BUKU TEKS ADDMATHS FORM 4 (ANSWERS)

3. !  5 x – ! 3 x = ! 7
x(! 5 – ! 3 ) = ! 7
! 7
x= ! 5 – ! 3

= ! 7 × ! 5 + ! 3
! 5 – ! 3 ! 5 + ! 3

= ! 7 (! 5 + ! 3 )

5–3

= ! 35 + ! 21
2
1
4. logx a + logx a =t

a×( )logx 1 =t
a
logx 1 = t
t=0

5. Let the radius of the smaller circle be r. 2
CD = ! 32 – 12 1

= ! 8 CD
ErF
A B

EF = CD

! (2 + r)2 – (2 – r)2 + ! (1 + r)2 – (1 – r)2 = ! 8

! 4 + 4r + r2 – 4 + 4r – r2 + ! 1 + 2r + r2 – 1 + 2r – r2 = ! 8

! 8r + ! 4r = ! 8

(! 8r + ! 4r )2 = (! 8 )2

8r + 2! 8r ! 4r + 4r = 8

12r + 2r! 32 = 8

r(12 + 2! 32 ) = 8 8
r= 12 + 8! 2

6. (a) T = 100(0.9)x
= 100(0.9)5

= 59.05ºC
(b) 100(0.9)x = 80

(0.9)x = 80
100
= 0.8

x log 0.9 = log 0.8

x= log 0.8
log 0.9

= 2.12 seconds 31

( )7. 7 n , RM 20 000
RM60 000 8

( )7 n , 20 000
8 60 000
7 1
n log 8 , log 3

n . log 1
log 3
7
8
n . 8.2274

n = 9 years

Thus, the number of years when the price of the car to be less than RM2 000 for the first

time is 9 years.

8. logx 3 = s log! y 9 = t
xs = 3
(! y )t = 9
1 2
y = 9t
x = 3 s

log9 x3y = log9 x3 + log9 y

( ) ( )= log9 1 3 + 2
log9
3s 9t

= 3 log9 3 + 2 log9 9
s t
log 3
( )= 3 log 9 + 2
s t
3 1 2
( )= s 2 + t

= 3 + 2
2s t

9. 3(9x) = 27y…1
log2 y = 2 + log2 (x – 2)…2

From 1, 3(32x) = 33y
1 + 2x = 3y…3

From 2, log2 y = log2 4 + log2 (x – 2)

log2 y = log2 4(x – 2)
y = 4(x – 2)
y = 4x – 8…4

Substitute 4 into 3.
1 + 2x = 3(4x – 8)

1 + 2x = 12x – 24

10x = 25
5
x= 2

32

Substitute x = 5 into 4.
2
4( 5 )
y = 2 – 8

=2

( ) 10. 2
x log10 1– y = log10 p – log10 q

( )x log101– 2 = log10 p
y q

( )1 –2 x= p
y q

( )1 – 2 x = 10 000
20 100 000
( )x log 18 10 000
20 = log 100 000

x= –1
–0.04576
= 21.85 years

33

CHAPTER 5 PROGRESSION

Inquiry 1 (Page 128)
2.

3. n=1 n=2 n=3 n=4 n=5 n=6
Polygon arrangement, n 180° 360° 540° 720° 900° 1 080°
Sum of interior angles

The consecutive terms for the sum of interior angles can be obtained by adding 180° to
the previous term.

The difference between any two consecutive terms is D fixed constant and the constant
value that UHODWHV WKH WZR WHUPV is 180°.

The sum of interLRU angles for the tenth polygon DUUDQJHPHQW is when n = 10, which is
1 080° + 180° + 180° + 180° + 180° = 1 800°

Self Practice 5.1 (Pages 129 & 130)

1. (a) Common difference, d = –21 – (–35)
= 14

Add 14 to the previous terms.

(b) Common difference, d = 5! 3 – 2! 3
= 3! 3

Add 3! 3 to the previous terms.
(c) Common difference, d = 2p – (p + q)

=p–q

Add (p – q) to the previous terms.
(d) Common difference, d = loga 24 – loga 2

= loga 24 – loga 21
= loga 23
Add loga 23 to the previous term.
2. (a) d1 = 13 – 9 = 4


(b) d1 = 1 – 1 = – 1
4 2 4
1 1 1
d2 = 6 – 4 = – 12

d1 ≠ d2, thus this sequence is not an arithmetic progression.

