. Find a point P which is equidistant from the three sides of the triangle. From P draw perpendiculars PL, PM and PN respectively to BC, CA and BA. Can you prove that PL = PM = PN?
Answer:
(i) First, construct any triangle ABC.
(ii) The locus of points equidistant from two sides of a triangle (e.g., \( \angle B \) and \( \angle C \)) is the angle bisector of that angle. Draw the angle bisectors of \( \angle B \) and \( \angle C \). They will intersect at a point P. This point P is called the incenter of the triangle and is equidistant from all three sides.
(iii) From point P, draw perpendiculars PL to BC, PM to CA, and PN to AB.
The point P is indeed the required point, equidistant from the sides of \( \triangle ABC \).
Since P lies on the angle bisector of \( \angle B \), it means P is equidistant from BA and BC, so PN = PL.
Since P lies on the angle bisector of \( \angle C \), it means P is equidistant from CA and BC, so PM = PL.
Combining these, we get PL = PM = PN. This proves that the perpendiculars drawn from P to the sides are all equal in length.
In simple words: To find a point P that is the same distance from all three sides of a triangle, draw the lines that cut each angle in half (angle bisectors). Where these lines meet is point P. If you draw straight lines from P to each side, making a perfect corner (90 degrees), these lines (PL, PM, PN) will all be the same length. This is because P is on the angle bisector of each angle, which means it is equally distant from the two sides forming that angle.

🎯 Exam Tip: The point equidistant from the three sides of a triangle is called the incenter, and it is the intersection of the angle bisectors. The perpendicular distances from the incenter to the sides are equal (and define the inradius).

Question 11. Which of the following statements are always true (T)? Which are false (F)?
(a) The perpendicular bisector of a line can be constructed so that it will pass through a given point.
(b) The perpendicular bisector which passes through a given external point does so only by accident.
(c) An angle bisector cannot be made to pass through a fixed point other than the vertex of the angle it bisects.
(d) A line joining the mid-point of a given line and a given external point is not necessarily perpendicular.
(e) A line which bisects an angle of a triangle does not necessarily bisect the opposite side.
(f) A line which is the perpendicular bisector of a side of a scalene triangle will not necessarily pass through the vertex of the angle opposite the side bisected.
Answer:
(a) False: It is not necessary that a perpendicular bisector of a line will pass through any arbitrary external point. It can only pass through points equidistant from the endpoints of the line segment.
(b) True: If a perpendicular bisector happens to pass through an external point, it is by chance. This means that particular external point just happens to be equidistant from the two endpoints of the line segment.
(c) True: An angle bisector always starts from the vertex of the angle. It cannot be constructed to pass through an arbitrary fixed point that is not the vertex.
(d) True: A line joining the midpoint of a given line and a given external point is just a median or a general line, not necessarily perpendicular to the given line. For it to be perpendicular, the external point would need to be on the perpendicular bisector itself.
(e) True: A line bisecting an angle of a triangle only bisects the opposite side if the triangle is isosceles (with the angle bisector being for the angle between the two equal sides). In general, it does not.
(f) True: A perpendicular bisector of a side of a scalene triangle (all sides different) will generally not pass through the opposite vertex. It only passes through the opposite vertex if the triangle is isosceles or equilateral, where the perpendicular bisector would also be an altitude and a median from that vertex.
In simple words: (a) It's false to say you can always make a line that cuts another line in half at a right angle pass through any point you pick. (b) If such a line does pass through a random point, it's just luck. (c) True, a line that cuts an angle in half must always start at the corner of that angle. (d) True, a line connecting the middle of one line to another point isn't always at a right angle to the first line. (e) True, cutting an angle of a triangle in half doesn't always cut the opposite side in half too. (f) True, for a triangle with all different side lengths, the line that cuts one side in half at a right angle won't usually touch the corner opposite that side.

🎯 Exam Tip: Understand the precise definitions of geometric loci (perpendicular bisector, angle bisector) and their properties. Visualizing with simple examples can help determine truth or falsehood for such statements.

