NSS Mathematics in Action 5A Full Solutions 82 4 Locus Review Exercise 4 (p. 4.5) 1. units 25 units 576 49 units )] 15 ( 9 [ ) 5 2 ( 2 2 AB 2. units) 2 4 (or units 32 units 16 6 1 units ) 14 10 ( )] 1 ( 3 [ 2 2 AB 3. units 5 units 25 units 16 9 units ) 0 4 ( ) 4 ( 2 2 2 2 2 a a a a a a a AB 4. Let (0, k) be the coordinates of A. 4 or 2 0 ) 4 )( 2 ( 0 8 6 1 ) 3 ( ) 0 3 ( ) 4 3 ( ) 3 ( ) 0 3 ( 2 2 2 2 2 2 k k k k k k k k BP AP The possible coordinates of A are (0, 2) or (0, 4). 5. Coordinates of M ) 2 , 4 ( 2 0 4 , 2 6 2 6. Coordinates of M ) 5 , 0 ( 2 ) 7 ( 3 , 2 5 5 7. (a) P is the mid-point of AB. Coordinates of P ) 3 , 8 ( 2 4 10 , 2 3 13 (b) Coordinates of the mid-point of AP 2 13 , 2 21 2 ) 3 ( 10 , 2 8 13 8. Coordinates of the required point ) 4 , 2 ( 2 1 ) 8 ( 1 ) 2 ( 2 , 2 1 ) 4 ( 1 ) 1 ( 2 9. Coordinates of the required point ) 0 , 13 ( 4 3 ) 12 ( 3 ) 9 ( 4 , 4 3 ) 21 ( 3 ) 7 ( 4 10. Coordinates of the required point ) 4 1 , 9 ( 1 4 ) 16 ( 4 ) 6 ( 1 , 1 4 2 23 4 ) 1 ( 1 11. Slope of AB 2 5 2 8 6 12. Slope of AB 2 5 10 6 3 7 13. Slope of AB 3 4 ) 3 ( ) 4 ( h h k k 14. The required slope 3 1 6 2 15. Slope of k k L 2 ) ( 2 1 Slope of 8 ) 8 ( 2 k k L Slope of L 1 = Slope of L 2 4 or 4 16 8 2 2 k k k k k 16. (a) L cut the x-axis at P. By substituting y = 0 into L: x – 2y = – 8, we have 8 8 ) 0 ( 2 x x Coordinates of P ) 0 , 8 ((b) Slope of AP a a 8 6 8 6 0
Transcript
NSS Mathematics in Action 5A Full Solutions
82
4 Locus
Review Exercise 4 (p. 4.5)
1.
units25
units57649
units)]15(9[)52( 22
AB
2.
units)24(orunits32
units1661
units)1410()]1(3[ 22
AB
3.
units5
units25
units169
units)04()4(
2
22
22
a
a
aa
aaaAB
4. Let (0, k) be the coordinates of A.
4or2
0)4)(2(
086
1)3(
)03()43()3()03(
2
2
2222
kk
kk
kk
k
k
BPAP
The possible coordinates of A are (0, 2) or (0, 4).
5. Coordinates of M
)2,4(
2
04,
2
62
6. Coordinates of M
)5,0(
2
)7(3,
2
55
7. (a) P is the mid-point of AB.
Coordinates of P
)3,8(
2
410,
2
313
(b) Coordinates of the mid-point of AP
2
13,
2
21
2
)3(10,
2
813
8. Coordinates of the required point
)4,2(
21
)8(1)2(2,
21
)4(1)1(2
9. Coordinates of the required point
)0,13(
43
)12(3)9(4,
43
)21(3)7(4
10. Coordinates of the required point
)41,9(
14
)16(4)6(1,
14223
4)1(1
11. Slope of AB
252
86
12. Slope of AB
2
5106
37
13. Slope of AB
3
4
)3(
)4(
hh
kk
14. The required slope
3
16
2
15. Slope of
k
kL
2
)(
21
Slope of
8
)8(2
k
kL
Slope of L1 = Slope of L2
4or4
16
8
2
2
kk
k
k
k
16. (a) L cut the x-axis at P.By substituting y = 0 into L: x – 2y = – 8, wehave
8
8)0(2
x
x
Coordinates of P )0,8(
(b) Slope of AP
a
a
8
68
60
4 Locus
83
Slope of L
2
1
)2(
1
Slope of AP = 2 slope of L
2
862
12
8
6
a
aa
Activity
Activity 4.1 (p. 4.7)1. (a)
(b) The locus of P is a circle with centre A and radius6 cm.