1

(c) d1 = 0.01 – 0.1 = – 0.09
d2 = 0.001 – 0.01 = – 0.009
d1 ≠ d2, thus this sequence is an arithmetic progression.
(d) d1 = 5 – (5 – x) = x
d2 = (5 + x) – 5 = x
d1 = d2, thus this sequence is not an arithmetic progression.

3. (a) 10 (b) p

59 13 2p
8 –p 3p 7p

(c) 17x 12p

12x

7x 9x 11x

x 5x

4. The sequence of the distance of each flag: 5, 10, 15, …
d1 = 10 – 5 = 5
d2 = 15 – 10 = 5
Common difference, d = 5, thus the arrangement of flag follows an arithmetic progression.

Inquiry 2 (Page 130) Method to obtain the value Formula
3.
Value of term of term (deduction method)
Term a
Does not have d T1 = a + 0d
T1

T2 a + d Add d at T1 term T2 = a + 1d

T3 a + d + d Add d at T2 term T3 = a + 2d

… ……

Tn Add d at Tn – 1 term Tn = a + (n – 1)d

4. (a) T20 = a + 19d
(b) The common difference for the nth term, Tn is (n – 1).
(c) Tn = a + (n – 1)d
2

Self Practice 5.2 (Page 132) 3x – 1, 4x, 6x – 2 T5 = 20 and
are three consecutive T4 = 2T2, findT10.
START terms. Find T6.

Find the 9th term
in the progression
9, 5, 1, …

–23 25 8x – 4 24 40

Given that Given that tGahniedvfeTinr1s7tth=tae5trm4T,2,f=ain.3d
Tn = 8 – 5n, Tn = 8 – 3n. – 4 –0.4
find T4. Find the common
difference, d.
–62
–18 3

FINISH Given that a = 10 Given that –17, –14,
and d = –4, –11, … 55, find
find T7. the number of
terms.

2. (a) The sequence of salary: RM36 000, RM37 000, RM38 000, …

This sequence is an arithmetic progression with a = RM36 000 and d = RM1 000

The salary for n years of Encik Muiz working, Tn = RM72 000
Tn = a + (n – 1)d

72 000 = 36 000 + (n – 1)(1 000)
36 = n – 1

n = 37 years

Thus, Encik Muiz needs to work for 37 years to receive twice his first annual

income.

(b) The salary on the 6th year , T6 = RM43 500
Tn = a + (n – 1)d
43 500 = 36 000 + (6 – 1)d

7 500 = 5d

d = 1 500

Thus, the yearly salary increment of Encik Muiz is RM1 500.

3

Inquiry 3 (Page 133)
5.

Sum of Number of grids based on the Formula of a rectangle
terms number of terms by deduction method

Diagram I S2 = T1 + T2 Diagram II S2 = [a + a + d]2
= a + (a + d) Diagram IV 2
S2 = 2a + d 2[2a + (2 – 1)d]
1 unit 1 unit = 2

Diagram III S2 = T1 + T2 + T3 [a + a + 2d]3
2
S3 = a + (a + d) + S3 =
(a + 2d)
= 3[2a + (3 – 1)d]
= 3a + 3d 2

S4 = T1 + T2 + T3 + T4 S4 = [a + a + 3d]4
= a + (a + d) + 2
(a + 2d) + 4[2a + (4 – 1)d]
= 2
S4 (a + 3d)
= 4a + 6d

 

Sn = T1 + T2 + T3 + T4 + … + Tn Sn = n[2a + (n – 1)d]
= a + (a + d) + (a + 2d) + (a + 3d) + 2
Sn … + [a + (n – 1)d]
n[2a + (n – 1)d]
= 2

Mind Challenge (Page 134)
S20 = 230 + 630 means the sum of the twentieth term is the first ten terms add to the
subsequent ten terms.

Mind Challenge (Page 135)
The value of n , –25.99 is ignored because the number of days cannot be negative.
Self Practice 5.3 (Page 136)

1. (a) a = –20, d = –15 – (–20)
=5

Tn­ = a + (n – 1)d
100 = –20 + (n – 1)(5)
120 = 5n – 5
5n = 125

n = 25

4

Thus, S25 = 25 [2(–20) + (25 – 1)(5)]
2

= 25 [–40 + 120]
2
25
= 2 [80]

