Solution Manual ENG - Buku Teks Add Math KSSM TKT 4

m1 = m2

– 1 = 3p
2
1
p = – 6

(b) x – y =1 py = 4x – 6
3 6
y
6 = x – 1 y= 4 x – 6
3 p p

y = 2x – 6 º m2 = 4
p

º m1 = 2
m1 = m2

2= 4
p

p=2

3. (a) 3x + 5y = 15 5x – ky = 2

5y = –3x + 15 ky = 5x – 2

y = –  3 x + 3 y= 5 x – 2
5 k k

º m1 = – 53 º m2 = 5
k

m1m2 = –1

(– 53 )( 5 ) = –1
k

y k=3
9
(b) x + =1 ky = 2x – 7
3
y x 2 7
9 = –  3 + 1 y= k x – k

y = –3x + 9 º m2 = 2
k

º m1 = –3

m1m2 = –1

(–3)( 2 ) = –1
k
k=6

4. mAB = 4–1 and mBC = a–4
–1 –1 5 – (–1)

= – 32 = a–4
6

OSilnechesAebBaibs ApeBrpbeenrsdeicreunlajarntog BdeCn,gtahnusBC, jadi

(–  3 )( a – 4 ) = –1
2 6

a–4=4

a=8

7

Self Practice 7.5 (Page 190)

1. 3x + 2y = 48 m1m2 = –1

2y = –3x + 48 ( )– 3 (m2) = –1
2
y = –  3 x + 24 2
2 m2 = 3

º m1 = – 23

PTehrseameqauaantijoenjaorif rAaBdiiuaslaAhB is

y – 12 = 2 (x – 8)
3

3y – 36 = 2x – 16

3y – 2x = 20

3 + 7 8+2
2 2
( )2.
(a) TiMtikidtpenoginath AB = ,

= (5, 5)

(b) PaLnejnagntghlofntgrkeanncgh = ! (7 – 5)2 + (8 – 5)2

= ! 13

= 3.606 units

Intensive Practice 7.2 (Page 190 – 191)

1. (a) mAB = 4 – 2 mCD = 3 – (–1)
3 – 6 –3 – 3

= – 32 = –  2
3

OBleechausseebaobf mAB = mCD, AB danand CD are parallel. i.

(b) mAB = 4 – (–3) mCD = 1–4
–3 – 4 –2 – 1

= –1 =1

OBelechausseebaobf mABmCD = –1, mthauksa AB daannd CD ardealpaehrpbeenrdseicruenlajra.ng.

2. mABmBC = –1

k–8
12 – 6
( )( )8 – 2 = –1
6–1

( )6 k–8 = –1
6
5

k – 8 = –5

k=3

3. (a) mQR = 4–2 = – 2
–1 – 2 3

The equation of straight line that passes through point P(7, 3) and parallel to QR is

y – 3 = – 2 (x – 7)
3y – 9 = 3 + 14 3y
–2x + 2x = 23… 1

8

(b) The equation of straight line that passes through point R(–1, 4) and perpendicular to QR is

y–4= 3 (x + 1)
2

2y – 8 = 3x + 3

2y – 3x = 11…2

1 × 3: 9y + 6x = 69…3

2 × 2: 4y – 6x = 22…4

3 + 4: 13y = 91
91
y= 13

y=7

Substitute y = 7 into 1: 3(7) + 2(x) = 23

2x = 23 – 1
2x = 22

x=1

Thus, coordinate S is (1, 7).

4. (a) mPQ = mQR

–12 – (–6) = 6 – (–12)
3 – (–1) e–3

– 3 = 18
2 e–3

e – 3 = –12

e = –9

(b) mPQmPR = –1

(– 32 )(e 12 ) = –1
+1

e + 1 = 18

e = 17

5. mPQ = –2 – 1 mRS = h–5
1 – (–6) –3 – 0

= – 37 = –  h – 5
3
3( )––  h – 5
7 3 = –1

h – 5 = –7

h = –2

6. (a) mAB = 5–0
0 – 10

= –  1
2
PTehresaemquaaatniognaorifsslturaruigshAt BliniealAahB is

y – 0 = –  1 (x – 10)
2
2y = –x + 10

2y + x = 10

The equation of straight line OC with gradient of 2 and passes through the origin

(0, 0) is y = 2x.

9

(b) 2y + x = 10…1
y – 2x = 0…2

1 × 2: 4y + 2x = 20…3
2 + 3: 5y = 20

y=4
Substitute y = 4 into 1: 2(4) + x = 10

8 + x = 10
x=2

ºC(2, 4)

DisJtanracke OC = ! (2 – 0)2 + (4 – 0)2

= ! 20
= 4.472 units

7. (a) 3y – x = 15
3y = x + 15

y= 1 x + 5
3

The gradient of DC is 1 .
3
Equation AB that paraellel to DC and passses through point A(1, 2) is

y – 2 = 1 (x – 1)
3
3y – 6 = x – 1

3y – x = 5

Because DE is the perpendicular bisector of DC,
1
3 m2 = –1

m2 = –3
Gradient of DE is –3.

Equation DE that perpendicular to DC and passes through point D(3, 6) is

y – 6 = –3(x – 3)

y – 6 = –3x + 9

y + 3x = 15

(b) The coordinate of E is the values of x and y of the simultaneous equation AB and DE.

3y – x = 5…1
y + 3x = 15…2
1 × 3: 9y – 3x = 15…3
2 + 3: 10y = 30

y=3
Substitute y = 3 into 1: 3(3) – x = 5

9–x=5

x=4
º The coordinate of E is (4, 3).
Since DE is the perpendicular bisector of AB, thus E is the midpoint of AB. Let the
coordinate of point B is (x, y).

10

1 + x 2+y
2 2
( )TJhaudsi, , = (4, 3)

W hicIahitius, 1+x = 4 adnadn 2+y = 3
2 2

1 + x = 8 2+y=6

x=7 y=4

º The coordinate of B is (7, 4).

8. (a) mAB = 10 – 6 mBC = 4 – 10
11 – 3 9 – 11

= 1 =3
2

mCD = 4 – 2 mAD = 6 –2
9 – 5 3 –5

= 1 = –2
2
Since mAB = mCD thus, AB and CD are parallel.

Since mABmAD = –1, thus AB and AD are perpendicular.

Since mCDmAD = –1, thus CD and AD are perpendicular.

(b) Equation of straight line AB

y–6= 1 (x – 3)
2
2y – 12 = x – 3

2y = x + 9

(c) Equation of straight line

y – 4 = –2(x – 9)

y – 4 = –2x + 18

y + 2x – 22 = 0

( )TitMikitdepnogianht AB = 3 + 11 , 6 + 10
2 2

= (7, 8)

Left hand side = 8 + 2(7) – 22

=0

= Left hand side

Thus, the line passes through the midpoint AB.

9. (a) (i) mAC = 9 – (–3)
9–1
3
= 2

9 + 1, 9 + (–3)
2 2
( )TitMikidtepnogianht AC =

= (5, 3)

TPehresaemquaaatniopneomfbpaehrapgeindiucaulsaarmbaisseecrteonrjAanCg AC

y – 3 = – 23 (x – 5)

3y – 9 = –2x + 10

3y + 2x = 19

11

(ii) 3y + 2x = 19…1
y – 8x = –63…2

1 × 4: 12y + 8x = 76…3
2 + 3: 13y = 13

y=1
Substitute y = 1 into 1: 3(1) + 2x = 19

2x = 16
x=8

Thus, coordinate B is (8, 1).

(b) (i) Let point D is (x, y)

y+1
2
( )x+ 8, = (5, 3)
2
x + 8 y+1
2 = 5 2 = 3

x + 8 = 10 y+1=6

x = 2 y=5

MakTah, uksoocordoirndaitnaDteiaDlaihs (2, 5).

(ii) AC = ! (9 + 3)2 + (9 – 1)2

= ! 208

= 4! 13

BD = ! (1 – 5)2 + (8 – 2)2

= ! 52

= 2! 13

AC = 4! 13
BD 2! 13

=2
AC = 2BD (tSehrtouwnjnu)k)

10. (a) 3y − x = 8 …1
y − x = 4 …2
1 − 2: 2y = 4

y=2
Substitute y = 2 into 1: 3(2) − x = 8

6−x=8

x = −2

Thus, coordinate P is (−2, 2). (Shown)

(b) The equation of straight line perpendicular to m = − 1 and passes through point P(−2, 2) is
2
−− x12−(2x
y − 2 = + 2)
2y − 4 =

2y + x = 2 (Stehrotuwnnju)k)

(c) The equation of straight line AB with m = 2 and passes through point P(−2, 2) is
y − 2 = 2(x + 2)

y − 2 = 2x + 4

y = 2x + 6

12

at x-axis, y = 0,

0 = 2x + 6

2x = −6

x = −3

and at y-axis, x = 0,

y = 2(0) + 6

y=6

Thus, coordinate A is (−3, 0) and coordinate B is (0, 6). (shown)

(d) Let P(−2, 2) divides AB in the ratio m : n and x-coordinate of P is −2.