2. (a)
(b) The locus of P is a pair of straight lines parallel toand at a distance 1 cm from L.
3. (a)
(b) The locus of P is the straight line parallel to andmidway between L1 and L2.
Classwork
Classwork (p. 4.18)(a) x = 2(b) y = 4 and y = –2
(c)2
1x
(d) y = x and y = –x(e) x2 + y2 + 4x – 2y + 1 = 0
Quick Practice
Quick Practice 4.1 (p. 4.10)(a)
(b) Length of a lap
m)60200(
m])30(2100100[
The man should run (200 + 60π) m to complete a lap.
Quick Practice 4.2 (p. 4.11)The required locus is the perpendicular bisector of AB.
Quick Practice 4.3 (p. 4.12)(a) The required locus is the two angle bisectors L3 and L4.
(b)
Let B1 and B2 be the possible points of B on L3 and L4
respectively.cm821 ABAB
Refer to the figure.Distance between B1 and L1 or L2
cm34
cm2
38
cm60sin8
Distance between B2 and L1 or L2
cm4
cm2
18
cm30sin8
The possible distance between B and the two
intersecting lines are 4 cm or 34 cm.
Quick Practice 4.4 (p. 4.13)(a) The circumference of the wheel is 2πr. Thus, when the
wheel rolls along AB, it turns a complete revolution and Ptouches the vertex B.Similarly, P touches C and A when the wheel rolls alongBC and CA.
(b)
NSS Mathematics in Action 5A Full Solutions
84
Quick Practice 4.5 (p. 4.19)Let (x, y) be the coordinates of P.
0941430
04707015055
)44(4)96(4)96(9)4914(9
])2()3[(4])3()7[(9
)2()]3([2)]3([)7(3
23
22
22
2222
2222
2222
yxyx
yxyx
yyxxyyxx
yxyx
yxyx
DPCP
The required equation is x2 + y2 – 30x + 14y + 94 = 0.
Quick Practice 4.6 (p. 4.20)Let (x, y) be the coordinates of P.
CPD = 90°CP PD
028
1612
)4)(4()4)(3(
14
4
)4(
)3(
22
22
yyx
xyy
xxyy
x
y
x
y
The required equation is x2 + y2 – y – 28 = 0.
Quick Practice 4.7 (p. 4.20)(a) Let (x, y) be the coordinates of P and A be a point on L
such that PA L.
34
034
16
9
2
3)44(
16
25
2
5
4
3)2(
4
5
4
3)2(
4
5
2
2
222
22
2
2
2
xxy
yxx
yyxxyy
yxy
yxy
PFPA
The required equation is y = x2 – 4x + 3.(b)
The above curve is a parabola.
Quick Practice 4.8 (p. 4.21)L1 :
2
1
012
x
x
L1 is vertical and L2 is horizontal.The required locus is the pair of angle bisectors of the anglesformed by L1 and L2. We name the angle bisectors as L3 and L4.Since L1 is perpendicular to L2, the acute angle between L2 andL3 is 45°. The inclination of L3 is also 45°. Hence, the slope ofL3 is tan 45°, i.e. 1.
Since the angle bisectors are mutually perpendicu lar, we have
slope of 11
14
L
The intersection of L1 and L2 is A
3,2
1. L3 and L4 also pass
through A
3,2
1.
The equation of L3 is
07222
13
1
21
3
yx
xy
x
y
The equation of L4 is
05222
13
1
21
3
yx
xy
x
y
The required equations are 2x – 2y + 7 = 0 and2x + 2y – 5 = 0.
Quick Practice 4.9 (p. 4.21)The required locus is the straight line parallel to and midwaybetween L1 and L2. We name the line as L3.
y-intercept of L1 10)2(
20
y-intercept of L22
35
)2(
)35(
y-intercept of L34
15
2
3510
2
1
Slope of L3 = slope of L1 = slope of L22
5
)2(
5
The equation of L3 is
0154104
15
2
5
yx
xy
The required equation is 10x – 4y – 15 = 0.
Quick Practice 4.10 (p. 4.23)The required locus is a pair of straight lines parallel to and at adistance 3 units from L. We name them as L1 and L2.Refer to the figure.