= 1 000

(b) a= 3 , d = 6 – 3
5 5 5
3
= 5

[ ( ) ( )]S23 =2323 3
2 5 + (23 – 1) 5

[ ] = 23 6 + 66
2 5 5
[ ] = 23 72
2 5
3
= 165 5

2. Horizontal:

(a) a = 38, d = 34 – 38 (e)3
= –4 00
18 (c)3
S18 = 2 [2(38) + (18 – 1)(– 4)] 1 1
5 0
= 9[8] 00
87
= 72 (a)7 (d)2 0
(b)2
(b) a = –10, d = 6

S100 = 100 [2(–10) + (100 – 1)(6)]
2
= 50[574]

= 28 700

(c) S42 = 5 838, Tn = –22
42
S42 = 2 [a + (– 22)]

5 838 = 21[a – 22]

278 = a – 22

a = 300

Vertical:

(c) n = 140, a = 2, T140 = 449
T140 = 449

2 + (140 – 1)d = 449

139d = 447
447
d= 139

[ ( )]S140 =140 447
2 2(2) + (140 – 1) 139

= 70[2 + 449]

= 70[451]

= 31 570

5

(d) a = –15, d = –3, Sn = –1 023
n
–1 023 = 2 [2(–15) + (n – 1)(–3)]

–2 046 = –27n – 3n2

3n2 + 27n – 2 046 = 0

n2 + 9n – 682 = 0

(n + 31)(n – 22) = 0

or n = 22

Thus, n = 22.0

(e) S200 = S2252500–[2S5500 50
= 2
+ 1] – [50 + 1]

= 31 375 – 1 275

= 30 100

3. The sequence of the length of line parallel to the y-axis: 1, 3, 5, …
a = 1, d = 3 – 1 = 2

T11 = 1 + (11 – 1)(2)
= 21

The length of the final line parallel to the y-axis is 21 units.

The progression of the length of the pattern: 1, 1.5, 2, …, 21

a = 1, d = 1.5 – 1 = 0.5

Tn = 1 + (n – 1)(0.5)
21 = 0.5n + 0.5
n = 41

S41 = 41 [2(1) + (41 – 1)(0.5)]
2
41
= 2 [22]

= 451

Thus, the sum of the length of the overall pattern is 451 units.

4. (a) 1, 3, 5, 7, …, Sn < 200
n [2(1) + (n – 1)2] < 200
2
n[1 + (n – 1)] < 200
n2 < 200
n < 14.14

º n = 14

S14 = 14 [2 + 13(2)]
2
= 7[28]

= 196

There are 14 complete wood pieces with the same colour and 4 remaining wood pieces.

(b) T14 = 1 + 13(2) = 27 wood pieces
If blue, 1, 5, 9, …, 27
Tn = 27 = 1 + (n – 1)4
26 = 4(n – 1)

30 = 4n
30
n= 4 not an integer

6

If white, 3, 7, 11, …, 27
Tn = 27 = 3 + (n – 1)4

24 = 4(n – 1)
n – 1 = 6

n = 7, an integer
Thus, the last wood piece is white and the number of white woods is 27.

Self Practice 5.4 (Page 138)

1. (a) The sequence of sales expectation of a book: 10, 14, 18, …
a = 10, d = 14 – 10

=4 Sn . 1 000

n [2(10) + (n – 1)(4)] . 1 000
2 n[16 + 4n] . 2 000

4n2 + 16n – 2 000 . 0
n2 + 4n – 500 . 0

n. – 4 + ! 2 016 oaortaru  n , – 4 – ! 2 016
2 2
. 20.4 , –24.4 ((AIgbnaoikrea)n)

Thus, n = 21, so Mr. Tong needs 21 days to sell all the books.
10
(b) S10 = 2 [2(10) + (10 – 1)d]

1 000 = 5[20 + 9d]
200 = 20 + 9d
9d = 180
d = 20

The increasing rate of the books to sell each day is 20 books.

2. (a) T15 = 30
S15 = 240

15 [a + 30] = 240
2
15a + 450 = 480

15a = 30

a=2

The length of the shortest wire is 2 cm.

(b) T15 = 30
2 + (15 – 1)d = 30

2 + 14d = 30

14d = 28

d=2

The difference between two consecutive wires is 2 cm.

Intensive Practice 5.1 (Page 138)

1. (a) d1 = –17 – (–32) = 15

d2 = –2 – (–17) = 15
d3 = 13 – (–2) = 15

The sequence is an arithmetic progression because the common difference, d, is the

same, which is 15. 7

(b) d1 = 5.7 – 8.2 = –2.5
d2 = 3.2 – 5.7 = –2.5
d3 = 1.7 – 3.2 = –1.5
The sequence is not an arithmetic progression because the common difference, d, is different.