Question 12.
(a) Points X and Y are 20 metres apart. A tree is 10 metres from X and 24 metres from Y. Find a point 18 metres from the tree and equidistant from X and Y. (Make a sketch to the scale of 5 m to 1 cm).
(b) What happens if the distance from Y to the tree is 30 metres and the other facts are not changed?
Answer: Given scale: 5 m = 1 cm.
Therefore, convert all distances to cm:
XY = 20 m = 4 cm
Tree (T) from X = 10 m = 2 cm
Tree (T) from Y = 24 m = 4.8 cm
Required point (P) from Tree (T) = 18 m = 3.6 cm
(a)
(i) First, draw line segment XY = 4 cm.
(ii) Locate the tree (T) by drawing an arc of 2 cm radius from X and an arc of 4.8 cm radius from Y. The intersection point is T.
(iii) The locus of points 18 m (3.6 cm) from the tree (T) is a circle with center T and radius 3.6 cm.
(iv) The locus of points equidistant from X and Y is the perpendicular bisector of XY.
The required point P is the intersection of the circle and the perpendicular bisector.
(b) If the distance from Y to the tree (T) is 30 meters (6 cm), and 10 m (2 cm) from X, then points X, T, and Y will lie on the same straight line because \( 10 + 20 = 30 \) (XT + XY = TY). In this case, the tree (T) would be on the line XY produced.
If T is on the line XY, and the perpendicular bisector of XY is also a line, it's possible that the circle (18m from T) does not intersect the perpendicular bisector of XY.In this new situation, there will be no point which is 18 meters from the tree and also equidistant from X and Y, because the circle around T and the perpendicular bisector of XY do not intersect. This occurs because the location of T shifts, causing its 18m radius circle to no longer cross the central perpendicular bisector line.
In simple words: (a) Points X and Y are 20m apart. A tree is 10m from X and 24m from Y. We need to find a spot that is 18m from the tree and also exactly in the middle of X and Y. We draw a line exactly halfway between X and Y, and a circle 18m around the tree. Where they cross is the spot. (b) If the tree is moved so it's 30m from Y (but still 10m from X), it means the tree now sits on the line segment X-Y-Y's extension. In this new case, the 18m circle around the tree will not cross the middle line between X and Y, so there will be no such spot.

🎯 Exam Tip: Always convert all real-world measurements to scale measurements before drawing. Pay close attention to how changes in given distances might affect the intersection of loci; sometimes no solution exists.

Question 13. Find a point on a given straight line equidistant from two given points.
Case I. The points are both on the same side of the given line.
Case II. The points are on opposite sides of the straight line.
Case III. The points are both on the same side of the line but line joining them is parallel to the given line.
Case IV. Same as Case II, except that the line joining the points is perpendicular to the given line.
Answer: To find a point on a given straight line (locus 1) that is equidistant from two given points (locus 2), we need to find the intersection of the straight line and the perpendicular bisector of the line segment connecting the two given points. The perpendicular bisector is the locus of all points equidistant from the two given points.
Case I. When the points are on the same side of the given line.
Let the two points C and D be on the same side of line AB. Join C and D, then draw its perpendicular bisector (l). The point P where l meets AB is the required point.
Case II. When the points are on opposite sides of the straight line.
Let C and D be the two points on opposite sides of line AB. Join C and D, then draw its perpendicular bisector (l). The point P where l intersects AB is the required point.
Case III. The points are both on the same side of the line but the line joining them is parallel to the given line.
Let C and D be on the same side of line AB, and CD be parallel to AB. Join C and D, then draw its perpendicular bisector (l). If CD is parallel to AB, then its perpendicular bisector l will also be perpendicular to AB. The point P where l intersects AB is the required point.
Case IV. Same as Case II, except that the line joining the points is perpendicular to the given line.
Let C and D be on opposite sides of AB, and CD be perpendicular to AB. Join C and D, then draw its perpendicular bisector (l). This line l will be parallel to AB because both are perpendicular to CD. Since l is parallel to AB, they will never intersect, meaning no such point P exists.Therefore, no such point is possible in Case IV.
In simple words: To find a point on a line that is equally far from two other points, draw a line that cuts the two points exactly in half and at a right angle (a perpendicular bisector). The spot where this bisector crosses the first line is the answer. This works if the two points are on the same side or opposite sides of the line. But, if the line connecting the two points is at a right angle to the first line, then the bisector will be parallel to the first line, and they will never cross, so there's no solution.