TJhaudsi, n(–3) + m(0) = −2
m + n −3n = −2m
− 2n

2m = n
m 1
n = 2

Thus, ratio AP is 1 . (Shown)
PB 2
8 – (–2)
11. mAB = 1 – (– 4)

=2

Equation of tangent BC

y – 8= – –12x(x+ – 1)
2y – 16 = 1

2y + x = 17

Inquiry 3 (Page 192)

3. (a) AD = 4 units, DE = 4 units, BE = 5 units and CD = 8 units
(b) Area ∆ACD = 16 units2, area trapezium BCDE = 26 units2 and area ∆ABE = 20 units2
(c) Area ∆ABC = 22 units2

4. Area ∆ABC = Area ∆ACD + area trapezium BCDE – area ∆ABE
= 16 + 26 – 20
= 22 units2

5. Yes, there are many ways to determine the area of a triangle, one of it is by using formula.

Self Practices 7.6 (Page 195)

1. (a) Area = 1 5 2 85
2 10 1 3 10

= 1 [(5 + 6 + 80) – (20 + 8 + 15)]
2
1
= 2 (48)

= 24 units2

13

(b) Area= 1 3 6 – 4 3
2 1 421

= 1 [(12 + 12 – 4) – (6 – 16 + 6)]
2

= 1 (24)
2

= 12 units2

(c) Area = 1 – 4 5 2 – 4
2 –3 1 6 –3

= 1 [(– 4 + 30 – 6) – (–15 + 2 – 24)]
2

= 1 (57)
2

= 28 1 units2
2

2. 1 3 1 k 3 = 10
2 4 –2 0 4

1 [(–6 + 0 + 4k) – (4 – 2k + 0)] = ±10
2
1
2 (6k –10) = ±10

1 (6k –10) = 10 1 (6k –10) = –10
2 2
6k – 10 = 20 6k – 10 = –20

6k = 30 6k = –10
k = – 35
k = 5

( )Thus, the possible coordinates of R are (5, 0) or – 5 , 0 .
3
1 8 2 –2 8
3. Area = 2 4 1 –1 4

= 1 [(8 – 2 – 8) – (8 – 2 – 8)]
2

= 0 (Shown)

4. 1 –2 2 10 –2 = 0
2 –1 p 5 –1
1
2 [(–2p + 10 – 10) – (–2 + 10p – 10)] = 0

1 (12 – 12p) = 0
2 12 – 12p = 0

12p = 12

p=1

5. (a) 1 – 4 5 –1 – 4 = 15
2 –1 3 k –1

1 [(–12 + 5k + 1) – (–5 – 3– 4k)] = ±15
2 1 (9k – 3) = ±15
2

14

1 (9k – 3) = 15 1 (9k – 3) = –15
2 9k – 3 = 30 2 9k – 3 = –30

9k = 33 9k = –27
11
k= 3 k = –3

(b) 1 5 3 1 5 = 10
2 k 7 3 k

1 [(35 + 9 + k) – (3k + 7 + 15)] = ±10
2
1
2 (22 – 2k) = ±10

1 1 (22 – 2k) = –10
2 2
22 – 2k = 20 22 – 2k = –20

2k = 2 2k = 42

k = 1 k = 21

(c) 11 k 1 1 = 12
2 –2 6 2 –2

1 [(6 + 2k – 2) – (–2k + 6 + 2)] = ±12
2
1
2 (4k – 4) = ±12

1 1 (4k – 4) = –12
2 2
4k – 4 = 24 4k – 4 = –24

4k = 28 4k = –20

k = 7 k = –5

(d) 1 3 4 1 3 =5
2 0 k 4 0

1 [(3k + 16 + 0) – (0 + k + 12)] = ±5
2
1
2 (2k + 4) = ±5

1 (2k + 4) = 5 1 (2k + 4) = –5
2 2
2k + 4 = 10 2k + 4 = –10

2k = 6 2k = –14

k=3 k = –7

Self Practice 7.7 (Page 196)

1. 1 1 –5 –2 2 1
(a) Area = 2 7 6 –4 –3 7

= 1 [(6 + 20 + 6 + 14) – (–35 – 12 – 8 – 3)]
2

= 1 (104)
2

= 52 units2

15

(b) Area = 1 2 –6 –1 8 2
2 9 4 –3 1 9

= 1 [(8 + 18 – 1 + 72) – (–54 – 4 – 24 + 2)]
2

= 1 (177)
2

= 88 1 units2
2

(c) Area = 1 0 –6 –3 –1 0
2 2 –2 –5 –3 2

= 1 [(0 + 30 + 9 – 2) – (–12 + 6 + 5 + 0)]
2

= 1 (38)
2

= 19 units2

(d) Area = 1 3 –2 2 5 3
2 4 0 – 4 1 4

= 1 [(0 + 8 + 2 + 20) – (–8 + 0 – 20 + 3)]
2

= 1 (55)
2

= 27 1 units2
2
k –2 4 2 k
2. 1 615 8 6 = 30
2

1 [(k – 10 + 32 + 12) – (–12 + 4 + 10 + 8k)] = 30
2
1
2 (32 – 7k) = 30

32 – 7k = 60

7k = –28

k = –4

Self Practices 7.8 (Page 197)

1. Area = 1 –2 3 2 0 –3 –2
2 –5 2 8 9 1 –5

= 1 [(– 4 + 24 + 18 + 0 + 15) – (–15 + 4 + 0 – 27 – 2)]
2

= 1 (93)
2

= 46 1 units2
2

16

2. Area = 1 0 2 1 –2 – 4 –3 0
2 –1 1 5 6 2 –1 –1

= 1 [(0 + 10 + 6 – 4 + 4 + 3) – (–2 + 1 – 10 – 24 – 6 – 0)]
2

= 1 (60)
2

= 30 units2

Self Practices 7.9 (Page 198)

( )1. –1 + 5, 6 + 2
(a) M = 2 2

= (2, 4)

–3 + x = 2 adnadn 0+y = 4
2 2

–3 + x = 4 y=8

x=7

º C(7, 8)

(b) Area ∆ABM = 1 –3 5 2 –3
2 0 2 4 0

= 1 [(–6 + 20 + 0) – (0 + 4 – 12)]
= 2

1 (22)
2

= 11 units2

Area ABCD = 1 –3 5 7 –1 –3
2 0 2 86 0

= 1 [(–6 + 40 + 42 – 0) – (0 + 14 – 8 – 18)]
= 2

1 (88)
2

= 44 units2

Thus, the ratio area ∆ABM to the area of parallelogram ABCD is 11 : 44 = 1 : 4.

2. (a) 1 0 8–k 0 0 = 12
2 0 2 8 0
k

1
2

1 1 (32 – 4k) = 12
2 32 – 4k = –24 2 32 – 4k = 24

4k = 56 4k = 8

k = 14 k=2

ºk=2
(b) P(3, 2)

17

3. (a) Area ∆ABC = 14 74 4
2 –2 5 7 –2

= 1 [(20 + 49 – 8) – (–14 + 20 + 28)]
2

= 1 (27)
2

= 13 1 units2
2

(b) 14 4 k 4 = 27
2 –2 7 0 –2 2

1 [(28 + 0 – 2k) – (–8 + 7k + 0)] = 27
2 2

36 – 9k = 27

9k = 9

4 7, k=1
+ –2 + 5
2 2
( )(c) TitMik itdepnogianht AB =

= ( 11 , 3 )
2 2

x+4 = 11 y+7 = 3
2 2 2 2

2x + 8 = 22 2y + 14 = 6

x=7 y = –4

º E(7, – 4)

(d) Area ACBE = 1 47 7 4 4
2 –2 – 4 5 7 –2

= 1 [(–16 + 35 + 49 – 8) – (–14 – 28 + 20 + 28)]
2

= 1 (54)
2
= 27 units2

Intensive Practice 7.3 (Page 199)
( )1.
Midpoint AC = –5 + 3 , 3 + 5
22

= (–1, 4)

º E(–1, 4)

x+0 = –1 y + (–2) =4
2 2
x = –2 y = 10

º D(–2, 10)

(b) Area ABCD = 1 –5 0 3 –2 –5
2 3 –2 5 10 3

= 1 [(10 + 0 + 30 – 6) – (0 – 6 – 10 – 50)]
2

= 1 (100)
2

= 50 units2 18

3 + (–5) 3 + (–1)
2 2
( )2. (a) TitiMk itdepnogianht PR = ,

= (–1, 1)

h+0 = –1 adnadn 3+k =1
2 2

h = –2 k = –1

(b) Area PQRS = 1 3 –2 –5 0 3
2 3 3 –1 –1 3

= 1 [(9 + 2 + 5 + 0) – (–6 – 15 + 0 – 3)]
2

= 1 (40)
2

= 20 units2

3. (a) Area ∆ABC = 1 –1 2 6 –1
2 –5 1 9 –5

= 1 [(–1 + 18 – 30) – (–10 + 6 – 9)]
2

= 1 (0)
2

= 0 unit2

(b) Points A, B and C are collinear.

