The availability of step-by-step S Chand Class 10 Maths Solutions ICSE Chapter 13 Loci Ex 13(b) can make challenging problems more manageable.
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.
Solution:
Let T be the tower and O be the oak tree. We have drawn two circles. First circle is drawn taking T as centre and m as radius and second circle is drawn taking O as centre and 40 mas radius. There can be three possibilities.
(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 say at P, then the treasures 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.
Thus only the point of Intersection is the place.
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.
Solution:
Let M be the mouth of cave and R is the pine tree
Mouth of cave is 20 m East of pine tree with the centre as M and 50 m as radius, draw a circle
Draw a peipendicular R which intersects the circle at P. Which is in North of pine tree
∴ P is the place whose the treasure is burried
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 ?
Solution:
There can be three possibilities,
(i) When distance between A and B is less than 5 + 3 = 8 cm
(ii) When the distance between A and B is equal to 8 cm
(iii) When the distance between A and B is more than 8 cm
(i) We see that when the distance between A and B is less then 8 cm than the locus of point will be 2, P and Q, where the two circles are intersecting each other.
(ii) When the distance between A and B is 8 cm, there will be one point of locus which is P, where the two circles are touching each other.
(iii) When the distance between A and B is more than 8 cm, then the circles will not touch or intersect each other then there will be no point of locus.
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?
Solution:
(i) Construct an equilateral triangle PQR with each side equal to 2.7 cm.
(ii) Draw a circle with P and radius 1.7 cm.
(iii) Draw parallel lines l and m of QR at a distance of 0.7 cm. From the line QR.
The line l intersects the circle at P, at A and B Then A and B are the required locus.
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.
Solution:
(i) Construct an equilateral △LMN with side 3.5 cm.
(ii) Draw two parallel lines l and m at a distance of 1.8 cm from MN. These lines intersect LM at P and LM produced at Q. Then P and Q are the required points.
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?
Solution:
(i) Construct a △FGH in which FG = 3 cm, GH = 4 cm and HF = 5 cm respectively.
(ii) Draw the perpendicular bisector n of FG.
(iii) Draw parallel lines l and m at distance of 2 cm from GH.
We see that the parallel lines l and m and the perpendicular bisector of FG do not intersect each other at any point.
∴ There is no such locus.
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.
Solution:
(i) Construct a triangle FGH such that FG = 3 cm, GH = 4 cm and HF = 5 cm respectively.
(ii) With centre H and radius 2.5 cm, draw a circle.
(iii) Draw the perpendicular bisector ‘n’ of FG and draw two parallel lines l and m at a distance of 3 cm.
(iv) The line l intersects the circle at P and Q. These are the required points which satisfy the given conditions.
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.
Solution:
(i) Construct △PQR such that PQ = 2.7 cm, QR = 1.9 cm and RP= 3 cm.
(ii) Draw the angle bisector l of ∠R.
(iii) With centre Q and radius 2 cm, draw a circle which intersects the angle bisector l at O. O is the required point which is equidistant from RP and RQ and at a distance of 2 cm from Q.
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.
Solution:
(i) Construct ABC = 65°.
(ii) Draw the angle bisector l of ∠ABC.
(iii) Draw a line m parallel to one arm BC of ∠ABC which intersects the angle bisector l at P.
P is the required point.
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?
Solution:
(i) Construct a △ABC.
(ii) Draw the angle bisectors of ∠B and ∠C intersecting each other at P.
(iii) Draw PL ⊥ BC, PM ⊥ CA and PN ⊥ AB.
Then P is the required point which is equidistant from the sides of the △ABC.
Now ∵P lies on the angle bisector of ∠B
∴PL = PN
∵P lies on the angle bisector of ∠C
∴PL = PM
From (i) and (ii)
PL = PM = PN
Hence proved.
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 to the line.
(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.
Solution:
(a)False : It is not necessarly that it will pass through the external point.
(b) True : The perpendicular bisector can pass through external points by chance.
(c) True : The angle bisector passes from the vertex of the angle but not necessarily from any other external point.
(d) True : It is not necessarily that a line joining the midpoint of a given line to an external point is perpendicular.
(e) True : The angle bisector is not necessarily will bisect the opposite side.
(f) True : It is not necessarily that a perpendicular bisector of a side of a scalene triangle.
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?
Solution:
(a) X and Y are apart from eachother at a distance of 20 m. A tree is 10 m from X and 24 m from Y.
∴ T is the position of the tree in the figure Now P is the required point which is equidistant from X and Y (at the perpendicular bisector of XY) and 18 m from the tree T.
(b) If the tree is 30 m from Y and 10 m from X, then tree and points X and Y will be in the same line
There will be no point which is 18 m from the tree and equidistant from X and Y.
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.
Solution:
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 the line AB
Join C and D and draw its perpendicular bisector which meets AB at P
Then P is the required point on the line AB
Case II.
When the points are on the opposite side of the straight line
Let C and D be the two points on the opposite of the straight line AB
Join C and D and draw its perpendicular bisector l which intersects AB at P Then P is the required point on the line AB
Case III.
The points are both on the same side of the line but line joining them is parallel to the given line
Join C and D and draw the perpendicular bisector l of CD which intersects AB at P Then P is the required point on the line AB
Case IV.
Same as case II except that the line joining the points is perpendicular to the given line
Join C and D and draw its perpendicular bisector l. Which is parallel to the given line AB Therefore no such point is possible.