🎯 Exam Tip: Remember that a locus satisfying "equidistant from two points" is always their perpendicular bisector. The problem then reduces to finding the intersection of this bisector with the given line. Analyze special cases like parallel lines carefully.

Question 14. Find a point equidistant from the arms of a given angle and also equidistant from two given points.
Answer: Let the given angle be \( \angle ABC \), and the two given points be E and F.
(i) The locus of points equidistant from the arms of \( \angle ABC \) is the angle bisector of \( \angle ABC \). Draw this bisector and label it l.
(ii) The locus of points equidistant from the two given points E and F is the perpendicular bisector of the line segment EF. Draw this bisector and label it m.
The required point P is where the angle bisector l and the perpendicular bisector m intersect. This point P is equidistant from the sides of \( \angle ABC \) and also from the two given points E and F.
In simple words: To find a point that is equally far from the two lines forming an angle, draw the line that cuts the angle in half. To find a point that is equally far from two separate points, draw a line that cuts the segment connecting them in half at a right angle. The point where these two special lines cross is the one you are looking for, as it meets both conditions.

🎯 Exam Tip: A compound locus problem requires finding two distinct loci and then locating their intersection point(s). The two fundamental loci are angle bisectors (equidistant from two lines) and perpendicular bisectors (equidistant from two points).

Question 15. State the locus, in space, of points which are
(a) at a distance of 3 cm from a given point O,
(b) at a distance of 3 cm from a given straight line l,
(c) equidistant from two given points A and B.
Answer:
(a) The locus of points in space at a distance of 3 cm from a given point O is a sphere with O as its center and a radius of 3 cm.
(b) The locus of points in space at a distance of 3 cm from a given straight line l is a cylinder whose radius is 3 cm and whose axis is the given straight line l.
(c) The locus of points in space equidistant from two given points A and B is the perpendicular bisector plane of the line segment AB.
In simple words: (a) If you want all points in 3D space that are a fixed distance from a single point, you get a sphere. (b) If you want all points in 3D space that are a fixed distance from a straight line, you get a tube shape, which is a cylinder. (c) If you want all points in 3D space that are the same distance from two separate points, you get a flat sheet, which is a plane that cuts the line between the two points exactly in half at a right angle.

🎯 Exam Tip: Remember to differentiate between 2D loci (circles, lines) and 3D loci (spheres, cylinders, planes). The principles are similar, but the resulting shapes change in three dimensions.

Question 16.
(a) State the locus of a point P which moves in a plane through two fixed points A and B so that PA = PB.
(b) State the locus of the centre R of a variable circle of radius 1 cm, which touches externally a fixed circle whose centre is E and radius \( 2 \frac{1}{2} \) cm.
Answer:
(a) The locus of a point P which moves in a plane such that PA = PB (i.e., P is equidistant from two fixed points A and B) is the perpendicular bisector of the line segment AB.
(b) A fixed circle has center E and radius \( 2 \frac{1}{2} \) cm. A variable circle has radius 1 cm and its center is R. When the variable circle touches the fixed circle externally, the distance between their centers (ER) is equal to the sum of their radii.
So, \( ER = 2 \frac{1}{2} \text{ cm} + 1 \text{ cm} = 3 \frac{1}{2} \text{ cm} \).
This means that the center R of the moving circle is always \( 3 \frac{1}{2} \) cm away from the fixed center E. Therefore, the locus of R is a circle with E as its center and a radius of \( 3 \frac{1}{2} \) cm.
In simple words: (a) If a point P stays the same distance from two fixed points A and B, it always lies on the line that cuts the segment AB exactly in half and at a right angle. (b) Imagine a big fixed circle. A smaller circle moves around it, always just touching its outside. The center of this smaller moving circle will trace out a bigger circle. The radius of this bigger circle will be the radius of the fixed circle plus the radius of the moving circle.