4. Area of polygon = 1 5 2 –3 – 4 –1 3 5
2 2 6 2 0 –3 –2 2

= 1 [(30 + 4 + 12 + 2 + 6) – (4 – 18 – 8 – 9 – 10)]
2

= 1 (95)
2

= 47 1 unit2s2
2

5. 1 5 3 –6 5 = 16
2 –1 3 p –1

1 [(15 + 3p + 6) – (–3 – 18 + 5p)] = ±16
2
1
2 [(42 – 2p) = ±16

1 1 [(42 – 2p) = –16
2 2

42 – 2p = 32 42 – 2p = –32

2p = 10 2p = 74

p = 5 p = 37

19

6. 1 2 r–1 r+3 2 =0
2 2r – 1 r+1 0 2r – 1

1 [2(r + 1) + 0 + (r + 3)(2r – 1) – (2r – 1)(r – 1) – (r + 1)(r + 3) – 0] = 0
2
2r + 2 + 2r2 + 5r – 3 – 2r2 + 3r – 1 – r2 – 4r – 3 = 0

–5 + 6r – r2 = 0

r2 – 6r + 5 = 0

(r – 1)(r – 5) = 0

8 –1 3 8 r = 1 or r = 5
a 2 10 a
7. (a) 1 =0
2

1 [(16 – 10 + 3a) – (–a + 6 + 80)] = 0
2

4a = 80

a = 20

(b) 1 8 –1 3 8 = 12
2 a 2 10 a

1 [(16 – 10 + 3a) – (–a + 6 + 80)] = ±12
2
1
2 (4a – 80) = ±12

1 (4a – 80) = 12 1 (4a – 80) = –12
2 2
4a – 80 = 24 4a – 80 = –24

4a = 104 4a = 56

a = 26 a = 14

8. (a) EF = FG

! (k – 4)2 + (0 – 4)2 = ! (4 – 8)2 + (4 – 7)2

k2 – 8k + 16 + 16 = 25

k2 – 8k + 7 = 0

(k – 1)(k – 7) = 0

k = 1 or k = 7

Thus, the value of k is 7.

(b) (i) EH = GH

! (x – 0)2 + (11 – 7)2 = ! (x – 7)2 + (11 – 8)2

x2 + 16 = x2 – 14x + 49 + 9

14x = 42

x=3

º H(3, 11)

(ii) Area ∆EFG = 1 0 4 7 0
2 7 4 8 7

= 1 [(0 + 32 + 49) – (28 + 28 + 0)]
2

= 1 (25)
2
1
= 12 2 units2

20

Area EFGH = 1 0 4 730
2 7 4 8 11 7

= 1 [(0 + 32 + 77 + 21) – (28 + 28 + 24 + 0)]
2

= 1 (50)
2

= 25 uunniitts22

NiRsbataiho lauraesa ∆EFG : alureaas EFGH = 25 : 25
2

=1:2

9. (a) 1 0 m + 1 2m + 1 m 0 = 34 1
2 0 m – 7 2m m+6 0 2

1 [(2m2 + 2m + 2m2 + 13m + 6) – (2m2 – 13m – 7 + 2m2] = ±34 1
2 2

1 (28m + 13) = ±34 1
2 2

1 (28m + 13) = 34 1 oatrau 1 (28m + 13) = –34 1
2 2 2 2

28m + 13 = 69 28m + 13 = –69

m = 2 m = – 4114

MThakuas, m = 2 (. 0) 0
0 3 2 0
(b) Area ∆OPR = 1 0 –5 8
2

= 1 [(0 + 24 + 0) – (0 – 10 + 0)]
2

= 1 (34)
2
= 17 uunniitts22

10. (a) AB = ! (0 – 9)2 + (7 – 0)2
= ! 130
= 11.402 units
= 11.402 × 100 m
= 1140.2 m
= 1.1402 km

(b) Area = 1 0 7 12 0
2 9 0 12 9

= 1 [(0 + 84 + 108) – (63 + 0 + 0)]
2

= 1 (129)
2
= 64.5 units22

1 unit = 100 m (0.1 km)
1 unit2 = 0.01 km2
64.5 unit2 = 0.645 km2
Thus, the area covered by the three stations is 0.645 km2.

21

Inquiry 4 (Page 200)
4. The shape of locus P formed is a circle.

5. (x – x1)2 + (y – y1)2 = r2

Inquiry 5 (Page 201)

4. Yes. The circle is locus of moving point P.
(x – x1)2 + (y – y1)2
PTerhseameqauaantiioanlaihs (x – x2)2 + (y – y2)2 = m2
n2

6. If PA : PB = 1 : 1, point P(x, y) always has the same distance from two fixed points

A(x1, y1) and B(x2, y2). Thus, locus P is the perpendicular bisector of line AB. The
equation is

PA = PB

(x – x1)2 + (y – y1)2 = (x – x2)2 + (y – y2)2

Self Practice 7.10 (Page 203)

1. (a) ! (x – 0)2 + (y – 0)2 = 3
x2 + y2 = 9

x2 + y2 – 9 = 0
(b) ! (x – 2)2 + (y – 3)2 = 3

x2 – 4x + 4 + y2 – 6y + 9 = 9
x2 + y2 – 4x – 6y + 4 = 0

(c) ! (x + 4)2 + (y – 5)2 = 3
x2 + 8x + 16 + y2 – 10y + 25 = 9
x2 + y2 + 8x – 10y + 32 = 0

(d) ! (x + 1)2 + (y + 6)2 = 3
x2 + 2x + 1 + y2 + 12y + 36 = 9
x2 + y2 + 2x + 12y + 28 = 0

2. ! (x + 2)2 + (y – 1)2 = 5
x2 + 4x + 4 + y2 – 2y + 1 = 25

x2 + y2 + 4x – 2y – 20 = 0

3. (a) PA = 1
PB 2

2PA = PB

2! (x + 2)2 + (y – 0)2 = 1! (x – 4)2 + (y – 0)2

4(x2 + 4x + 4 + y2) = 1(x2 – 8x + 16 + y2)
4x2 + 16x + 4y2 + 16 = x2 – 8x + y2 + 16

3x2 + 3y2 + 24x = 0
x2 + y2 + 8x = 0

22

(b) PC = 1
PD 3

3PC = PD

3! (x + 3)2 + (y – 0)2 = 1! (x – 2)2 + (y – 5)2

9(x2 + 6x + 9 + y2) = 1(x2 – 4x + 4 + y2 – 10y + 25)
9x2 + 54x + 81 + 9y2 = x2 – 4x + y2 – 10y + 29
8 x2 + 58x + 8y2 + 10y + 52 = 0
4x2 + 4y2 + 29x + 5y + 26 = 0

(c) PE = 3
PF 2

2PE = 3PF

2! (x – 0)2 + (y – 2)2 = 3! (x + 2)2 + (y – 4)2

4(x2 + y2 – 4y + 4) = 9(x2 + 4x + 4 + y2 – 8y + 16)
4x2 + 4y2 – 16y + 16 = 9x2 + 36x + 36 + 9y2 – 72y + 144
5x2 + 36x + 5y2 – 56y + 164 = 0
5x2 + 5y2 + 36x – 56y + 164 = 0

(d) PR = 2
PS 1

PR = 2PS

! (x – 1)2 + (y – 2)2 = 2! (x – 4)2 + (y + 1)2

(x2 – 2x + 1 + y2 – 4y + 4) = 4(x2 – 8x + 16 + y2 + 2y + 1)
x2 – 2x + y2 – 4y + 5 = 4x2 – 32x + 4y2 + 8y + 68

