Locus ML Aggarwal Solutions ICSE Maths Class-10

Concept of Locus with Construction Chapter-14 APC Understanding Solved Exercise/ Questions

Locus ML Aggarwal Solutions ICSE Maths Class-10 Chapter-14 . We Provide Step by Step Answer of Exercise-14 Locus , with  Chapter-Test Questions  / Problems related  for ICSE Class-10 APC Understanding Mathematics  .  Visit official Website CISCE  for detail information about ICSE Board Class-10.

Locus ML Aggarwal Solutions ICSE Maths Class-10 Chapter-14

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Exe – 14

Chapter – Test

How to Solve Locus Problems/Questions / Exercise of ICSE Class-10 Mathematics

Before viewing Solution of Chapter -14 Locus of ML Solution read the Chapter Carefully then solve all example of your text book.

For more practice on Locus related problems /Questions / Exercise try to solve Locus  exercise of other famous publications also such as Goyal Brothers Prakshan (RS Aggarwal ICSE)  / Concise Selina Publications ICSE  Mathematics. Get the formula of Locus  for ICSE Class 10 Maths  to understand the topic more clearly in effective way.

Chapter – 14, “Locus” ML Aggarwal Solution ICSE Maths Class 10

(Page 304)

Question 1.

A point moves such that its distance from a fixed line AB is always the same. What is the relation between AB and the path traveled by P ?

Let point P moves in such a way that
it is at a fixed distance from the fixed line AB.

∴ It is a set of two lines l and m parallel to AB

drawn on either side of it at equal distance from it.

Question 2.

A point P moves so that its perpendicular distance from two given lines AB and CD are equal. State the locus of the point P.

(i) When two lines AB and CD are parallel,
then the locus of the point P which is equidistant
from AB and CD is a line (l)
in the midway of AB and CD and parallel to them

(ii) If AB and CD are intersecting lines,
then the locus of the point P will be a
pair of the straight lines l and m which bisect
the angles between the given lines AB and CD.

Question 3.

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 ?

∴ P is a fixed point and Q is a moving point
such that it is always at an equidistant from P.

∴ P is the centre of the path of Q which is a circle.
The distance between P and Q is the radius of the circle.
Hence locus of point Q is a circle with P as centre.

Question 4.

(i) AB is a fixed line. State the locus of the point P so that ∠APB = 90°.
(ii) A, B are fixed points. State the locus of the point P so that ∠APB = 60°.

(i) AB is a fixed line and P is a point
such that ∠APB = 90°.

The locus of P will be the circle whose diameter is AB.
We know that the angle in a semi-circle is always equal to 90°.
∠APB = 90°
(ii) AB is a fixed line and P is a point such that ∠APB = 60°.
The locus of P will be a major segment of a circle whose AB is a chord.

Question 5.

Draw and describe the locus in each of the following cases :

(i) The locus of points at a distance 2.5 cm from a fixed line.
(ii) The locus of vertices of all isosceles triangles having a common base.
(iii) The locus of points inside a circle and equidistant from two fixed points on the circle.
(iv) The locus of centers of all circles passing through two fixed points.
(v) The locus of a point in rhombus ABCD which is equidistant from AB and AD. (1998)
(vi) The locus of a point in the rhombus ABCD which is equidistant from points A and C.

1. Draw a given line AB.
2. Draw lines of l and m parallel to AB at a distance of 2.5 cm.
Lines l and m are the locus of point P which is at a distance of 2.5 cm.

(ii)

∆ABC is an isosceles triangle in which AB = AC.
From A, draw AD perpendicular to BC.
AD is the locus of the point A the vertices of ∆ABC.
In rt. ∆ABD and ∆ACD
Hyp. AB = AC (given)
∴ ∆ABD = ∆ACD (R.H.S. Axiom)
∴ BD = DC (c.p.c.t.)
Hence locus of vertices of isosceles triangles
having common base is the perpendicular bisector of BC.

(iii)

(i) Draw a circle with centre O.

(ii) Take points A and B on it and join them.
(iii) Draw a perpendicular bisector of AB
which passes from O and meets the circle at C.
CE the diameter, which is the locus of a point inside the circle
and equidistant from two points A and B at the circle.
(iv) Let C1, C2, C3 be the centers of the circle
which pass through the two fixed points A and B.

Draw a line XY passing through these centres C1, C2, C3.
Hence locus of centres of circles passing through two points A and B

is the perpendicular bisector of the line segment joining the two fixed points.

(v)

In rhombus ABCD, Join AC.