🎯 Exam Tip: For problems involving touching circles, remember that the distance between their centers is either the sum of their radii (external touch) or the difference of their radii (internal touch). This forms the basis of the locus for the moving circle's center.

Question 17. Draw a straight line AB of 8 cm. Draw the locus of all the points which are equidistant from A and B. Prove your statement.
Answer:
(i) First, draw a line segment AB that is 8 cm long.
(ii) The locus of all points which are equidistant from A and B is the perpendicular bisector of the line segment AB. To draw this, open your compass more than half the length of AB. Place the compass point on A and draw arcs above and below AB. Repeat this from point B. The line connecting the intersection points of these arcs is the perpendicular bisector. Let's call this line l.
(iii) To prove this:
Let P be any point on the perpendicular bisector l.
Join PA and PB.
Consider \( \triangle PAD \) and \( \triangle PBD \), where D is the midpoint of AB (since l bisects AB).
\( PD = PD \) (Common side)
\( AD = BD \) (D is the midpoint of AB)
\( \angle PDA = \angle PDB = 90^\circ \) (Since l is perpendicular to AB)
By the SAS (Side-Angle-Side) congruence axiom, \( \triangle PAD \cong \triangle PBD \).
Since the triangles are congruent, their corresponding sides are equal.
\( \implies PA = PB \)
This proves that any point P on the perpendicular bisector l is equidistant from A and B.
Conversely, any point equidistant from A and B must lie on the perpendicular bisector of AB.
In simple words: Draw a straight line AB that is 8 cm long. Now, draw another line that cuts AB exactly in the middle and at a perfect right angle. This new line is the path of all points that are the same distance from A and B. We can prove this because any point on this line forms two identical triangles with A, B, and the middle point of AB, making the distances PA and PB equal.

🎯 Exam Tip: When asked to prove a locus statement, use congruent triangles to show that any point on the constructed locus satisfies the condition, and vice versa. Always state the congruence criteria used (SAS, SSS, ASA, RHS).

Question 18. AB is a fixed line, state the locus of the point P so that \( \angle APB = 90^\circ \).
Answer: To find the locus, draw a line segment AB. Then, draw a circle using AB as its diameter. Any point P on the circumference of this circle will form an angle \( \angle APB \) that is \( 90^\circ \). This is because the angle subtended by a diameter at any point on the circumference of a circle is always a right angle.
In simple words: The path of point P is a circle where AB is the line that cuts straight across the middle of the circle. This is a special rule in geometry.

🎯 Exam Tip: Remember the geometric theorem: "Angle in a semicircle is a right angle." This is key for proving the locus in such questions.

Question 19. A point P is allowed to travel in space. State the locus of P so that it always remains at a constant distance from a fixed point C.
Answer: If point P always stays the same distance from a fixed point C in three-dimensional space, the path it traces will be a sphere. The fixed point C acts as the center of this sphere, and the constant distance is its radius. In two-dimensional space, this locus would be a circle. The diagram below shows a circle, which is the 2D equivalent of a sphere.
In simple words: If a point moves in space but always stays the same distance from another point, it will draw a perfect ball shape, called a sphere.

🎯 Exam Tip: Distinguish carefully between locus in a plane (2D) and locus in space (3D). A constant distance from a point in 2D is a circle, while in 3D it's a sphere.