3x2 – 30x + 3y2 + 12y + 63 = 0
x2 + y2 – 10x + 4y + 21 = 0

4. QJ = 2
QK 3

3QJ = 2QK

3! (x + 1)2 + (y – 3)2 = 2! (x – 4)2 + (y – 6)2

9(x2 + 2x + 1 + y2 – 6y + 9) = 4(x2 – 8x + 16 + y2 – 12y + 36)

9x2 + 18x + 9y2 – 54y + 90 = 4x2 – 32x + 4y2 – 48y + 208

5x2 + 5y2 + 50x – 6y – 118 = 0

5. RA = 2RB
! (x – 6)2 + (y – 0)2 = 2! (x + 3)2 + (y – 0)2
x2 – 12x + 36 + y2 = 4(x2 + 6x + 9 + y2)
x2 – 12x + 36 + y2 = 4x2 + 24x + 36 + 4y2
3x2 + 3y2 + 36x = 0
x2 + y2 + 12x = 0

23

6. PO = 1
PA 4

4PO = PA

4! x2 + y2 = ! (x – 2)2 + (y – 0)2

16(x2 + y2) = x2 – 4x + 4 + y2

15x2 + 15y2 + 4x – 4 = 0

7. (a) PA = PB

! (x + 2)2 + y2 = ! x2 + (y – 4)2

x2 + 4x + 4 + y2 = x2 + y2 – 8y + 16
4x + 8y – 12 = 0
x + 2y – 3 = 0

(b) PA = PB

! (x + 3)2 + (y – 5)2 = ! (x – 2)2 + (y + 4)2

x2 + 6x + 9 + y2 – 10y + 25 = x2 – 4x + 4 + y2 + 8y + 16
10x – 18y + 14 = 0
5x – 9y + 7 = 0

(c) PA = PB

! (x – 2)2 + (y – 3)2 = ! (x – 6)2 + (y – 8)2

x2 – 4x + 4 + y2 – 6y + 9 = x2 – 12x + 36 + y2 – 16y + 64
8x + 10y – 87 = 0

Self Practice 7.11 (Page 204)

1. PA = 4
! (x – 3)2 + (y – 4)2 = 4

x2 – 6x + 9 + y2 – 8y + 16 = 16

x2 + y2 – 6x – 8y + 9 = 0

2. (a) PQ = PR

! (x – 8)2 + (y – 7)2 = ! (x – 11)2 + (y – 4)2

x2 – 16x + 64 + y2 – 14y + 49 = x2 – 22x + 121 + y2 – 8y + 16

6x – 6y – 24 = 0

x–y–4=0

(b) ST = 5

! (x – 7)2 + (y – 8)2 = 5
x2 – 14x + 49 + y2 – 16y + 64 = 25

x2 + y2 – 14x – 16y + 88 = 0 (Shown)

(c) y = x – 4…1
x2 + y2 – 14x – 16y + 88 = 0…2

24

Substitute 1 into 2
x2 + (x – 4)2 – 14x – 16(x – 4) + 88 = 0

x2 + x2 – 8x + 16 – 14x – 16x + 64 + 88 = 0
2x2 – 38x + 168 = 0
x2 – 19x + 84 = 0
(x – 7)(x – 12) = 0

x = 7 or x = 12

Substitute x = 7 into 1: y = 7 – 4
=3

Substitute x = 12 into 1: y = 12 – 4
=8

Thus, the intersection point is (7, 3) or (12, 8).

3. (a) PA = 2
PB 1

PA = 2PB

! (x + 2)2 + y2 = 2! (x – 1)2 + y2
x2 + 4x + 4 + y2 = 4(x2 – 2x + 1 + y2)
x2 + 4x + 4 + y2 = 4x2 – 8x + 4y2 + 4
3x2 + 3y2 – 12x = 0

x2 + y2 – 4x = 0 (Shown)

(b) Substitute point C(2, 2) into the equation of circle:

22 + 22 – 4(2) = 0

= right hand side

Thus, point C(2, 2) is on the circle.

Intensive Practice 7.4 (Page 205)

1. (a) 2RA = RB

2! (x + 1)2 + (y – 10)2 = ! (x – 2)2 + (y – 6)2

4(x2 + 2x + 1 + y2 – 20y + 100) = x2 – 4x + 4 + y2 – 12y + 36

4x2 + 8x + 4y2 – 80y + 404 = x2 – 4x + y2 – 12y + 40

3x2 + 3y2 + 12x – 68y + 364 = 0

(b) At y-axis, x = 0

3y2 – 68y + 364 = 0

(3y – 26)(y – 14) = 0

( )y =26 or y = 14
3
The coordinate of locus R that touches y-axis is 0, 26 and (0, 14).

2. mASmBS = –1 3

y–9( )( )y – 1 = –1
x–7x–1

(y – 1)(y – 9) = –(x – 1)(x – 7)

y2 – 10y + 9 = –x2 + 8x – 7

x2 + y2 – 8x – 10y + 16 = 0

25

3.(a) RJeadjaiuris bouf lcaitraclne = ! (8 – 5)2 + (2 – 6)2
= ! 25
=5

! (x – 6)2 + (y – 5)2 = 5
x2 – 12x + 36 + y2 – 10y + 25 = 25

x2 + y2 – 12x – 10y + 36 = 0
(b) When S(k, 2),

k2 + (2)2 – 12k – 10(2) + 36 = 0
k2 – 12k + 20 = 0
(k – 2)(k – 10) = 0
k = 2 or k = 10

4. PS = PR
x + 1 = ! (x – 1)2 + y2

(x + 1)2 = (x – 1)2 + y2
x2 + 2x + 1 = x2 – 2x + 1 + y2

y2 = 4x

5. (a) a2 + b 2 = 92

a2 + b 2 = 81

(b) x = 2(0) + 1(a) y= 2(b) + 1(0)
x= a3 y=
3 2 3
3
b

a = 3x b= 3 y
2

( )Substitute a = 3x and b = 3 y into a2 + b 2 = 81
2
(3x)2 + 3 y 2 = 81
2

9x2 + 9 y2 = 81
4

36x2 + 9y2 = 324

4x2 + y2 = 36

6. 2 1 m A1m D 1
P 2 2
m
1m S

B CR

1 1 m
2

Q

The locus consists of curves of a quadrant of 3 circles:

(i) APQ which is a quadrant with center A and radius of 2 1 m.
2

(ii) BQR which is a quadrant with center B and radius of 1 1 m.
2

(iii) CRS which is a quadrant with center C and radius of 1 m.
2
26

Mastery Practice (Page 207-209)

1. (a) h+7 = 2 1+k = 3
2 2

h + 7 = 4 1+k=6

h = –3 k=5
5–1
(b) mAB = 7 – (–3)

= 4
10

= 2
5

(c) m = – 52 , (2, 3)

y – 3 = –  5 (x – 2)
2

2y – 6 = –5x + 10

2y + 5x = 16

2. (a) x = 3(–2) + 2(8) y= 3(6) + 2(– 4)
5 5
10 10
= 5 = 5

= 2 = 2

º P(2, 2)
6 – (– 4)
(b) mAB = –2 – 8

= –1
m = 1, (2, 2)
y – 2 = 1(x – 2)
y–2=x–2

y=x

3. mPQ = mQR

2 – (–1) = n+3–2
n–1 n2 – n

3 = n+1
n–1 n2 – n

3(n2 – n) = (n + 1)(n – 1)

3n2 – 3n = n2 – 1

2n2 – 3n + 1 = 0

(2n – 1)(n – 1) = 0

n= 1 or n = 1
2

27

4. Let coordinate T is (0, h).
 1
2 –3 3 0 –3   = 13.5
4 –1 h 4

1 [(3 + 3h + 0) – (12 – 0 – 3h)] = ±13.5
2

1 (3 + 3h – 12 + 3h) = ±13.5
2

1 (6h – 9) = ±13.5
2

1 (6h – 9) = 13.5 1 (6h – 9) = –13.5
2 2

6h – 9 = 27 6h – 9 = –27

6h = 36 6h = –18

h = 6 h = –3

Thus, the possible coordinates of T is (0, 6) or (0, –3).

5. PA = 3PB
! (x – 2)2 + (y – 0)2 = 3! (x + 4)2 + (y – 0)2

x2 – 4x + 4 + y2 = 9(x2 + 8x + 16 + y2)
x2 – 4x + 4 + y2 = 9x2 + 72x + 144 + 9y2
8x2 + 8y2 + 76x + 140 = 0
2x2 + 2y2 + 19x + 35 = 0

6. x= 1(–3) + 2(6) y= 1(–1) + 2(5)
3 3
9 9
= 3 = 3

= 3 = 3

Both birds meet at coordinate (3, 3).