Let θ be the inclination of L.
4 Locus
85
5
343
3sin
4
3
ofslopetan
22
L
5
3
5
3
3sin
aa
a
x-intercept of L 33
)9(
x-intercept of L1
2
53
x-intercept of L2
8
53
Slope of L1 = slope of L2 = slope of L4
3
The equation of L1 is
0643
4
3
)2(
0
yx
x
y
The equation of L2 is
024434
3
8
0
yxx
y
The required equations are 3x – 4y + 6 = 0 and3x – 4y – 24 = 0.
Further Practice
Further Practice (p. 4.14)1. (a) The required locus is two circles each of radius 2 cm,
with centres A and B respectively.
(b) The required locus is a semi-circle below AB withcentre M and radius 2 cm, excluding points A and B.
2. The required locus is chord AB.
Further Practice (p. 4.24)1. Let (x, y) be the coordinates of P.
022
016816
4496442510
)2()3()2()5(
)2()]3([)]2([)5(
2222
2222
2222
yx
yx
yyxxyyxx
yxyx
yxyx
DPCP
The required equation is 2x – y – 2 = 0.
2. Let (x, y) be the coordinates of P.
06
9)3(
3)]3([)0(
22
22
22
yyx
yx
yx
OAPA
The required equation is x2 + y2 + 6y = 0.
3.
1
01
02222
21212
2)1()1(
2)1()0()0()1(
2
2
2
2
22222
22222
22
222
22
222
xxy
yxx
yxx
yyyxyxx
yyxyx
yyxyx
PQPBPA
The required equation is y = x2 – x + 1.
Exercise
Exercise 4A (p. 4.15)Level 1
1.
The locus of P is made up of:(i) two parallel line segments, each at a distance of 3 cm
from MN; and(ii) two semi-circles each of radius 3 cm, with centres M
and N respectively.
2.
The locus of P is the angle bisector of CAB.
3.
The locus of P is a straight line parallel to and midwaybetween the two given lines.
NSS Mathematics in Action 5A Full Solutions
86
4.
The locus of P is the perpendicular bisector of the linesegment joining A and B.
5.
6.
The locus of the centre of the rolling circular coin is acircle with radius 5 cm. Its centre is the same as that of thefixed coin.
7. (a)
The locus of P is a line segment parallel to andmidway between line segments AD and BC.
(b)
The locus of P is the diagonals of the square.
8. (a)
The locus of vertex A is a circle with centre C andradius AC.
(b)
The locus of E is a circle with centre C and radius EC.
9.
The locus of P is two circles, each with the same centre asthe given circle, and with radii 1 cm and 3 cm respe ctively.
10. The condition for the locus is:P maintains equal distances from AB and AC and P liesinside or on the triangle.(or any other reasonable answers)
11. The condition for the locus is:P maintains a fixed distance of 3 cm above L.(or any other reasonable answers)
Level 212. (a)
The locus of P is made up of:(i) two parallel line segments, each of a distance of
2 cm from AB; and(ii) two semi-circles each of radius 2 cm, with
centres A and B respectively.(b)
The locus of P is a pair of straight lines parallel toand at a distance 2 cm from AB.
13.
The locus of C is a circle with diameter AB.
14. (a)
The required locus is the angle bisector of R1MR2.(b) Another condition for the locus is:
P is always equidistant from MR1 and MR2.(or any other reasonable answers)
locus of P
4 Locus
87
15.
The required locus is a straight line perpendicular to L atA.
16. (a)
Let 'P be the lowest point of P.The required locus is the perpendicular bisector ofAB and below AB.
(b)
cm12'
cm)513('
theorem)(Pyth.''2222
222
MP
MP
APMPAM
The length of the locus is 12 cm.
17.
The locus of vertex A is made up of:
(i)
1AA with centre C and radius 7 cm,
(ii)
21AA with centre B2 and radius 6 cm.
18.
The locus of vertex D is the three arcs with centres B, Pand R respectively.
19. (a)
(b)
20. (a)
(b) Total length of the locus
cm)2(20
cm)22212423)1(2(
Exercise 4B (p. 4.24)Level 11. L is a vertical line with x-intercept 3.
The required locus is a pair of vertical lines at adistance of 3 units from L.
x-intercepts of the straight lines
6and0
33and33
The required equations are x = 0 and x = 6.