Question 14.
Find a point equidistant from the arms of a given angle and also equidistant from two given points.
Solution:
Let E and F be the two points and ∠ABC be the angle
Draw the bisector l of ∠ABC and perpendicular bisector of line joining the points EF. which intersects the angle bisector l at P
P is the required point which is equidistant from the sides of the angle ∠ABC and also from the two given points E and F.
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.
Solution:
In space,
(a) The locus of the points which are at a distance of 3 cm from a given point is a sphere with centre O and radius will be 3 cm.
(b) The locus of the points which are at a distance of 3 cm from a given line l, is a cylinder whose radius is 3 cm and whose axis is the given straight line l.
(c) The locus of the points is the right bisector plane of the line AB.
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.
Solution:
(a) Take two points A and B and join them. Draw the perpendicular bisector l of AB. Then l is the locus of point P such that PA = PB
(b) A circle with centre E and radius 2\(\frac { 1 }{ 2 }\) cm is fixed. Another circle with radius 1 cm and centre R is moving such that it touches the given fixed circle externally.
∴ The locus of R, the centre of moving circle is a circle with E as centre and radius
= 2\(\frac { 1 }{ 2 }\) + 1 = 3\(\frac { 1 }{ 2 }\) cm as radius.
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.
Solution:
Draw a line segment AB = 8 cm
Draw the perpendicular bisector ‘l’ of AB at D
Take a point P on l
Join PA and PB
In △PAD and △PBD,
PD = PD (common)
AD = BD (∵ D is midpoint of AB)
∠PDA = ∠PDB (each 90°)
∴△PAD ≅ △PBD (SAS axiom)
∴ PA = PB
∴ P is equidistant from A and B
Similarly we can take any other point on l This will also be equidistant from A and B
∴ The locus of point of P is the perpendicular bisector of AB
Question 18.
AB is a fixed line, state the locus of the point P so that ∠APB = 90°.
Solution:
Draw a line AB and AB as diameter, draw a circle
Take a point P on the circumference of the circle and join AP, BP
Then ∠APB = 90° (Angle in a semicircle is 90°)
∴ The locus of P will be the circle described on AB as diameter
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.
Solution:
∵ The point P is at a constant distance from a fixed point C
∴ The locus of P will be a sphere
The fixed point C is the centre and the constant distance is its radius
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?
Solution:
∵ P is a fixed point and Q moves in such a way that PQ remains constant
∴ The locus of Q will be a circle whose centre will be P and the PQ as radius
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?
Solution:
∵ P moves in such a way that it is at a constant distance from the given line AB
∴ The path of P will be a line parallel to AB
∵ The distance between parallel lines always remain same
Question 22.
What is the locus of the mid-points of all equal chords in a circle?
Solution:
∵ The mid-points of all equal chords of a circle will be equidistant from the centre of the circle
∴ The locus of P, the midpoint of the chord will be another circle which is 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?
Solution:
∵ P is equidistant from two fixed points A and B
∴ P will be on the perpendicular bisector of the line joining the fixed points $A$ and B Hence locus of P will be the perpendicular bisector of the line joining the fixed points A and B
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.
Solution:
AB and CD are two parallel lines
P is a point such that it is equidistant from AB and CD
∴ A line parallel to the given parallel lines drawn through the mid way of AB and CD is the required line
∴ The locus of P will the line which is parallel to AB and CD drawn through the midway of AB and CD
Question 25.
Two straight lines PQ and PK cross each other at P at 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 roads.
Solution:
PQ and PR two roads such that they meet each other at P making an angle of 75°
S is a stone on PQ such that PS = 800 m (8 cm taking 100 m = 1 cm)
∵ Flagstaff X is equidistant from P and S
∴ It will be on the perpendicular bisector of PS
∵ It is equidistant from the two roads PQ and PR
∴ It will be on the bisector of ∠RPQ which intersects the perpendicular bisector of PS at X
∴ X is the required position of the flagstaff
Question 26.
State the locus of a point in rhombus ABCD which is equidistant from
(i) AB and AD, and
(ii) A and C.
Solution:
ABCD is a rhombus
(i) ∵ The point is equidistant from AB and AD
∴ It will lie on the angle bisector of ∠BAD which is a diagonal AC of rhombus ABCD Hence the locus of the point is diagonal AC
(ii) ∵ The point is equidistant from A and C
∴ It will be on the perpendicular bisector AC which is the other diagonal BD Hence the locus will be the diagonal BC
Questions 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.
Solution:
A, B and C are three towers such that AB = 60 m, BC = 73 m and CA = 52 m
Taking 10 m = 1 cm, the length of the lines AB = 6 cm, BC = 7.3 cm and CA = 5.2 cm Now we construct a △ABC with this given data
Draw the perpendicular bisector of AB and AC which intersect each other at P
Then P is the required point which is equidistant from A, B and C
Join PA, PB and PC on measuring PA = 3.6 cm i.e. 36 m
Question 28.
D, F are fixed points ; DEFG is a variable rhombus. Find the locus of E.
Solution:
DEFG is a rhombus in which D and F are fixed points
The locus of E will be the perpendicular bisector of the line joining D and F. Which is the diagonal ED
∵ Diagonals DF and EG bisect each other at right angles
Hence locus of E is the diagonal EG of the rhombus