7. (a) Area ∆ ABC = 10

1 ×4×t= 10
2
t=5

x= 2+6 y=2–5
2 = –3
= 4

º C(4, –3)

(b) x+4 =6 y + (–3) = 2
2 2

x + 4 = 12 y–3=4

x=8 y=7

º D(8, 7)

(c) (i) mAC = –3 – 2
4–2

= –  5
2

28

k–7 = – 5
11 – 8 2

2(k – 7) = –5(3)

2k – 14 = –15

2k = –1
1
k = –  2

(ii) mEC = – 1 – (–3)
2
11 – 4

= 5
14

mDE = – 25 5 5
14 2
mEC × mDE = × –

= – 2258

≠ –1

º CED is not a right-angled triangle.

8. (a) y = 2x – 5…1
Sy u=bs31titxu…te22 into 1

1 x = 2x – 5
3
x = 6x – 15

5x = 15
x=3

Substitute x = 3 into 2

y= 1 (3)
3

=1

º P(3, 1)

(b) m = –3, R(11, 7)

Equation QR: y – 7 = –3(x – 11)

y – 7 = –3x + 33

y + 3x = 40

m= 1 , R(11, 7)
3
Equation SR: y – 7 = 1 (x – 11)
3
3y – 21 = x – 11

3y – x = 10

29

(c) y + 3x = 40…1
1
y= 3 x…2

Substitute 2 into 1
1
3 x + 3x = 40

x + 9x = 120

10xx == 11220

Substitute x = 12 into 2

y= 1 (12)
3

=4

º Q(12, 4)

y – 2x = –5…1

3y – x = 10…2

2 × 2: 6y – 2x = 20…3

3 – 1: 5y = 25
y=5

Substitute y = 5 into 1
5 – 2x = –5
2x = 10
x=5

º S(5, 5)

(d) Area PQRS = 1 3 12 11 5 3
2 1 4 7 5 1

= 1 [(12 + 84 + 55 + 5) – (12 + 44 + 35 + 15)]
2

= 1 (50)
2

= 25 units2

Area ∆PQR = 1 3 12 11 3
2 1 4 7 1

= 1 [(12 + 84 + 11) – (12 + 44 + 21)]
2

= 1 (30)
2

= 15 units2

Area ∆PRS = 1 3 11 5 3
2 1 7 5 1

= 1 [(21 + 55 + 5) – (11 + 35 + 15)]
2

= 1 (20)
2

= 10 units2 2

30

PQ = ! (12 – 3)2 + (4 – 1)2 SR = ! (11 – 5)2 + (7 – 5)2
= ! 90 = ! 40
= 3! 10 = 2! 10

LAuraesa ∆ PQR = 15
ALureaas ∆ PRS 10

= 3
2
3! 10
PQ = 2! 10
SR

= 3
2

LAuraesa ∆ PQR = PQ (t(eSrthuonwjunk))
ALureaas ∆ PRS SR

9. (a) Area ∆ JKL = 1 2 11 5 2
21 5 91

= 1 [(10 + 99 + 5) – (11 + 25 + 18)]
2

= 1 (60)
2

= 30 units2

(b) Area ∆ JKP = 1 2 11 h 2
2 1 5 k 1

= 1 [(10 + 11k + h) – (11 + 5h + 2k)]
2

= 1 (9k – 4h – 1)
2

= 9k – 4h – 1
2

Area ∆ KLP = 1 11 5 h 11
2 5 9 k 5

= 1 [(99 + 5k + 5h) – (25 + 9h + 11k)]
2

= 1 (74 – 6k – 4h)
2

= 37 – 3k – 2h

(c) 9k – 4h – 1 = 10
2

9k – 4h – 1 = 20

9k – 4h = 21…1

37 – 3k – 2h = 10

3k + 2h = 27…2

2 × 2: 6k + 4h = 54…3

1 + 3: 15k = 75

k=5

31

Substitute k = 5 into 1

9(5) – 4h = 21

45 – 4h = 21

4h = 24

h=6

º P(6, 5)

(d) J(2, 1), P(6, 5)

mJP = 5–1
6–2

=1

Equation JP

y – 1 = 1(x – 2)

y–1=x–2

y=x–1
9–5
(e) (i) mKL = 5 – 11

= 4
–6

= –  2
3

Equation KL:
y – 5 = – 23 (x – 11)

3y – 15 = –2x + 22

3y + 2x = 37…1
y – x = –1…2

2 × 2: 2y – 2x = –2…3
1 + 3: 5y = 35

y=7
Substitute y = 7 into 1
3(7) + 2x = 37

2x = 16

x=8

º Q(8, 7)

(ii) n(11) + m(5) = 8
m + n

11n + 5m = 8m + 8n

3m = 3n

m = 3
n 3

= 1
1

º KQ : QL = 1 : 1

10. (a) OR = ! 45

! x2 + y2 = ! 45

x2 + y2 = 45…1

y = –2x…2

32

Substitute 2 into 1
x2 + (–2x)2 = 45

5x2 = 45
x2 = 9
x = ±3
Substitute x = –3 into 2

y = –2(–3)
=6

º R(–3, 6)

mOR = 6–0
–3 – 0

= –2

Equation RS

y–6 = 1 (x + 3)
2y – 12 = x2 +3

2y = x + 15

At y-axis, x = 0
2y = 0 + 15

y = 15
2
S(0, 15 )
º 2

Equation ST

y – 15 = –2(x – 0)
2

2y + 4x = 15…1

y = 2x…2

SGuabnstiktuatne 2 kinetodalam 1

2(2x) + 4x = 15

8x = 15

x = 15
8

( )ApWabhielna x= 15 , y = 2 15
8 8

= 15
4
( )º 15 15
T 8 , 4

(b) Area ORST = 1 0 15 0 –3 0
2 0 8
15 15 60
4 2

( )= 1 225 + 45
2 16 2

= 18 9 units2
32

33

11. (a) y = 8
x
At point P(h, 8)

8= 8
h

h=1

At point Q(k, 2)
8
2= k

k=4
(b) P(1, 8), Q(4, 2)

m = 2 – 8
4 – 1

= –2

Equation PQ

y – 8 = –2(x – 1)

y + 2x = 10

(c) y = –2x + 8

y = –2x – 8

12. (a) 5y – x = 33

y= 1 x + 33
5 5

m= 1
5

mBP = –5

Equation BP is
y – 1 = –5(x + 2)

y + 5x + 9 = 0
(b) 5y – x = 33

x = 5y – 33…1
y + 5x + 9 = 0…2

Substitute 1 into 2
y + 5(5y – 33) + 9 = 0

26y – 156 = 0
y = 6

Substitute y = 6 into 1
x = 5(6) – 33
= –3

Thus, the coordinate of P is (−3, 6).

2(–8) + x = –3 2(5) + y =6
3 3

–16 + x = –9 10 + y = 18

x=7 y=8

Thus, the coordinate of D is (7, 8).

34

mDC = mAB = 5–1
–8 – (–2)

= –  2
3

Equation BC

y – 8 = –  2 (x – 7) y – 1 = x51+(x2+ 2)
3 5y – 5 =

3y – 24 = –2x + 14

3y = –2x + 38 x = 5y – 7

3y = –2x + 38…1
x = 5y – 7…2

Substitute 2 into 1
3y = –2(5y – 7) + 38

13y = 52
y=4

Substitute y = 4 into 2
x = 5(4) – 7
= 13

Thus, coordinate C is (13, 4).

(c) Area ABCD = 1 –8 –2 13 7 –8
2 5 1 4 8 5

= 1 [(–8 – 8 + 104 + 35) – (–10 + 13 + 28 – 64)]
2

= 1 (156)
2
= 78 units22

( )13. –1 + 7, –2 + 4
(a) E = 2 2

= (3, 1)

0 + x = 3 5 + y = 1
2 x = 6 2 y = –3

º B = (6, −3)

(b) AB = ! (–1 – 6)2 + (–2 + 3)2
= 5! 2

BC = ! (7 – 6)2 + (4 + 3)2
= 5! 2

CD = ! (0 – 7)2 + (5 – 4)2
= 5! 2

AD = ! (0 + 1)2 + (5 + 2)2
= 5! 2

Since all sides have equal length, thus quadrilateral ABCD is a square.