2.
2
1i.e.
012:
y
yL
L is a horizontal line with y-intercept2
1 .
The required locus is a pair of horizontal lines at a
distance of2
5 units from L.
y-intercepts of the straight lines
2and32
5
2
1and
2
5
2
1
The required equations are y = –3 and y = 2.
3. The required locus is a straight line parallel to and midwaybetween L1 and L2.
y-intercept of the straight line 42
)1(7
The required equation is y = –4.
4.
2
3i.e.
032:1
x
xL
4
9i.e.
094:2
x
xL
The required locus is a straight line parallel to and midwaybetween L1 and L2.
NSS Mathematics in Action 5A Full Solutions
88
x-intercept of the straight line8
3
249
23
The required equation is
0388
3
x
x
5. Let (x, y) be the coordinates of the point.
01422
161212
16)1()1(
4)1()1(
22
22
22
22
yxyx
yyxx
yx
yx
The required equation is x2 + y2 – 2x – 2y – 14 = 0.
6. Let (x, y) be the coordinates of the point.
02
4
9
4
1
4
9
2
1
2
3)0(
2
1
22
22
22
2
2
xyx
yxx
yx
yx
The required equation is x2 + y2 + x – 2 = 0.
7. Let (x, y) be the coordinates of the point P.(a)
012
08816
122510251096
)1()5()5()3(
)1()5()5()]3([
2222
2222
2222
yx
yx
yyxxyyxx
yxyx
yxyx
BPAP
The required equation is 2x – y + 1 = 0.
(b)
02512
020089688
)122510(9251096
])1()5[(9)5()3(
)1()5(3)5()]3([
3
22
22
2222
2222
2222
yxyx
yxyx
yyxxyyxx
yxyx
yxyx
BPAP
The required equation isx2 + y2 – 12x – y + 25 = 0.
9. Let P(x, y) be a point on the locus.Let A be a point on L such that AP L.
2
2
222
222
22
8
1
08
4444
)2()2(
2)]2([)0(
xy
yx
yyyyx
yyx
yyx
PAPF
The required equation is .8
1 2xy
10. Let P(x, y) be a point on the locus.Let A be a point on L such that AP L.
54
054
16
9
2
3
16
25
2
544
4
3
4
5)2(
4
3
4
5)]2([
2
2
222
222
22
xxy
yxx
yyyyxx
yyx
yyx
PAPF
The required equation is y = x2 + 4x + 5.
11. Let (x, y) be the coordinates of P.
(a) Slope ofx
y
x
yOP
0
0
Slope of4
1
08
02
OA
OP OA
04
4
14
1
yx
xyx
y
The required equation is 4x + y = 0.
(b) Slope ofx
yOP
Slope of8
2
x
yAP
OP AP
4 Locus
89
028
82
)8()2(
18
2
22
22
yxyx
xxyy
xxyyx
y
x
y
The required equation is x2 + y2 – 8x – 2y = 0.
12. Let C(x, y) be the centre of a circle passing through A(3, 7)and B(6, –5).
0182
03246
61101258146
)5()6()7()3(
)]5([)6()7()3(
2222
2222
2222
yx
yx
yxyxyxyx
yxyx
yxyx
BCAC
The required equation is 2x – 8y – 1 = 0.
Level 213. The required locus is the pair of angle bisectors of the
angles formed by L1 and L2, namely L3 and L4.
Since L1 L2, the acute angle between L2 and L3 is 45°.Slope of L3 = tan 45° = 1L3 L4
Slope of L4 11
1
L3 and L4 also pass through the intersection of L1 and L2,i.e. (2, 2).The equation of L3 is
xy
xyx
y
22
12
2
The equation of L4 is
04
22
12
2
yx
xyx
y
The required equations are y = x and x + y – 4 = 0.
14. The required locus is the pair of angle bisectors of theangles formed by L1 and L2, namely L3 and L4.
The slopes of L1 and L2 are 1 and –1 respectively.L3 is horizontal and L4 is vertical.
L3 and L4 also pass through the intersection of L1 and L2,i.e. (0, 0).The equation is L3 is y = 0.The equation of L4 is x = 0.
The required equations are x = 0 and y = 0.
15. The required locus is a pair of straight l ines parallel to and
at a distance 2 units from L, namely L1 and L2.