AC is the diagonal of rhombus ABCD
∴ AC bisect ∠A.
∴ Any point on AC, is the locus which is equidistant from AB and AD.
(vi) ABCD is a rhombus. Join BD

BD is the locus of a point in the rhombus which is equidistant from A and C.
Diagonal BD bisects  ∠B and ∠D.
Any point on BD will be equidistant from A and C.

Question 6.

Describe completely the locus of points in each of the following cases :

(i) mid-point of radii of a circle.
(ii) centre of a ball, rolling along a straight line on a level floor.
(iii) point in a plane equidistant from a given line.
(iv) point in a plane, at a constant distance of 5 cm from a fixed point (in the plane).
(v) centre of a circle of varying radius and touching two arms of ∠ADC.
(vi) centre of a circle of radius 2 cm and touching a fixed circle of radius 3 cm with centre 0.

(i) The locus of midpoints of the radii of a circle
is another concentric circle with radius is
half of the radius of the given circle.

(ii)

AB is the straight line on the ground and the ball is rolling on it
∴ locus of the centre of the ball is a line parallel A lo the given line AB.

(iii)

AB is the given line and P is a point in the plane.

From P, draw a line CD and another line EF from P’ parallel to AB.
Thus CD and EF are the lines which are the locus of the point equidistant from AB

(iv)

Take a point O and another point P such that OP = 5 cm.
with centre O and radius equal to OP, draw a circle.
Thus this circle is the locus of point P
which is at a distance of 5 cm from O, the given point.

(v)

Draw the bisector BX of ∠ABC.
This bisector of an angle is the locus of the centre of a circle with different radii.
Any point on BX, is equidistant from the arms BA and BC of the ∠ABC.

(vi)

(a) If the circle with 2 cm as radius touches the given circle
externally then the locus of the centre of the circle
will be a concentric circle with radius 3+2 = 5 cm.

If the circle with 2 cm as radius touches the given circle with 3 cm as radius internally,
then the locus of the centre of the circle will be a concentric circle with radius 3-2 = 1 cm.

(Page 305)

Question 7.

Using ruler and compasses construct :    (2009)
(i) a triangle ABC in which AB = 5.5 cm, BC = 3.4 cm and CA = 4.9 cm.
(ii) the locus of points equidistant from A and C.

(i) Draw BC = 3.4 and mark the arcs of 5.5 add 4.9 cm from B and C.
Join A, B and C.
ABC is the required triangle.
(ii) Draw ⊥ bisector of AC.
(iii) Draw an angle of 90° at AB at A which intersects ⊥ bisector at O.
Draw circle taking 0 as centre and OA as the radius.

Question 8.

Construct triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ABC = 60°. Locate by construction the point P such that :

(i) P is equidistant from B and C and
(ii) P is equidistant from AB and BC
(iii) Measure and record the length of PB. (2000)

(i) Take BC = 8 cm a long line segment. At B,
draw a ray BX making an angle of 60° with BC.
Cut off BA = 7 cm. and join AC.

(i) Draw the perpendicular bisector of BC.
(ii) Draw the angle bisector of ∠B which intersect
the perpendicular bisector of BC at P. P is the required point.
(iii) On measuring the length of BP = 4.6 cm (approx.)

Question 9.

A straight line AB is 8 cm long. Locate by construction the locus of a point which is :

(i) Equidistant from A and B.
(ii) Always 4 cm from the line AB.
(iii) Mark two points X and Y, which are 4 cm from AB and equidistant from A and B. Name the figure AXBY.(2008)

Steps of construction:

(i) Draw a line segment AB = 8 cm.
(ii) With the help of compasses and ruler,
draw the perpendicular bisector l of AB which intersects AB at O.
(iii) Then any point on l, is equidistant from A and B.
(iv) Cut off OX = OY = 4 cm. The X and Y are the required loci,
which is equidistant from AB and also from A and B.
(v) Join AX, XB, BY and YA.
The figures so formed AXBY is the shape of a square because
its diagonals are equal and bisect each other at right angles.

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 arid record the length of PB. (2010)

In ∆ABC, AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°

Steps of construction :

(a) (i) Draw a line segment BC = 6 cm
(ii) At, B draw a ray BX making an angle of 60° and cut off BA = 3.5 cm.
(iii) Join AC.
The ∆ABC is the required triangle.
(b) Draw the bisector BY of ∠ABC.
(c) Draw the perpendicular bisector of BC which intersects BY at P.
P is the required point P which is equidistant from BC and BA
and also equidistant from B and C.
On measuring PB it is 3.4 cm (approx.)

Question 11.

Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°. Hence:
(i) Construct the locus of points equidistant from BA and BC.
(ii) Construct the locus of points equidistant from B and C.
(iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC. (2015)

Steps of construction :

Construct the triangle ABC with AB = 5.5 cm
∠BAC = 105° and AC = 6 cm
(i) Points which are equidistant from BA and BC lies on the bisector of ∠ABC.
(ii) Points equidistant from B and C lies on the perpendicular bisector of BC.
Draw perpendicular bisector of BC.
The required point P is the point of intersection of the bisector of
∠ABC and the perpendicular bisector of BC.
(iii) Required length of PC = 4.8 cm.

Question 12.

Points A, B and C represent position of three towers such that AB = 60 m, BC = 73m 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 the towers.

AB = 60 mm = 6.0 cm, BC = 73 mm = 7.3 cm
and CA = 52 mm = 5.2 cm.

(i) Draw a line segment BC = 7.3cm
(ii) With Centre B and radius 6cm and with centre C
and radius 5.2 cm, draw two arcs intersecting each other at A
(iii) Joining AB and AC.
(iv) Draw the perpendicular bisector of AB, BC and CA respectively,
which intersect each other at point P. Join PB.
P is equidistant from A, B and C on measuring PB = 3.7 cm.
Actual distance = 37 m.

Question 13.

Draw two intersecting lines to include an angle of 30°. Use ruler and compasses to locate points which are equidistant from these lines and also 2 cm away from their point of intersection. How many such points exist ?

(i) Two lines AB and CD intersect each other at O.

(ii) Draw the bisector of ∠BOD and ∠AOD
(iii) With centre O and radius equal to 2 cm.
marks points on the bisector of angles at P, Q, R and S respectively.
Hence there are four points which are equidistant
from AB and CD and 2 cm from 0, the point of intersection of AB and CD.

Question 14.

Without using set square or protractor, construct the quadrilateral ABCD in which ∠BAD = 45°, AD = AB = 6 cm, BC = 3.6 cm and CD = 5 cm.
(i) Measure ∠BCD.
(ii) Locate the point P on BD which is equidistant from BC and CD. (1992)

(i) Take AB = 6 cm long
(ii) AT A, draw the angle of 45° and cut off AD = 6 cm

(iii) With centre D and radius 5 cm and with centre B,
and radius 3.5 cm draw two arcs intersecting each other at C.
(iv) Join CD and CB and join BD
(v) On measuring ∠BCD = 65°.
(vi) Draw the bisector of ∠BCD which intersects BD at P.
P is the required point which is equidistant from CD and CB.

Question 15.

Without using set square or protractor, construct rhombus ABCD with sides of length 4 cm and diagonal AC of length 5 cm. Measure ∠ABC. Find the point R on AD such that RB = RC. Measure the length of AR.

(i) Take AB = 4 cm
(ii) With centre A, draw an arc of 5 cm radius
and with B draw another arc of radius 4 cm intersecting each other at C.
(iii) Join AC and BD.
(iv) Again with centre A and C,
draw two arcs of radius 4 cm intersecting each other on D.
ABCD is the required rhombus and on measuring the ∠ABC, it is 78°.
(vi) Draw perpendicular bisector of BC intersecting AD at R.
On measuring the length of AR, it is equal to 1.2 cm.

(Page 306)

Question 16.

Without using set-squares or protractor construct :

(i) Triangle ABC, in which AB = 5.5 cm, BC = 3.2 cm and CA = 4.8 cm.
(ii) Draw the locus of a point which moves so that it is always 2.5 cm from B.
(iii) Draw the locus of a point which moves so that it is equidistant from the sides BC and CA.
(iv) Mark the point of intersection of the loci with the letter P and measure PC.

Steps of Construction :
(i) Draw BC = 3.2 cm long.
(ii) With centre B and radius 5.5 cm
and with centre C and radius 4.8 cm
draw arcs intersecting each other at A.

(iiii) Join AB and AC.
(iv) Draw the bisector of ∠BCA.
(v) With centre B and radius 2.5 cm,
draw an arc intersecting the angle bisector of ∠BCA at P and P’.
P and P’ are two loci which satisfy the given condition.
On measuring CP and CP’
CP = 3.6 cm and CP’ =1.1 cm.

Question 17.

By using ruler and compasses only, construct an isosceles triangle ABC in which BC = 5 cm, AB = AC and ∠BAC = 90°. Locate the point P such that :

(i) P is equidistant from the sides BC and AC.
(ii) P is equidistant from the points B and C.