35

14. (a) P = 6x – 2x – 400
= 4x – 400
º P = 4x – 400

(b)

P

2000 100 200 300350400 500 x
1600
1500
1000

500

0
− 500

(i) RM1 600
(ii) 350 copies

13.

y

B(6, 7)

A (1, 2) P
0 C(7, 2)
x

36

CHAPTER 8 VECTOR

Mind Challenge (Page 212)

Scalar Quantity Vector Quantity Not a Scalar or
Vector
Time Weight
Volume Force Conductivity of metal
Electric charge Impulse Elasticity
Density Momentum
Energy Radio frequency

Self Practice 8.1 (Page 214)

1. (a) Scalar quantity because the quantity only consists of magnitude.
(b) Vector quantity because the quantity consists of magnitude and direction.
(c) Scalar quantity because the quantity only consists of magnitude.
(d) Scalar quantity because the quantity only consists of magnitude.
(e) Vector quantity because the quantity consists of magnitude and direction.

Self Practice 8.2 (Page 216-217) (b) SN
1. (a)

4 cm RS

5N Y 1 cm represents 10 km
X

1 cm represents 1N

R

(c) (d)

∼v 3.5 cm N

4 cm a
1 cm represents 5 km h–1
1 cm represents 2 kg m s–1

2. Magnitude~f = !2_2__+ 42 Direction : tan q = 2
= ! 20 N q = 63.43°

3. Distance = ! 902 + 752 Direction f : 90° – 63.43° = 026.57°

= 117.15 km B
75

O 90 A

1

˜˜
4. ~a = ~d MN = CD
~c = ~f ˜˜
EF = KL
˜˜
~b = ~e GH = AB

˜ ˜
5. (a) (i) ED (b) (i) DC
˜
(ii) FE ˜
˜ (ii) CB
(iii) AF
˜
(iii) BA

Self Practice 8.3 (Page 218)

˜ 1 a~ ~x = – _3 ~a ~y = – 47 ~a ˜ 5 ~a
1. PQ = 2 2 RS = 4

Inquiry 1 (Page 218)

1. (a) ˜AB  = ! 32 + 42

= 5 units

˜CD = ! 62 + 82

= 10 units

(b) ˜AB  : ˜CD = 5 : 10

=1:2

(c) Gradient of AB = 4 , Gradient of CD = 4
3 3
The straight lines AB and CD are parallel.
˜ ˜
(d) AB = 1 CD
2

2. a~ = k~b, with k as a constant.

Mind Challenge (Page 219)
˜˜
XY = a XZ
˜˜
XY = b YZ
˜
XZ =lY˜Z with a, b and l as constants.

Self Practice 8.4 (Page 220)

1. ˜AB  = 250~a~a
˜PQ 
˜ ˜
AB = 1 PQ
4
˜LM
2. M˜N = 168~x~x

˜ 1 ˜
LM = 3 MN

2

˜˜
Thus, LM and MN are parallel. Because of M is a common point, then L, M and N are

collinear.

3. (a) 4m + 3 = 0 n–7=0
m = – 34 n=7

(b) m + n = 1 …1

m – 2n = 10…2

1 – 2: 3n = –9

n = –3…3

Substitute 3 into 1

m = 1 – (–3)

=4

4. ˜VW = 21
˜XY  6

˜ 7 ˜
VW = 2 XY

5. a= 1 (k – 2)a
2
1
2 (k – 2) = 1

k=4

6. PQT and PRS are similar triangles.

PS = 8
PT 5

RS = 8
QT 5
˜ ˜
Thus, SR = – 8_ QT
5

Intensive Practice 8.1 (Page 220)

1. ˜DC = 2 cm
˜
DC = u~
= 6 cm
˜AB 
˜
Thus, AB = 3~u
˜˜

2. (a) AB = 3DC
˜AB  = 3 × 4 cm

= 12 cm

(b) The triangles ECD and EAB are similar triangles
˜
(i) EC = 2~a
˜
(ii) BE = 6~b

3

3. ˜AB  = 64~~xx
˜AC 
˜ ˜
AB = 2 AC
3
˜ ˜
Thus, AB and AC are parallel. Because of A is a common point, then A, B and C are

collinear.

4. h + k = 0 …1

h – k + 1 = 0 …2

1 – 2: 2k – 1 = 0

k= 1 …3
2

Substitute 3 into 1
˜ ˜ h = – 12
5. PQ and QR are parallel. Thus,

(k + 2)~x + 4~y = l (h~x +~y )

Comparing both sides of the equation,
k + 2 = lh …1
4 = l …2

Substitute 2 into 1
k = 4h – 2

Inquiry 2 (Page 221)

3.

Ranjit Tan Mia Dayang

Finish Start Start Finish

Finish

Start Start
Finish

4. The path taken by them produce a displacement which is a resultant vector.

Self Practice 8.5 (Page 224)

1. (a) (b)

2 ∼u+∼v 1_ ∼v + 2 ∼u
2

4

(c) (d)

∼u − 2 ∼v 2 _3
2
∼u − ∼v

________
2. Magnitude = ! 702 + 802

= 106.30 km h−1 ␪ ~p +~q
70 80
Direction: tan q = 80
70
q = 48.81°



= 131.19°

3. (a) ˜ 2 ˜
AB = 3 DC

= 2 ~y
3
˜˜˜
(b) AC = AD + DC

= –~x + ~y
˜˜˜˜
(c) BC = BA + AD + DC

= – 32 ~y – ~x + ~y

= 1 ~y – ~x
3
˜˜˜
(d) BD = BA + AD
_2
= – 3 ~y – ~x

4. (a) Initial velocity of the plane 160
600
= ! 6002 – 1602
␪
= 578.27 km h−1

(b) tan q = 160
600

q = 14.93°

Original direction of the plane

= 360° – 14.93°

= 345.07°

Self Practice 8.6 (Page 225)

˜˜˜ ˜˜˜
1. XY = XO + OY XZ = XO + OZ

= ((k– 4–~x4+)~x2+~y)~y+ (k~x – ~y) = (– 4~x 7+~y2~y) + (6~x + 5~y)
= = 2~x +

X, Y and Z are colinear.
˜˜
XY = l XZ

5

(k – 4)~x + ~4y = 2ll(2 …~x +17~y)
k – =

1 = 7l …2

l= 1
7
Substitute the value of l into 1
1
k–4= 2× 7

= 30
˜˜ 7˜
2. (a) BD = BA + AD

= – 24~x + 20~y
˜
˜
BE = 3ED

˜ 3 ˜
BE = 4 BD

˜˜˜
AE = AB + BE
˜ ˜ ˜
AE = AB + 3 BD
4
3
= 24~x + 4 (– 24~x + 20~y)

= 24~x –  18~x + 15~y ˜
= 6~x + 15~y AB
˜˜˜
(b) BC = BD + DC

= (– 24~x + 20~y) + 4
3

= (– 24~x + 20~y) + 4 (24~x)
3

= (– 24~x + 20~y) + 32~x
˜ = 8~x + 20~y
AE = 6~x + 15~y

˜AE  = 6 = 15 = 3
˜BC  8 20 4

It is found that, ˜ = 3_ ˜ hence alley AE and BC are parallel.
AE 4 BC ,

Intensive Practice 8.2 (Page 226)

˜˜˜
1. (a) AC = AB + BC

= ~˜y + ~x ˜
˜
(b) QR = QP + PR

= –~y + ~x

6

˜˜˜
(c) PR = PT + TR
˜ ˜
= PT + 1 TQ
2

= 2~x + 1 (–2~x + ~y)
2

= ~x + 1 ~y
2
2. (a) 3~x + ~y
(b) ~y – 2~x
(˜c) –~y + 2~x
˜ ˜
3. BQ = BA + AQ
˜ ˜
= BA + 1 AC
4

= –~a + 1 (4b~)
4

= –~a + b~ k)b~ …1
4. r = h3a~p + (h +
r = – 4q

r = (3–6(1~a20~aa~–1++63~a1~b3))b~+–(49…(~b42~a+ –4~bb~))
=
=

Comparing 1 and 2

h = –10,

h + k = 13

–10 + k = 13

5. k = 23 ! 702 + 402
Distance PQ =

= 80.62 m

Final velocity of the boat = 80.62
12

= 6.718 m s−1

The speed of Hamid rowing the boat = ! 6.7182 – 1.82

˜˜˜ = 6.47 m s−1
6. (a) (i) BA = BO + OA

= –b~ + ~a
˜ ˜
(ii) BX = 3 BA
5
3
= 5 (–~b + a~)