2. (a) a = –12, d = –9 – (–12)

=3

T9 = –12 + (9 – 1)(3)
= 12

(b) a = 1 , d = – 31 – 1
3 3
= – 23
( )T15 =1 –  23
3 + (15 – 1)

= –9

3. (a) a = –0.12, d = 0.07 – (–0.12)
= 0.19
Tn = 1.97

–0.12 + (n – 1)(0.19) = 1.97
0.19n – 0.19 = 2.09
0.19n = 2.28
n = 12

(b) a = x, d = 3x + y – x
= 2x + y
Tn = 27x + 13y

x + (n – 1)(2x + y) = 27x + 13y
2nx – x + ny – y = 27x + 13y
(2n – 1)x + (n – 1)y = 27x + 13y

2n – 1 = 27
2n = 28
n = 14

4. (a) a = –23, d = –17 – (–23)
=6

S17 = 17 [2(–23) + (17 – 1)(6)]
2
17
= 2 [50]

= 425

(b) S2n = 2n [2(–23) + (2n – 1)(6)]
2
= n[12n – 52]

= 4n[3n – 13]

(c) Tn­ = 121
–23 + (n – 1)(6) = 121

6n – 6 = 144

6n = 150

n = 25

8

S25 = 25 [–23 + 121]
2
= 1 225

5. (a) Sn­= 2n2 – 5n
S1 = 2(1)2 – 5(1)
= –3

(b) T9 = S9 – S8
= [2(9)2 – 5(9)] – [2(8)2 – 5(8)]

= 117 – 88

= 29

(c) T4 + T5 + … + T8 = S8 – S3
= [2(8)2 – 5(8)] – [2(3)2 – 5(3)]

= 88 – 3

= 85

6. (a) T2 = 1 , S14 = –70
2
1
a+d= 2

2a + 2d = 1…1

S14 = –70

14 [2a + 13d] = –70
2 = –10…2
2a + 13d
2 – 1: 11d = –11

d = –1
Thus, the common difference is –1.

(b) T14 = 3 + (14 – 1)(–1)
2
= – 223

7. The sequence of monthly income of company A: RM3 500, RM3 520, RM3 540, …
The sequence of yearly income of company B: RM46 000, RM47 000, RM48 000, …

The total income of company A within 3 years:
S36 = 326[2(3 500) + 35(20)]

= RM138 600

The total income of company B within 3 years:

S3 = 3 [2(46 000) + 2(1 000)]
2

= RM141 000

The difference of the total income = RM141 000 – RM138 600

= RM2 400

Yui Ming needs to choose company B with an additional total income of RM2 400.

9

Inquiry 4 (Page 139) 4 = 22 8 = 23
3. 1 64 = 26 128 = 26
2 1 024 = 210 2 048 = 211
16 = 24 32 = 25 16 384 = 214 32 768 = 215
256 = 28 512 = 29 262 144 = 218 524 288 = 219
4 096 = 212 8 192 = 213
65 536 = 216 131 072 = 217
1 048 576 = 220 2 097 152 = 221

4. 20 minutes × 22 = 440 minutes
= 7 hours 20 minutes

5. Times two to the previous time.
6.

Self Practice 5.5 (Page 141)

1. (a) r1 = 40 = 1
120 3
40
3 1
r2 = 40 = 3

r1 = r2, therefore the sequence is a geometric progression.

(b) r1 = 0.003 = 0.1
0.03
0.0003
r2 = 0.003 = 0.1

r1 = r2, tthheerreeffoorree tthheesperqougernecsseioisnaisgeaogmeeotmricetprircogpreosgsrieosns.ion.
2x
(c) r1 = x+1

r2 = 5x + 12
2x + 1
r1 ≠ r2, tmthheaerkreeaffoojurreejuttkhhaeenspebrqouugkeranenscsejaioinsnjaninosgtnaogtegoaemogmeotermtir.iectrpicropgrroegsrseiossni.on.