Let a and b be the y-intercepts of L1 and L2 respectively.Slope of L = –1The acute angle between L and the y-axis is 45°.
2
245sin
aa
Similarly, b = –2.Slope of L1 = slope of L2 = slope of L = –1The equation of L1 is
02
2
yx
xy
The equation of L2 is
02
2
yx
xy
The required equations are x + y + 2 = 0 andx + y – 2 = 0.
16. The required locus is a pair of straight lines parallel to andat a distance 3 units from L, namely L1 and L2.
Let a units be the vertical distance between L1 and L.
Slope of L 3The acute angle between L and the y-axis is 30°.
6
330sin
aa
Also, y-intercept of L = –3y-intercept of L1 = –3 + 6
= 3y-intercept of L2 = –3 – 6
= –9
Slope of L1 = slope of L2 = slope of 3LThe equation of L1 is
33 xyThe equation of L2 is
93 xy
The required equations are 33 xy and
93 xy .
NSS Mathematics in Action 5A Full Solutions
90
17. (a) L1 :
7
8
7
2
0872
xy
yx
Slope of7
21 L
L2 :
7
12
7
2
01272
xy
yx
Slope of7
22 L
Slope of L1 = slope of L2
L1 and L2 are parallel.(b) The required locus is the straight line parallel to and
midway between L1 and L2.
y-intercept of the locus
7
102
712
78
The required equation is
010727
10
7
2
yx
xy
18. (a) L1 :
1
01
xy
yx
Slope of L1 = 1L2 :
1
01
xy
yx
Slope of L2 = –1Slope of L1 slope of L2
L1 and L2 are not parallel.(b) The required locus is the pair of angle bisectors of the
angles formed by L1 and L2, namely L3 and L4.
Inclination of L1 = 45° and L1 L2
L3 is horizontal and L4 is vertical.
)2(01:
)1(01:
2
1
yxL
yxL
(1) + (2) :
1
22
x
x
By substituting x = 1 into (1), we havey = 0The intersection of L1 and L2 is (1, 0) whichmust lie on L3 and L4.The equation of L3 is y = 0.The equation of L4 is x = 1.The required equations are x = 1 and y = 0.
19. (a) L1 :
2
02
xy
yx
Slope of L1 = 1L2 :
2
5
0522
xy
yx
Slope of L2 = 1Slope of L1 = slope of L2
L1 and L2 are parallel.(b) The required locus is the straight line parallel to and
midway between L1 and L2.
y-intercept of the locus
4
12
25
2
The required equation is
01444
1
yx
xy
20. (a) L1 :
xy
yx
3
1
03
Slope of L13
1
L2 :
xy
xy
3
03
Slope of L2 3Slope of L1 ≠ slope of L2
L1 and L2 are not parallel.(b) The required locus is the pair of angle bisectors of the
angles formed by L1 and L2, namely L3 and L4.
The inclinations of L3 and L4 are 45° and 135°respectively.Slope of L3 = tan 45° = 1Slope of L4 = tan 135° = –1
The intersection of L1 and L2 is (0, 0) which must lieon L3 and L4.
The equation of L3 is y = x.The equation of L4 is y = –x.The required equations are y = x and y = –x.
21. (a) (i) The required locus is the perpendicular bisectorof AB.
Slope of AB 2)5(3
15
Slope of the locus2
1
4 Locus
91
Coordinates of the mid-point of AB
)3,4(
2
51,
2
)3(5
The required equation is
022
462
)]4([2
13
yx
xy
xy
(ii) The required locus is the perpendicular bisectorof BC.
Slope of BC 2)3(3
57
Slope of the locus2
1
Coordinates of the mid-point of BC
)1,0(
2
)7(5,
2
33
The required equation is
022
)0(2
1)1(
yx
xy
(b) (i) D is equidistant from A and B.D lies on L1 : x + 2y – 2 = 0D is equidistant from B and C.D also lies on L2 : x – 2y – 2 = 0D is the intersection of L1 and L2.
)2(022
)1(022
yx
yx
(1) + (2) :
2
042
x
x
By substituting x = 2 into (1), we have
0
0222
y
y
Coordinates of D )0,2(
(ii) Let (x, y) be the coordinates of P.
0464
5044
50)2(
)01()25()0()2(
22
22
22
2222
xyx
yxx
yx
yx
ADPD
The required equation isx2 + y2 – 4x – 46 = 0.