Steps of Construction :

(i) Take BC = 5.0 cm and bisect it at D.

(ii) Taking BC as diameter, draw a semicircle.
(iii) At D, draw a perpendicular intersecting the circle at A
(iv) Join AB and AC.
(v) Draw the angle bisector of C intersecting
the perpendicular at P. P is the required point.

Question 18.

Using ruler and a compass only construct a semi-circle with diameter BC = 7 cm. Locate a point A on the circumference of the semicircle such that A is equidistant from B and C. Complete the cyclic quadrilateral ABCD, such that D is equidistant from AB and BC. Measure ∠ADC and write it down.

Steps of Construction :

1. Draw a line segment BC = 7 cm.
2. Draw its perpendicular bisector 1 and let it intersect BC in M.
3. With M as centre and radius equal to BM or CM, draw a semi-circle and let the semi-circle intersect the perpendicular bisector of line segment BC in A. Join BA.
4. Draw the angle bisector of ∠ABC and let it intersect the semi-circle in D.

Question 19.

Using ruler and compasses only, construct a quadrilateral ABCD in which AB = 6 cm, BC = 5 cm, ∠B = 60°, AD = 5 cm and D is equidistant from AB and BC. Measure CD.

Steps of Construction :
(i) Draw AB = 6 cm

(ii) At B, draw angle of 60° and cut off BC = 5 cm.
(iii) Draw the angle bisector of ∠B.
(iv) With centre A and radius 5 cm. draw an arc
which intersects the angle bisector of ∠B at D
On measuring CD, it is 5.3 cm (approx.).

Question 20.

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 (2001)

Steps of Construction :
(i) Draw a line AB = 6 cm

(ii) With centre A and B and radius 4 cm,
draw two arcs intersecting each other at C.
(iii) Join CA and CB
(iv) Draw the bisector of ∠C and cut off CP = 5 cm
(v) Draw a line XY parallel to AB at a distance of 5 cm.
(vi) From P, draw arcs of radius 5 cm each intersecting
the line XY at Q and R. Hence Q and R are the required points.

Question 21.

Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of length 6 cm and 5 cm respectively.

(i) Construct the locus of points, inside the circle, that are equidistant from A and C. Prove your construction.
(ii) Construct the locus of points, inside the circle, that are equidistant from AB and AC.

Steps of Construction :
(i) With centre 0 and radius 4 cm draw a circle.
(ii) Take point A on this circle.

(iii) With centre A and radius 6 cm draw an arc cutting the circle at B.
(iv) Again with radius 5 cm, draw another arc cutting the circle at C.
(v) Join AB and AC.
(vi) Draw the perpendicular bisector of AC.
Any point on it, will be equidistant from A and C.
(vii) Draw the angle bisector of ∠A intersecting
the perpendicular bisector of AC at P. P is the required locus.

Question 22.

Ruler and compasses only may be used in this question. All construction lines and arcs must be clearly shown, and be of sufficient length and clarity to permit assessment.

(i) Construct a triangle ABC, in which BC = 6 cm, AB = 9 cm. and ∠ABC = 60°.
(ii) Construct the locus of all points, inside ∆ABC, which are equidistant from B and C.
(iii) Construct the locus of the vertices of the triangles with BC as base, which are equal in area to ∆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.

Steps of Construction :
(i) Draw AB = 9 cm
(ii) At B draw an angle of 60° and cut off BC = 6 cm.
(iii) Join AC

(iv) Draw perpendicular bisector of BC.
All points on it will be equidistant from B and C.
(v) From A, draw a line XY parallel to BC.
(vi) Produce the perpendicular bisector of BC to meet XY in Q.
(vii) Join QC and QB.
∆QBC will be the triangle equal in area to ∆ABC
because these are on the same base BC and between the same parallel lines.
On measuring, the length of CQ is 8.2 cm (approx.).

Chapter – Test , “Locus” ML Aggarwal Solutions ICSE Maths Class 10

(Page 307)

Question 1.

Draw a straight line AB of length 8 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.

(i) Draw a line segment AB = 8 cm.

(ii) Draw the perpendicular bisector of AB intersecting AB at D.
∴ Every point P on it will be equidistant from A and B.

(iii) Take a point P on the perpendicular bisector.

(iv) Join PA and PB.
PD = PD (common)
AD = BD (D is mid-point of AB)
∠PDA = ∠PDB (each 90°)
∴ ∆ PAD ≅ ∆ PBD (SAS axiom of congruency)
∴PA = PB (c.p.c.t.)
Similarly, we can prove any other point on the
perpendicular bisector of AB is equidistant from A and B.
Hence Proved.