10

2. (a) 32 (b) 11 1
1 28 1 6 12 24
3 1
4 2 12 Formulae
1 a
6
ar = ar2 – 1
2 1 11 1 ar2 = ar3 – 1
2 12 24 48 ar3 = ar4 – 1
ar4 = ar5 – 1
3. r1 = r2
x+1 4x + 4 
x–2 = x+1 arn – 1

(x + 1)(x + 1) = (4x + 4)(x – 2)
x2 + 2x + 1 = 4x2 – 4x – 8
3x2 – 6x – 9 = 0
x2 – 2x – 3 = 0

(x + 1)(x – 3) = 0

x = –1 or x = 3
The positive value of x is 3.
a = T1 = 3 – 2 = 1
T2 = 3 + 1 = 4

Common ratio, r = 4
Therefore, the first three terms are 1, 4, 16 with a common ratio of 4.

Inquiry 5 (Page 141)
2. Value of term Method to obtain the value of term

Term

T1 2 2(3)1 – 1 = 2(3)0
T2 6 2(3)2 – 1 = 2(3)1
T3 18 2(3)3 – 1 = 2(3)2
T4 54 2(3)4 – 1 = 2(3)3
T5 162 2(3)5 – 1 = 2(3)4

 

Tn 2(3)n – 1
3. The formula for the nth term = arn – 1

11

Self Practice 5.6 (Page 143) If T1 = 4 and FINISH
T3 = T2 + 24, find
START the positive value 0.01
of r.
T1 = 12 and
T3 = 27, find T5. 3

GDivbenri a = 50 adnadn Given that –3, 6, Given that 0.12,
T5 = 202125, cfianrdi rr.. –12, …, −192, 0.0012, 0.000012.
find the number Find r.

of terms

3 12
5

Given that T2 = 8 Given that r = 1 and Find T5 for the
and T6 9 2 sequence x + 1,
9 1 x + 3, x + 8, …
= 2 . Find T5. T3 = 6 . Find T10.

2. The sequence of the height of the ball: 3, 3(0.95), 3(0.95)2, …
Tn , 1

3(0.95)n – 1 , 1
1
0.95n – 1 , 3

(n – 1) log 0.95 , log 1
3
1
n – 1 . log 3

log 0.95
n – 1 . 21.4

n . 22.4
Thus, the height of the ball is less than 1 m on the 23rd bounce.

Self Practice 5.7 (Page 145)

1. (a) a = 0.02, r = 0.04 = 2
0.02
0.02(212 – 1)
S12 = 2–1

= 81.9 p3
p
(b) a = p, r = = p2

T n = p(p2)n – 1
p21 = p(p2)n – 1
p20 = p2n – 2

20 = 2n – 2

2n = 22

n = 11 12

S11 = p[(p2)11 – 1]
p2 – 1
p[p22 – 1]
= p2 – 1

(c) S15 = 1 [315 – 1]
2
2
= 3 587 226.5

2. a = 3 500, r = 700 = 1
3 500 5
Sn = 4 368
[ ( ) ]3 500 1 – 1n
1 5 = 4 368
5
1–

[ ( ) ]3 500 1 –1n = 3 494.4
5
( )1 – 1 n = 0.9984
5
( )1 n = 0.0016
5
1
n log10 5 = log10 0.0016

n = log10 0.0016
= 1
4 log10 5

The number of terms is 4.

3. (a) The sequence of the number of squares: 1, 4, 16, …
4
r1 = 1 =4

r2 = 16 =4 tjhuejuskeaqnuemnceemibseantguekomjaentjraincgprgoegormesestiroi.n.
r1 = 4 MThauksa,,
r2.
(b) a = 1
1(46 – 1)
S6 = 4–1

= 1 365

13

Inquiry 6 (Page 146)

2. n rn Sn
1
1 64
2

2 1 96
4

3 1 112
8

4 1 120
16

5 1 124
32

10 1 127.875
1 024

20 1 127.999
1 048 576

3. When n increases, the value of rn approaches zero and the value of Sn is increasing.
a a
4. When n increases to infinity, the value of Sn is approaching 1–r daannd S∞ = 1 – r.