22. (a) AB is a horizontal line segment.Length of AB
units3
units)25(
Refer to the figure.
Area of ABP = 3 sq. units
2
3)3(2
1
h
h
The locus of P is a pair of horizontal lines 2units above or below AB, namely L1 and L2.The equation of L1 is y = 5.The equation of L2 is y = 1.The required equations are y = 1 and y = 5.
(b) ABP is not an obtuse-angled triangle.x-coordinate of P lies between 2 and 5.y-coordinate of P is 1 or 5.The possible integral coordinates of P are (2, 1),(3, 1), (4, 1), (5, 1), (2, 5), (3, 5), (4, 5) and(5, 5).
23. Let (x, y) be the coordinates of M.Let (a, 0) and (0, b) be the coordinates of A and B.
M is the mid-point of AB.
2
0
ax and
2
0 by
2
ax and
2
by
units10AB
25
2522
2544
100
10)0()0(
22
22
22
22
22
yx
ba
ba
ba
ba
The required equation is x2 + y2 = 25.
Revision Exercise 4 (p. 4.27)Level 11. The required locus is a circle with centre at the nose cone
and with the length of the blade as radius.
2. The required locus is the straight line parallel to the laneand at a distance r from the lane, where r is the radius ofthe bowling ball.
3.
The locus of P is made up of:(i) two pairs of parallel line segments, each at a distance
of 1 cm from the rhombus; and(ii) four arcs of a circle, each of radius 1 cm, centred at A,
B, C and D respectively.
NSS Mathematics in Action 5A Full Solutions
92
4. Let (x, y) be the coordinates of P.
0365
4
499
4
2565
4
49)3(
2
5
2
7)3(
2
5
22
22
22
2
2
yxyx
yxyx
yx
yx
The required equation is x2 + y2 + 5x – 6y + 3 = 0.
5. Let (x, y) be the coordinates of P.
0133108
1212264163612168
)11()8()6()4(
)11()]8([)6()]4([
2222
2222
2222
yx
yyxxyyxx
yxyx
yxyx
PBPA
The required equation is 8x – 10y + 133 = 0.
6. Let P(x, y) be a point on the locus.Let A be a point on L such that AP L.
22
022
16
9
2
3
16
25
2
512
4
3
4
5)1(
4
3
4
5)1(
2
2
222
222
22
xxy
yxx
yyyyxx
yyx
yyx
APFP
The required equation is y = x2 – 2x + 2.
7. Let P(x, y) be a point on the locus.Let A be a point on L such that AP L.
342
0324
1689644
)4()3()2(
4)3()]2([
2
2
222
222
22
xxy
yxx
yyyyxx
yyx
yyx
APFP
The required equation is 2y = –x2 – 4x + 3.
8. L1 and L2 are horizontal lines.The required locus is a horizontal line midwaybetween L1 and L2.
y-intercept of the locus
22
)6(2
The required equation is y = –2.
9. L3 :
3
1
013
x
x
L4 :
3
8
083
x
x
L3 and L4 are vertical lines.The required locus is a vertical line midway betweenL3 and L4.
x-intercept of the locus
2
32
3
8
3
1
The required equation is2
3x .
10. The required locus is the pair of angle bisectors of theangles formed by L2 and L4, namely L5 and L6.
Since L2 L4, the acute angle between L2 and L5 is 45°.Slope of L5 = tan 45° = 1L5 L6
Slope of L6 11
1
L5 and L6 also pass through the intersection of L2 and L4,
i.e.
6,3
8.
The equation of L5 is
010333
86
1
38)6(
yx
xy
x
y
The equation of L6 is
026333
86
1
3
8
)6(
yx
xy
x
y
The required equations are 3x – 3y – 10 = 0 and3x + 3y + 26 = 0.
11. Let (x, y) be the coordinates of P.
Coordinates of M
)0,2(
2
22,
2
15
094
13)2(
)02()]2(5[)0()]2([
22
22
2222
xyx
yx
yx
AMPM
The required equation is x2 + y2 + 4x – 9 = 0.
4 Locus
93
12. Let (x, y) be the coordinates of P.
0561012
0112202422
2098141434610
20)7()7()3()5(
)37()57(
)7()7()3()5(
22
22
2222
2222
222
222
222
222
yxyx
yxyx
yxyxyxyx
yxyx
yxyx
ABBPAP
The required equation is x2 + y2 – 12x – 10y + 56 = 0.