Question 2.

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.

The point P is moving in the space and
it is at a constant distance from a fixed point C.
∴ Its locus is a sphere.

Question 3.

Draw a line segment AB of length 7 cm. Construct the locus of a point P such that area of triangle PAB is 14 cm².

Base of ∆PAB = 7 cm
and its area = 14 cm²

Now draw a line XY parallel to AB at a distance of 4 cm.
Now take any point P on XY
Join PA and PB
area of ∆PAB = 14 cm.
Hence locus of P is the line XY
which is parallel to AB at a distance of 4 cm.

Question 4.

Draw a line segment AB of length 12 cm. Mark M, the mid-point of AB. Draw and describe the locus of a point which is
(i) at a distance of 3 cm from AB.
(ii) at a distance of 5 cm from the point M.

Mark the points P, Q, R, S which satisfy both the above conditions. What kind of quadrilateral is PQRS ? Compute the area of the quadrilateral PQRS.

Steps of Construction :

(i) Take a line AB = 12 cm
(ii) Take M, the midpoint of AB.
(iii) Draw straight lines CD and EF parallel to AB at a distance of 3 cm.
(iv) With centre M and radius 5 cm,
draw areas which intersect CD at P and Q and EF at R and S.
(v) Join QR and PS.
PQRS is a rectangle where the length PQ = 8 cm.
Area of rectangle PQRS = PQ x RS = 8 x 6 = 48 cm²

Question 5.

AB and CD are two intersecting lines. Find the position of a point which is at a distance of 2 cm from AB and 1.6 cm from CD.

(i) AB and CD are the intersecting lines which intersect each other at O.

(ii) Draw a line EF parallel to AB and GH parallel to CD intersecting each other at P
P is the required point.

Question 6.

Two straight lines PQ and PK cross each other at P at an angle of 75°. 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 road.

1 cm = 100 cm
800 m = 8 cm.
Steps of Construction :
(i) Draw the lines PQ and PK intersecting each other
at P making an angle of 75°.

(ii) Take a point S on PQ such that PS = 8 cm.
(iii) Draw the perpendicular bisector of PS.
(iv) Draw the angle bisector of ∠KPS intersecting
the perpendicular bisector at X.
X is the required point which is equidistant from P and S
and also from PQ and PK.

Question 7.

Construct a rhombus PQRS whose diagonals PR, QS are 8 cm and 6 cm respectively. Find by construction a point X equidistant from PQ, PS and equidistant from R, S. Measure XR.

Steps of Construction :

(i) Take PR = 8 cm and draw the perpendicular bisector
of PR intersecting it at O.
(ii) From O, out. off OS = OQ = 3 cm
(iii) Join PQ, QR, RS and SP.
PQRS is a rhombus. Whose diagonal are PR and QS.
(iv) PR is the bisector of ∠SPQ.
(v) Draw the perpendicular bisector of SR intersecting PR at X
∴ X is equidistant from PQ and PS and also from S and R.
On measuring length of XR = 3.2 cm (approx.)

Question 8.

Without using set square or protractor, construct the parallelogram ABCD in which AB = 5.1 cm. the diagonal AC = 5.6 cm and the diagonal BD = 7 cm. Locate the point P on DC, which is equidistant from AB and BC.

Steps of Construction :

(i) Take AB = 5.1 cm
(ii) At A, with radius  $\\ \frac { 5.6 }{ 2 }$ = 2.8 cm
and at B with radius $\\ \frac { 7.0 }{ 2 }$ = 3.5 cm,
draw two arcs intersecting each other at O.
(iii) Join AO and produce it to C such that
OC = AD = 2.8 cm and
join BO and produce it to D such that
BO = OD = 3.5 cm
(iv) Join BC, CD, DA
ABCD is a parallelogram.
(v) Draw the angle bisector of ∠ABC intersecting CD at P.
P is the required point which is equidistant from AB and BC.

Question 9.

By using ruler and compass only, construct a quadrilateral ABCD in which AB = 6.5 cm, AD = 4cm and ∠DAB = 75°. C is equidistant  to from the sides  if AB and AD,  if also C is equidistant from the points A and B.

Steps of Construction :

(i) Draw a line segment AB = 6.5 cm.
(ii) At A, draw a ray making an angle of 75° and cut off AD = 4 cm.
(iii) Draw the bisector of ∠DAB.
(iv) Draw perpendicular bisector of AB intersecting the angle bisector at C.
(v) Join CB and CD.

–: End of  Locus ML Aggarwal Solutions :–

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