Self Practice 5.8 (Page 148)

1. Horizontal: 1 (a)2 (c)2 5 0
3
(a) a = 1 500, r = 2
(d)5 4
S∞ = 1 500 (b)3 0 0 0 0
1
1– 3

= 2 250

(b) The sequence of the loan repayment:
RM15 000, RM7 500, RM3 750, RM1 875, …
1
a = 15 000, r = 2

S∞ = 15 000
1
1– 2

= RM30 000

Vertical:

(c) r= 1
2
S∞ = 4 480
a
1 = 4 480
1 – 2

a = 2 240

14

(d) 4.818181… = 4 + 0.81 + 0.0081 + 0.000081 + …

= 4 + S∞ 0.81
– 0.01
= 4 + 1

=4+ 0.81
0.99
9
= 4 + 11

= 53
11
Thus, h = 53

Self Practice 5.9 (Page 149)

1. (a) 4x + 20 = 3x – 10
10x 4x + 20
(4x + 20)2 = (3x – 10)(10x)

16x2 + 160x + 400 = 30x2 – 100x

14x2 – 260x – 400 = 0

7x2 – 130x – 200 = 0

(7x + 10)(x – 20) = 0
x = – 170 or x = 20
Sequence: a, 200, 100, 50
Thus, the longest piece is 400 cm.

(b) S∞ = 400
1
1– 2

= 800 cm

=8m

2. (a) The sequence of the radius: j, j(1.4), j(1.4)2, …

The sequence of the circumference: πj, πj(1.4), πj(1.4)2, …
π(2)(1.415 – 1)
(b) S15 = 1.4 – 1

= 772.8π

= 2 428.8 cm

= 24.28 m

Intensive Practice 5.2 (Page 149)

1. (a) a = –1, r = –3

Tn = 2 187
–1(–3)n – 1 = 2 187

(–3)n – 1 = –2 187

(–3)n – 1 = (–3)7

n–1=7

n=8

S8 = –1(1 – 38)
1 – (–3)
= 1 640

15

(b) a = log x–1, r = 2

Tn = log x–64
log x–1(2)n – 1 = log x–64

2n – 1 log x–1 = 64 log x–1

2n – 1 = 64

2n – 1 = 26

n–1=6

n=7

S7 = log x–1(27 – 1)
2–1
= –127 log x

= log x–127

(c) a = 0.54, r = 0.01

Tn = 5.4 × 10–17

0.54(0.01)n – 1 = 5.4 × 10–17

0.54(10–2)n – 1 = 0.54 × 10–16

10–2n + 2 = 10–16

–2n + 2 = –16

2n = 18

n=9

S9 = 0.54(1 – 0.019)
1– 0.01
6
= 11

(d) a = 3, r = 1
2
( )Tn = 3 1 n–1
2
( )3 1 n–1
=3 2
64
( ) ( ) 21 6 =1 n–1
2
6=n–1

n=7

[ ( ) ]S7 =31– 17
2
1
2
= 5 61
64

2. DGiibveerni tjhanatjatnhge geometric4p.5ro, g–r9e,ss1i8o,n…4.5d,a–n9h, a1s8i,l .t.a.manbdahth7e6s9u.m5. is 769.5.

r= –9 = –2
4.5
Sn = 769.5

4.5[1 – (–2)n] = 769.5
3
1 – (–2)n = 513

(–2)n = –512

(–2)n = (–2)9

TheBniulamngbaenr osfebteurtmans, n = 9. 16

3. Consecutive terms x, 2x + 3, 10x – 3
2x + 3 10x – 3
(a) x = 2x + 3

(2x + 3)(2x + 3) = x(10x – 3)

4x2 + 12x + 9 = 10x2 – 3x

6x2 – 15x – 9 = 0

2x2 – 5x – 3 = 0

(2x + 1)(x – 3) = 0

2x + 1 = 0   , x – 3 = 0
x = – 21
(b) JikIfa x , 0, the consecutive x=3 1 , 2, –8.
a = – 21 ,r = –4 terms are – 2

T6 = – 21 (–4)5
= 512

4. The area of the third triangle = 36 cm2, T3 + T4 = 54 cm2

(a) ar2 = 36…1
ar2 + ar3 = 54

ar3 = 54 – 36
ar3 = 18…2

2 ÷ 1: r= 1
2

Substitute r = 1 into 1.
2
( )a1 2 = 36
2
1
4 a = 36
a = 144

[ ( ) ] [ ( ) ]144 1 –1 10 12
(b) S10 – S2 = 1 2 144 1 – 2
–1
22

= 287.72 – 216

= 71.72 cm2

5. (a) Given Tn = 38 – n
T1 = 37, T2 = 36
36
r= 37

= 3–1

= 1
3
(b) 35 + 34 + 33 = 243 + 81 + 27

= 351 cm

17

6. a + ar + ar2 = 7(ar2)

a(1 + r + r2) = 7(ar2)

1 + r + r2 = 7r2

6r2 – r – 1 = 0

(2r – 1)(3r + 1) = 0

2r – 1 = 0    , 3r + 1 = 0
1 1
r= 2 r = – 3 (Ignore)

If a = 14.5,
( )14.5 1
2 = 7.25

Thus, the mass of the 2nd heaviest child is 7.25 kg.