13. (a) The required locus is a pair of straight lines parallelto and d units from L.The required equations are y = c + d and y = c – d.
(b) The equations of the locus are y = – 10 andy = 4.
)2(4
)1(10
dc
dc
[Note that d > 0.](1) + (2) :
3
62
c
c
By substituting c = –3 into (1), we have
7
103
d
d
14.
The locus of P is made up of:(i) two pairs of parallel line segments, each at a distance
of 1 cm from one of the line segments; and(ii) two semi-circles, each of radius 1 cm, with the
vertices as the centres respectively; and(iii) a quarter of radius 1 cm centred at the intersection of
the two line segments.
15.
The locus of P is the straight line that joins A and P asshown.
16.
The required locus is the straight line passing through O, Aand .O
17. (a) The condition for the locus is:a moving point which is equidistant from AB and AD,and lies inside or on rhombus ABCD.(or any other reasonable answers)
(b) The condition for the locus is:a moving point which maintains equal distances fromBC and AD, and lies inside or on rhombus ABCD.
(c) The condition for the locus is:a moving point which maintains a fixed distancefrom MN, and lies on rhombus ABCD.
18. The condition for the locus is:P is always 5 cm from M.(or any other reasonable answers)
Level 219. Let R and R be the radii of the larger circle and the
smaller circle respectively.
(a)
The locus of O is a circle with radius RR andcentred O.
(b)
The locus of O is a circle with radius RR andcentred .O
20.
The locus of P is an arc passing through A, B and P. The
angle subtended by
ABP is 100°.
21.
Let M be the mid-point of the fixed-length chord.The locus of M is a circle with radius OM and centred O.
22.
NSS Mathematics in Action 5A Full Solutions
94
23. L1 :
63
063
xy
yx
Slope of L1 = –3L2 :
2
93
0926
xy
yx
Slope of L2 = –3Slope of L1 = slope of L2
L1 and L2 are parallel.The required locus is a straight line parallel to and midwaybetween L1 and L2.
y-intercept of the locus
4
32
2
96
The required equation is
034124
33
yx
xy
24. The required locus is a pair of straight lines parallel to andat a distance 1 unit from L, namely L1 and L2.
Let θ be the inclination of L and d and –d be they-intercepts of L1 and L2 respectively.Note that L passes through the origin, andslope of L1 = slope of L2 = slope of L.
2
ofslopetan
L
5
1
12
1cos
2
Refer to the figure.
5
1cos
1)90sin(
d
d
d
The equation of L1 is
52 xyThe equation of L2 is
52 xy
The required equations are 52 xy and
52 xy .
25. (a) Let (x, y) be the coordinates of P.
)96(542
])3[()2()1(
)0()3()2()]1([
22222
22222
2222
xyxkyxyx
yxkyx
yxkyx
kBPAP
0954)62()1()1( 222222 kyxkykxk
The required equation is
0954)62()1()1( 222222 kyxkykxk .
(b) (i) The locus is a straight line.The coefficients of x2 and y2 are zero.
(rejected)1or1
1
012
2
kk
k
k
(ii) For k = 1,The equation of the locus is
012
0448
0)1(954])1(62[)11()11( 222222
yx
yx
yxyx
26. (a)
(b)
27. (a) The locus of P is the perpendicular bisector of AB.
(b) The locus of Q is a circle with AC as diameter,excluding A and C.
4 Locus
95
(c) The possible locations of a point M are theintersections of the loci of P and Q.
28.
The locus of P is a curve composed of two arcs of radii3 cm and 4 cm respectively.
29.
The locus of P is a larger circle with the same centre as thefixed circle.
30.
31. (a) Let (x, y) be the coordinates of P and B be a point onL such that BP L.
14844
4
12
4
12
2
1)(
2
1)0()(
22
22
2222
222
22
aaxxy
yaaxx
yyyaaxx
yyax
yyax
BPFP
The required equation is 4y = 4x2 – 8ax + 4a2 – 1.(b) By substituting (–1, 6) into 4y = 4x2 – 8ax + 4a2 – 1 ,
we have
2
3or
2
7
0)32)(72(
02184
14)1(8)1(4)6(42
22
aa
aa
aa
aa
32. (a) Let (x, y) be the coordinates of P and A be a point onL such that AP L.