˜˜˜
(iii) OX = OB + BX

= b~ – 3 b~ + 3 a~
5 5

= 2 ~b + 3 ~a
5 5

7

˜˜˜
(iv) BY = BO + OY

= –~b + 3 a~
= 4
( )˜ 2 3
(b) (i) l 5 b~ + 5 ~a
OP

= 2 l~b + 3 l~a
5 5
˜˜˜
(ii) OP = OB + BP
˜˜
= OB + µ BY

( )=~b + µ –~b + 3 a~
4

= (1 – µ)~b + 3 µa~
4

(c) 2 l~b + 3 l~a = (1 – µ)~b + 3 µ~a
5 5 4

2 l = 1 – µ… 1
5

3 l = 3 µ
5 4

l= 5 µ …2
4

Substitute 2 into 1

( )25 µ =1–µ
4
5
1
2 µ = 1 – µ

µ= 2
3

l = 5 ( 2 )
4 3

= 5
6

Inquiry 3 (Page 227)

3.

y

10 B(6,7)
8
6 C(4,1) x
4 46
2 A(1,3)

2

8

4. Beluran

( )5. 5
4
6. Arding's Distance = ! (6 – 1)2 + (7 – 3)2

= 6.403 units
Timan's distance = ! (6 – 4)2 + (7 – 1)2

= 6.325 units

Self Practice 8.7 (Page 229)

1. (a) ˜ = 2~i + 2~j ˜ = –8~i ˜ = –10~i + ~j
OA OF BC
˜ ˜ ˜
FA = 10~i + 2~j DE = 14~i DO = –~j

( )˜ = 2 ( ) ˜= –8 ( ) ˜= –10

(b) OA OF 0 BC 1
2

˜ ˜ ( )˜
DE
( ) ( FA DO =
= ) 2 010 = 14 0

2. –1
˜ 5
(a) Position vector OB =
)˜ ˜ ˜ 8

(b) AB = AO + OB
˜˜
= – OA + OB

( ) ( ) = 2 + 5
–3 8

( )= 7
5

˜AB  = ! 72 + 52

= 8.602 units

˜˜˜
3. (a) (i) AB = 4~i + ~j (ii) BA = – 4~i – ~j (iii) BC = –~i – 5~j
˜ ˜ ˜
(iv) DC = 2~i (v) AC = 3~i – 4~j (vi) DE = 4~i + ~j
˜ ˜ ˜ gradient.
(b) AB is parallel ˜ because AB = DE and has the same
to DE

˜ is the negative vector to DE because ˜BA  = ˜DE and the direction of ˜ is opposite
(c) BA BA

˜
to the direction of DE.

4. (a) p = 3~i – 4~j
q = –5~i – 7~j

(b) rP(=3~,i –+ 4)5~j

Q(–5, –7)

R(1, 5)
! 32 + (– 4)2
(c) |  ~p | = 5 units
=

9

__________

| q~| = ! (–5)2 + (–7)

= 8_.6_0_2__u_nits

|~r| = ! 12 + 52

= 5.099 units

Inquiry 4 (Page 230)

3. The unit vector that is obtained will change because the change in values of x1 and y1
c. auses the magnitude of the vector to change.

4. The method to find the unit vector in the direction with vector r = x~i – y~j is r =___x_~_i__+____y__~j___
! x2 + y2
Self Practice 8.8 (Page 231)

1. (a) ! 32 + 22 (b) ! (– 4)2 + (–7)2 (c) 4 units
= 8.062 units 7

(d) ! (–12)2 + (–5)2 (e) 6 unit

= 13 units ______

2. (a) Magnitude of the vector =! 3_2+__ 22
= ! 13
Unit vector =__3_~i_+___2_~j__
! 13 __________
(b) Magnitude of the vector = !(_-1_)_+ (–9)2

= ! 82

Unit vector = -~i - 9 ~j
! 82 _______

(c) Magnitude of the vector = ! (4)2 + 02

= 4

Unit vector = 4~i = ~i
4

____________
(d) Magnitue of the vector = ! (–8)2 + (–15)2

= 17

Unit vector = –8~i – 15~j
17

3. (a) ! (–1)2 = 1 (unit vector)

  – !12!( ) ( )(b)2+1 2 + = 1 (unit vector)
! 2

(c) ! (–0.6)2 + (–0.8)2 = 1 (unit vector)

( ) ( )!(d) 72+ 24 2= 1 (unit vector)
25 25

!( ) ( )(e)22+ !7 2 = 1.106 (not unit vector)
3 3

10

4. (a) ! k2 = 1 (b) ! k2 = 1
k = ±1 k = ±1

(c) ! k2 + 1 = 1 (d) ! k2 + k2 = 1
k2 = 0 ! 2k2 = 1
k = 0 k = ± !1 2

(e) ! (0.5)2 + k2 = 1 ( )!(f) 13 2
 k2 + 84 = 1

(0.5)2 + k2 = 1 ( )k2 +13 2 = 1
84
( )k2 = 1 –13 2
k2 = 0.75 84

k = ±0.866 k = ±0.988

 !( ) ( )5.p 2+ 8 2 = 1 or p2 + 82 = 73
! 73 ! 73 p2 = 9
p = ±3
( ) ( )p 2+ 8 2= 1
! 73
! 73

p2 + 64 = 73

p2 = 9

p = ±3

6. ! (1 – k)2 + h2 = 1
1 – 2k + k2 + h2 = 1
h2 = 2k –__k_2___
h = ±! 2k – k2

Self Practice 8.9 (Page 233)

( ) ( ) ( )1. 2 –3
(a) 2a~ – ~b + ~c = 5 – 4 + 1
–12 8
( )= – 9
30

( ) ( ) ( )(b) –3
–3a~ + 2~b – ~c = –3 + 2 4 – 1
5 –12 8
( ) = 16
– 47

( ) ( ) ( )(c)
1 ~b + ~c – 3~a = 1 4 + 1 – 3 –3
2 2 –12 8 5

( ) = 12
–13

( ) ( ) ( )(d) – –3 + 3 1
1 ~b – ~a + 3~c = 1 4 5 8
4 4 –12

( )= 7
16

11

( ) ( ) ( )2. 3
(a) ~u – 2~v + w~ = 6 – 2 –2 + 3
–8 – 4
( ) = 10
18

= 10~i + 18~j
( ) ( ) ( )(b)
3~u + 2~v – w~ = 3 3 + 2 –2 – 3
6 –8 – 4
( )= 2
6

= 2~i + 6~j

( ) ( ) ( )(c)1~v–w~– 3u~ = 1 –2 + 3 – 3 3
2 2 –8 –4 6

( )= – 7
–26

= –7~i – 26~j
( ) ( ) ( )(d)
1 ~v – w~ + 3~u = 1 –2 – 3 + 3 3
4 4 –8 –4 6

( ) = 5.5
20

= 5.5~i + 20~j
Self Practice 8.10 (Page 235)

˜( ) ( )1. = –3 , ~v = 2
OA –3
–2

Position vector after 2.5 seconds

( ) ( ) = –3 + 2.5 2
–2 –3

( )= 2
–9.5

˜ ) 15t ˜
OA( ) ( )Velocity of OB
( ) ( 2. = 30t , = 50 + 10t

boat A 5 + 10t of boat B = 10
= 30 , Velocity 10
15


Both boats can only collide if the position vector at time t is the same,

˜˜ 15t = 5 + 10t
OA = OB t = 1 hour


_

Since the value of t is not the same, boats A and B will not collide.

12

Intensive Practice 8.3 (Page 235)

) ) -34 5
= 3
)_____8__

(b) Magnitude = ! 32 + 82
= 8.544 N
( ) ( )2. k – 3 = l 1
14 k–8
…1
k – 3 = l

14 = l(k – 8)…2

Substitute 1 into 2

14 = (k – 3)(k – 8)

14 = k2 – 8k – 3k + 24

0 = k2 – 11k + 10

k = 1 or 10

~u( ) ( )3.= 5– 3 =( ) ( ) ~v m– 5
–2 1 –6 –2
( )= 2 ( ) = m – 5
–3 – 4

Since u~ and ~v are parallel, hence

( ) ( ) m – 5 = k 2
– 4 –3
m – 5 = 2k … 1

−4 = −3k
4
k= 3 …2

Substitute 2 into 1

m = 2( 34) + 5

= 23
3

 ~u  = ! 22 + (–3)2
= ! 13 units

( ) ( )!~v 8 8 2 + (–4)2
= 3 ,  ~v  =   3
–4
!  208
= 9

 ~u  :  ~v  = ! 13 : ! 2908

= 13 : 208
9

= 117 : 208

= 9 : 16 13

˜˜˜
4. (a) BC = BA + AC

( ) ( ) = –2 + 10
1 5
( )= 8

˜ 6
B˜BCC ==8~!i + 6j
(b) 82 ~

+ 62

= 10
Unit vector = _8_~i _1+_0_6_~j
4~i + 3~j
= ˜ 5
˜ ˜
(c) AR = AB + BR
˜ ˜
AB 1 BC
( ) ( )= + 2