Mastery Practice (Page 150)

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

When x = 2, the consecutive terms are –5, 8, 21.
Thus, d = 8 – (–5)

= 13
(b) T3 = 8

a + 2(13) = 8
a + 26 = 8

a = –18

2. a + 8d = 21 + 3p…1
a + a + d + a + 2d = 9p
3a + 3d = 9p
a + d = 3p…2
1 – 2: 7d = 21
d=3

3. (a) a + a + 2d = 24
2a + 2d = 24
a + d = 12…1
a + 4d = 36…2
2 – 1: 3d = 24

d=8
SGuabnsttiiktuatne d = 8 iknetoda1lam 1

a + 8 = 12

a = 12 − 8

=4

The volume of the smallest cylinder is 4 cm3.
9
(b) S9 = 2 [2(4) + 8(8)]

= 324 cm3

4. (a) ar2 = 30…1
ar2 + ar3 = 45…2

18

GSanutbisktaitnut1e 1ke idnatolam22.
ar3 = 15…3

3 ÷ 1: r= 1
2
1
SGubanstitkuatne r = 2 kientdoa1lam. 1.

( )a1 2 = 30
2
( )a1
4a = 30
a = 120
–
(b) S∞ = 1 r

= 120
1
1– 2

= 240

5. The height of an arrangement of chairs: 80, 84, 88, ...

(a) Tn = 300
80 + (n – 1)(4) = 300
n – 1 = 55

n = 56

The maximum number of chairs is 56.

(b) 56, 54, 52, … 13th term
13
S13 = 2 [2(56) + 12(–2)]

= 572

6. The sequence of savings: 14 000, 14 000(1.05), 14 000(1.05)2 , …
(a) T18 = 14 000(1.05)17
= RM32 088.26

Yes, a savings of RM30 000 can be obtained.

(b) T10 = 14 000(1.05)9
= RM21 718.60

After 10 years, the amount of money in the bank is RM21 718.60.

The sequence of money in the bank: 21 718.60, 21 718.60(1.03), …

T8 = 21 718.60(1.03)7
= RM26 711.14

The savings will not reach RM30 000.

7. (a) a(r4 – 1) = 10a(r2 – 1)
r–1 r–1
r4 – 1 = 10r2 – 10

r4 – 10r2 + 9 = 0 (Shown)

r4 – 10r2 + 9 = 0

(r2 – 1)(r2 – 9) = 0
r2 – 1 = 0 , r2 – 9 = 0
r2 = 9
r2 = 1 r = ±3

r = ± 1 (Ignore) 19

The positive value of r is 3.

(b) 2, 6, ..

S6 = 2(36 – 1)
3–1
= 728
JumTloathalpeexrpbenladnijtauaren = RM7.50 × 728

= RM5 460

20

CHAPTER 6 LINEAR LAW

Inquiry 1 (Page154)
1. (a)

x −3 −2 −1 0 1 234
y 41 26 15 8 5 6 11 20
(b)
−2 −1 0 123
x −4 −3 2 3 4 567

y0 1

2.

y

40 y
35
30 1234 x 7
25 6
20 5
15 4
10 3
2
5 1

–3 –2 –1 0 –4 –3 –2 –1 0

123 x

3. The graph 1(a) is a curve while graph 1(b) is a straight line. A linear relationship form a
straight line whereas a non-linear relationship does not form a straight line.

Self Practice 6.1 (Page155)

1. The graph of linear relation is Diagram 1(b). The graph in Diagram 1(a) represents
the relation of non-linear because the shape of the graph obtained is a curve while the
graph in Diagram 1(b) represents a linear relation because a straight line is obtained.

1

2. (a) (b)

X1 3 5 7 9 11 X 2 4 6 10 12 14
1.7
Y 3.16 5.50 9.12 16.22 28.84 46.77 Y 0.5 0.7 0.9 1.3 1.5
X
Y Y
1.8 Graph of Y against X
50 Graph of Y against X 1.6
45 1.4
40 1.2
35 1.0
30 0.8
25 0.6
20 0.4
15 0.2
10

5

0 2 4 6 8 10 12 X 0 246 8 10 12 14



The graph (b) which is a straight line is a graph of linear relation.