33. (a) (i) The locus of P is a circle with diameter AC,excluding A and C.
(ii) Let (x, y) be the coordinates of P.
Slope of AP4
2
x
y
Slope of CP4
3
)4(
3
x
y
x
y
AP CP
0105
1665
)4)(4()3)(2(
14
3
4
2
22
22
yyx
xyy
xxyyx
y
x
y
The required equation isx2 + y2 – 5y – 10 = 0.
NSS Mathematics in Action 5A Full Solutions
96
By substituting (2, 6) into x2 + y2 – 5y – 10 = 0,we have
R.H.S.
0
10)6(562L.H.S. 22
B lies on the locus.By substituting (–2, –1) intox2 + y2 – 5y – 10 = 0, we have
R.H.S.
0
10)1(5)1()2(L.H.S. 22
D lies on the locus.(b) Let (x, y) be the coordinates of Q.
Slope of BQ2
6
x
y
Slope of DQ2
1
)2(
)1(
x
y
x
y
BQ DQ
0105
465
)2)(2()1)(6(
12
1
2
6
22
22
yyx
xyy
xxyyx
y
x
y
The required equation is x2 + y2 – 5y – 10 = 0.(c) Both AC and BD are diameters of the circle
x2 + y2 – 5y – 10 = 0. AC = BD and ABC, BCD,CDA and DAB are equal to 90°.
ABCD is a rectangle.
34. (a) (i) The required locus is the angle bisector ofBAD. It is also the diagonal AC of the square.
(ii) The required equation is
,077
)1(707
10
0
1
yx
xyx
y
for 0 ≤ x ≤ 7
(b) The required locus is the pair of angle bisectors of L1
and L2, namely L3 and L4.L3 passes through B and D.
The equation of L3 is
0247
)4(7443
43
4
4
yx
xyx
y
ABC = 90°, i.e. L1 L2
L3 L4
Slope of L47
1
L4 passes through B(3, –3).The equation of L4 is
0187
)3()3(77
1
3
)3(
yx
xyx
y
The required equations are 7x – y – 24 = 0 andx + 7y + 18 = 0.
Multiple Choice Questions (p. 4.32)1. Answer: C
Radius of the circle = 1 cmCentre of the circle : M
The answer is C.
2. Answer: AIn ABC,
cm3
cm60cos6
60cos
ABBC
The answer is A.
3. Answer: DThe required locus is a pair of straight lines parallel to andat a distance of 3 units from the vertical line L.
The required equations are x = 0 and x = 6.
4. Answer: ALet (x, y) be the coordinates of P.
02024
25)1()2(
)]1(1[)]3(2[)]1([)2(
22
22
2222
yxyx
yx
yx
ABAP
The required equation is x2 + y2 – 4x + 2y – 20 = 0.
5. Answer: C
6. Answer: D
7. Answer: D
8. Answer: B
9. Answer: BThe required locus is the perpendicular bisector of AB.
Mid-point of AB
)0,0(
2
)(,
2
)2(2
hhhh
Slope of AB
2
122
hh
hh
Slope of the locus of P = –2The required equation is
02
20
0
yxx
y
10. Answer: DL1 :
32
032
xy
yx
Slope of L1 = –2L2 :
52
052
xy
yx
Slope of L2 = –2Slope of L1 = slope of L2
L1 is parallel to L2.The required locus is a straight line parallel to andmidway between L1 and L2.
4 Locus
97
y-intercept of the locus of P
12
)5(3
The required equation is
012
12
yx
xy
11. Answer: CThe required locus is an angle bisector of the angle formedby L1 and L2, and lies in the shaded region. We name th isstraight line as L.Refer to the figure.
3
1tan
3tan
1and
3
1tan
ofslope)90tan(andofslopetan 21
LL
452
90 ( L is the angle bisector.)
Slope of L
1
135tan
)45180tan(
L passes through O(0,0).The equation of L is
0
yx
xy
Investigation Corner (p. 4.35)
1. The locus of P is a circle of radius R units centred at(0, –h).
2. (a) Consider .900 Let (x, y) be the coordinatesof P.
From the figure, we have
and
hRy
RyhR
yh
RxR
x
sin
sin
sin
cos
cos
In fact, the results can be extended to the cases when 90 or 0 by using the properties of
trigonometric ratios.The coordinates of P )sin,cos( hRR