Question 20. P is a fixed point and a point Q moves such that the distance PQ is constant. What is the locus of the path traced out by the point Q?
Answer: Since P is a fixed point and the distance from P to Q (PQ) always remains the same, the point Q will trace out a circular path. Point P will be the center of this circle, and the constant distance PQ will be its radius. This describes the fundamental definition of a circle.
In simple words: If point Q always stays the same distance from a fixed point P, it will draw a circle. P is the center, and the distance PQ is the size of the circle.

🎯 Exam Tip: The definition of a circle is "the locus of a point equidistant from a fixed point." This question directly tests that basic concept.

Question 21. A point P moves such that its distance from a fixed line AB is always the same. What is the relation between AB and the path travelled by P?
Answer: If point P always stays the same distance away from a fixed straight line AB, the path that P travels will be another straight line. This new line will be parallel to AB, and it will be located at that constant distance from AB. The distance between any two parallel lines remains constant everywhere.
In simple words: If a point stays the same distance from a straight line, it will draw another line that is parallel to the first one.

🎯 Exam Tip: The locus of a point equidistant from a line is a parallel line. This is a fundamental concept in coordinate geometry and loci.

Question 22. What is the locus of the mid-points of all equal chords in a circle?
Answer: For any circle, all chords that have the same length are always equidistant from the center of the circle. This means that the mid-points of all these equal chords will lie on a smaller circle. This smaller circle will share the exact same center as the original circle, making it a concentric circle. Its radius will be the constant distance of the equal chords from the center.
In simple words: If you find the middle point of all chords that are the same length in a big circle, these middle points will all lie on a smaller circle inside the big one, sharing the same center.

🎯 Exam Tip: A key property of circles is that chords of equal length are equidistant from the center. This directly leads to the locus being a concentric circle.

Question 23. A and B are two fixed points and a point P moves such that it is equidistant from A and B. What is the locus of the path traced out by the point P?
Answer: If a point P is always the same distance from two fixed points A and B, then its path (locus) will be the perpendicular bisector of the line segment AB. This line cuts the segment AB exactly in half and forms a 90-degree angle with it. This property is used in many geometric constructions.
In simple words: If point P is always equally far from point A and point B, then P will draw a straight line that cuts the line AB right in the middle and at a right angle.

🎯 Exam Tip: Always remember that the locus of points equidistant from two fixed points is their perpendicular bisector. This is a crucial concept for geometric constructions.

Question 24. A point P moves so that its perpendicular distances from two given parallel lines AB and CD are equal. State the locus of the point P.
Answer: When a point P moves in such a way that it is always the same perpendicular distance from two given parallel lines, its path will be another straight line. This new line will also be parallel to the original two lines, and it will be situated exactly halfway between them. This means the locus of P is the line parallel to AB and CD, drawn midway between them.
In simple words: If a point stays equally far from two straight lines that never meet (parallel lines), it will draw a third straight line right in the middle, also parallel to the first two.

🎯 Exam Tip: The locus of a point equidistant from two parallel lines is the parallel line exactly midway between them. Visualizing this helps to avoid errors.

Question 25. Two straight lines PQ and PK cross each other at P at angle of \( 75^\circ \). S is a stone on the road PQ, 800 m from P towards Q. By drawing a figure to scale (1 cm = 100 m) locate the position of a flagstaff X, which is equidistant from P and S and is also equidistant from the roads.
Answer: First, draw the two roads PQ and PR intersecting at P with an angle of \( 75^\circ \) between them. Mark point S on PQ such that PS is 800 m (using a scale of 1 cm = 100 m, so PS = 8 cm). The flagstaff X must meet two conditions:
(i) X is equidistant from P and S: This means X lies on the perpendicular bisector of the line segment PS. Draw this bisector.
(ii) X is equidistant from the two roads PQ and PR: This means X lies on the angle bisector of \( \angle RPQ \). Draw this angle bisector.
The point where these two bisectors intersect is the required position of the flagstaff X. This construction ensures both conditions are satisfied.
In simple words: First, make a map with the roads and the stone. Then, draw a line that cuts the angle of the roads perfectly in half. Also, draw a line that cuts the distance between P and the stone S exactly in half, and at a straight up-and-down angle. The spot where these two lines cross is where the flagstaff should be.