= 2 + 1 8
–1 2 6
( )= 6
2 or 6~i + 2~j

( ) ( )5.~v= 2.4 , a~ = 0.5 2.9
1.5 –2.1 θ
( ) ( )~v
+ ~a = 2.4 + 0.5 0.6
1.5 –2.1
( )= 2.9
–0.6_____________
Magnitude = ! (2.9)2 + (–0.6)2
= 2.96 km j−1

tan q = 0.6
2.9

q = 11.69°
Direction of resultant velocity = 90° + 11.69°

= 101.69°
6. (a) ~r + ~s = ((22~i+–m5~)j~i) + (m~i – 3~j)
= – 8~j

 ~r + ~s  = ! (2 + m)2 + (–8)2
100 = (2 + m)2 + (−8)2

(2 + m)2 = 36

2 + m = 6 , 2 + m = −6

m = 4 , m = −8

14

(b) Since ~r is parallel to ~s
m~i – 3m~j = k2(k2 ~i…–15~j)
=

−3 = −5k
3
k= 5 …2

Substitue 2 into 1

m= 2× 3
5

= 6
5
( ) ( )(c)
2~r – ~s = 2 2 – m
–5 –3
( ) ( ) = 4 – m
–10 –3
( ) = 4 – m
–7

Since (=2~r0 – ~s) parallel to y-axis,
4–m

m=4

( )!7. 1 2 =1
 k2 + ! 2

k2 + 1 = 1
2

k2 = 1
2

k = ±  1 _
– ~j !2 !2
2
8. Let ~a = 2~i or +

~a^ = 2~i –~j
! 5

Thus, ~v = 5(–2~i +~j)
! 5

= –2! 5 ~i + ! 5~j

9. If vector ~p and vector ~q are perpendicular, then the gradient of vector ~p × gradient of vector q~ =
–1

(_m_2_–_1__)__×n___n8_ = –1
4m – 4 = –1

n = – 4m + 4

m= 4–n
4

15

˜
10. (a) ON = t(– 4~i + 4~j) + 50~i + 20~j
= (50 – o4vt)e~irt+ak(e20bo+at4Nt)jif
(b) Ship M will both are at the same position at the same time.

˜˜
OM = ON

6t~i + 8t~j = (50 – 4t)i + (20 + 4t)~j
~

It is found that if 6t = 50 – 4t then t = 5

and if 8t = 20 + 4t then t = 5

Therefore, ship M will overtake ship N after 5 hours of sailing.

Mastery Practice (Page 237 – 238)

1. (a) ~a + b~
(b) a~ – ~c
2. 3ka~ – 43b~k = l4l(4 a~…+18~b )
=

– 4 = 8l

l = –  1 …2
2

Substitute 2 into 1

( )3k = 4 × –  1
2
2
k = –  3

3. ! m2 + (–n)2 = 1

m2 + n2 = 1

m2 = 1 – n2

m = ! 1 – n2

4.  u~ + ~v  = ! k2 + h2 …1
 ~u + ~v = (k~i + h~j) + ( ~i – 4~j)

 ~u + ~v  = ! (k + 1)2 + (h – 4)2 …2
1=2

k2 + h2 = k2 + 2k + 1 + h2 – 8h + 16

8h = 2k + 17

h= 2k + 17
8
˜˜˜
5. (a) AC = AB + BC

( ) ( ) = 5 + 10
12 –3
( ) = 15
9

16

 ˜ = !__1_5__2__+_ 92

AC

= ! 306
Unit vector = _1_5_~i___+___9_~j_
! 306
˜˜˜
(b) OC = OA + AC

( ) ( ) = 3 + 15
4 9

( ) = 18
13

C = (18, 13)

˜ ˜
RS = RQ
( )6. 2
5
˜˜
RP + PQ
( )=2
5

= 2 (3~i – 2~j)
5
˜˜˜˜
7. BC = BE + ED + DC
˜ ˜
BC = –~v + 1 BC + u~
˜ = ~u – ~v 2
1 BC
2 ˜

BC = 2(~u – ~v)
˜˜˜ ˜˜˜
8. (a) (i) AB = AF + FB (ii) FO = FB + BO

= –~a + ~b = ~b – a~
˜˜ ˜˜˜
(iii) FC = 2FO (iv) BC = BF + FC

= 2(~b – ~a) = –b~ + 2(~b – a~)
= ~b – 2a~
˜˜˜ ˜˜
(v) FD = FO + OD (vi) AD = 2 BC
= 2(~b – 2a~)
˜˜
= FO + BC

= ~b – a~ + b~ – 2~a
˜= 2~b – 3~a
FC
˜ 1
(b) AB = 2

˜˜˜
(c) AC = AB + BC

= –2~ab~ + 3b~~a+ ~b – 2a~
= –
˜
FD = 2˜b~~ – 3a
˜ ˜ ˜
Since AC = FD, thus AC is parallel to FD.

17

˜˜˜
9. (a) AB = AO + OB
˜˜
= – OA + OB

( ) ( ) = 10 + 10
–10 –11
( )= 20
–21 ˜A_B___. _________
(b) The distance between town A and town B =

= ! (20)2 + (–21)2

˜˜˜ = 29 km
(c) OC = OA + AC

˜˜ ˜
OC = OA + 2 AB

( ) ( ) = –10 + 2 20
10 –21

( ) = 30
–32

˜˜˜ ˜ ˜˜
10. (a) (i) AC = AO + OC (ii) OM = OA + AM

= –3u~ – 2~v + 9~u + 2~v = ˜ + 1 ˜
= 6~u OA 2 AC

= 3~u + 2~v + 1 (6~u)
2

(b) (i) ˜ 3 ˜ = ˜6~u + 2˜~v
OB = 2 OM ˜
(ii) OB = OC + CB

( )= 3 = 9~u + 2~v + 3k~v
2 6~u + 2~v = 9~u + (2 + 3k)~v

˜ = 9u~ + 3~v

OB = 9~u + 3~v …1
˜
OB = 9~u + (2 + 3k)~v …2
Equation 1 = 2, therefore

2 + 3k = 3
1
k= 3

11. (a) (i) ˜˜˜ (ii) ˜ 3 ˜
OB = OA + AB OD = 4 OB

= 4~a + 4~c = 3 (4~a + 4~c)
4

˜˜˜˜ = 3a~ + 3~c
(iii) OY = OA + AB + BY ˜˜˜
(iv) ED = EO + OD

= 4~a + 4~c + 1 (4~c) = 1 (– 4a~) + (3~a + 3~c)
2 2
= 4a~ + 6~c = – 2~a + 3~a + 3~c
= a~ + 3~c
18

˜˜˜ ˜
(b) DY = DO + OY ED = ~a + 3~c

˜˜
= –OD + OY

= E˜~a–D(3+~a=3~+c˜D3Y~c) + (4~a + 6~c) point, E, D and Y are located on a straight line.
= and D is a common

Since

( )12.(a) + 1__ ~j
Velocity of Arul's boat = (3~i +~j ) + ~i 3

= 4~i + 34~j
( )!Magnitude of the vector of Arul's boat =
42 + 42
3

= 4.216 m s−1
( )Resultant velocity of Ben's boat = (6~i + 2~j) + i + 31_~j
7_
= 7~i + 3 ~j

( )!Magnitude of the vector of Ben's boat = 72
72 + 3

= 7.379 m s−1

Difference in speed = 7.379 – 4.216

= 3.163 m s−1
( ) ( )(b)Resultant 1_
velocity of Raju's boat = 32~i~i ––~j_34 ~j ~i + 3 j
= ~

Magnitude of the vector of Raju's boat = !_3_+_(–1)2
= ! 10

The unit vector of the direction of the vector is 3_~_i __-__~j_
! 10

19

CHAPTER 9 SOLUTION OF TRIANGLES
Inquiry 1 (Page 242)

a b c
sin A sin B sin C
(a) 11.3 11.3 11.3
(b) 11.76 11.76 11.76

Conjecture: c
a b sin C
sin A = sin B =

Mind Challenge (Page 243)

A

bh

C aB

h = a Ú sin A = a
sin 90° sin A h
h b b
sin 90° = sin B Ú sin B = h

Mind Challenge (Page 244)

Diagram (a) and Diagram (b) are acute-angled triangle and obtuse-angled triangle respectively.
CD is perpendicular to AB represented by h.

B c B
ah ha

c

CD A
b
D C bA

Diagram (b) Obtuse-angled triangle

1