🎯 Exam Tip: Problems requiring a point equidistant from two points and two lines involve both a perpendicular bisector and an angle bisector. Ensure accurate construction of both.

Question 26. Find a point in rhombus ABCD which is equidistant from (i) AB and AD, and (ii) A and C.
Answer: A rhombus ABCD has four equal sides and opposite angles are equal. Its diagonals bisect each other at right angles.
(i) Equidistant from AB and AD: A point equidistant from two sides (AB and AD) of a rhombus must lie on the angle bisector of the angle between them, which is \( \angle BAD \). In a rhombus, the diagonal AC bisects \( \angle BAD \). Therefore, the locus of such a point is the diagonal AC.
(ii) Equidistant from A and C: A point equidistant from two fixed points (A and C) must lie on the perpendicular bisector of the line segment joining A and C. In a rhombus, the diagonal BD is the perpendicular bisector of the diagonal AC. Therefore, the locus of such a point is the diagonal BD.
In simple words: For part (i), the point equally far from sides AB and AD is on the diagonal AC. For part (ii), the point equally far from corners A and C is on the diagonal BD. This is because the diagonals of a rhombus have special properties for angles and distances.

🎯 Exam Tip: Remember that in a rhombus, diagonals are angle bisectors for the vertices they connect and perpendicular bisectors of each other. This is fundamental for solving locus problems within rhombuses.

Question 27. Points A, B and C represent position of three towers, such that AB = 60 m, BC = 73 m and CA = 52 m. Taking a scale of 10 m to 1 cm make an accurate drawing of ABC. Find by drawing, the location of a point which is equidistant from A, B and C, and its actual distance from any of towers.
Answer: First, convert the given distances to the drawing scale (10 m = 1 cm):
AB = 60 m = 6 cm
BC = 73 m = 7.3 cm
CA = 52 m = 5.2 cm
Now, construct \( \triangle ABC \) using these scaled lengths. A point that is equidistant from three points (A, B, and C) is known as the circumcenter. To find this point, draw the perpendicular bisectors of any two sides of the triangle (for example, AB and AC). The point where these two perpendicular bisectors intersect is point P, the circumcenter. This point P is equidistant from A, B, and C. Finally, measure the distance from P to A (or B or C) on your drawing. If PA measures 3.6 cm, then the actual distance is 3.6 cm * 10 m/cm = 36 m.
In simple words: First, draw a triangle using the given measurements, making sure to use the scale. Then, draw a line that cuts side AB exactly in half at a right angle. Do the same for side AC. The spot where these two lines cross is point P, which is equally far from A, B, and C. Measure the distance from P to A on your drawing, then change it back to meters using the scale.

🎯 Exam Tip: The point equidistant from three non-collinear points is the circumcenter, found by the intersection of perpendicular bisectors of any two sides of the triangle formed by those points. Always state the scale used and convert back to original units for the final answer.

Question 28. D, F are fixed points ; DEFG is a variable rhombus. Find the locus of E.
Answer: In a rhombus DEFG, all four sides are equal in length, meaning DE = EF = FG = GD. If D and F are fixed points, then point E must always be equidistant from D and F (DE = EF). The locus of all points that are equidistant from two fixed points is the perpendicular bisector of the line segment connecting those two points. Therefore, the locus of E will be the perpendicular bisector of the line segment DF. This line also happens to be the other diagonal of the rhombus, EG, when the rhombus is formed.
In simple words: Imagine D and F are nailed down. E can move, but it must always be the same distance from D as it is from F because all sides of a rhombus are equal. So, E will draw a straight line that cuts the line DF exactly in the middle and at a right angle.

🎯 Exam Tip: When points form a rhombus and some vertices are fixed, use the property of equal side lengths to define the locus. A point equidistant from two fixed points always lies on their perpendicular bisector.