Inquiry 2 (Page 156)

1.

y

11

10

9

8
7

6

5

4

3

2

1
0 1 2345 67 x

2

3.



y x
Graph of y against x

30
25
20
15
10

5

0 5 10 15 20 25 30

3

2.

L(cm)
Graph of L against m

4

3

2

1

0 20 40 60 80 100 120 m(g)

Self Practice 6.3 (Page 159)

1. m= 0.6 – 0.02
0.32 – 0
29
= 16

c = 0.02
1
= 50

t = 29 x + 1
16 50
2. (a)

y

40 Graph of y against x

35

30

25

20

15

10

5

0 10 20 30 40 50 60 x

(b) y-intercept = 12.5 4
Gradient = 35 – 12.5
60 – 0
= 0.375

(c) y = 0.375x + 12.5

Self Practice 6.4 (Page 160)
1. (a) y

30 Graph of y against x

25

20

15

10

5

0 2 4 6 8 10 12 14 16 x

(b) (i) y-intercept = 4.0
(ii) When x = 12, y = 22

(iii)KGecraedruiennatn,, m = 26.5 – 4
15 – 0
3
= 2

(iv) When y = 15, x = 7.4

(c) y= 3 x + 4
2
3
AWWahbheielnna x = 28, y = 2 (28) + 4

= 46

Intensive Practice 6.1 (Page 161) (b) Graph of y2 against x–1
1. (a) y2
7
y 6
Graph of y 5
against x 8 4
3
6 2
1
4
1–x
2

–4 –2 0 2 x
–2

–4

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

Graph (a) is a non-linear graph while graph (b) is a linear graph. The shape of graph (a)
is a curve while the shape of graph (b) is a straight line.

5

2. Y Graph of Y against X
120

100

80

60

40

20

X
0 10 20 30 40 50 60 70

Gradient, m = 119 – 108
70 – 20
11
= 50

Y = mX + c

c = Y – mX

= 108 – 11 (20)
50
518
= 5

EPqeurastaimonaaonf sgtarariisghlut rluinse: Y = 11 X + 518
(a) 50 5
3.

log10 y

Graph of log10 y against log10 (x + 1)
0.9

0.8

0.7

0.6
0.5

0.4
0.3
0.2
0.1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 log10 (x + 1)

6

(b) (i) GKraedcieernutn, amn, m = 0.75 – 0.15
= 1 0.6 – 0

(ii) y-intercept = 0.15

(iii) When log10 y = 0.55, log10 (x + 1) = 0.4
100.4 = x + 1

2.512 = x + 1

x = 1.512

(c) (i) log10 y = log10 (x + 1) + 0.15
= log10 (3.5) + 0.15
= 0.6941

y = 4.9442

(ii) log10 (1.5) = log10 (x + 1) + 0.15
log10 (x + 1) = 0.0261
x + 1 = 100.0261

x + 1 = 1.0619

x = 0.0619

4. (a)

xy

45 Graph of xy against x2

40

35

30

25

20

15

10

5

0 x2
5 10 15 20 25 30 35 40

(b) (i) KGecrearduiennatn,, m = 45 – 12
42 – 5
33
= 37

(ii) Y- intercept = 7.5

(iii) When xy = 16.5, x2 = 10

(iv) When x = 2.5, x2 = 6.25

When x2 = 6.25, xy = 13
2.5y = 13

y = 5.2

7

(c) EqPueartsiaomn aoafnstgrairgishtlulirnues: xy = 33 x2 + 7.5
37
33
AWpahbeilna xy = 100, 100 = 37 x2 + 7.5

x2 = 103.7

x = 10.18

Self Practice 6.5 (Page 165)

1. (a) y = px2 – q

y =p– q
x2 x2

Through comparison,
y 1
Y= x2 , X = x2 , m = –q, c = p

(b) y = hx2 + x

y = hx + 1
x

Through comparison,
y
Y= x , X = x, m = h, c = 1

(c) y= p +q
x2
yx2 = p + qx2

Through comparison,

Y = yx2, X = x2, m = q, c = p

2. (a)

–1y Graph of y–1 againstͱසx

2.5

2

1.5

1

0.5

0 0.5 1 1.5 2 ͱසx
–0.5

–1

(b) (i) q = –0.75

(ii) p = 2.75 – 1.20
1.8 – 1.00
31
= 16

8

(iii) When x = 1.21, thus !wx = 1.10
1
y = 1.40

y= 5
7

Intensive Practice 6.2 (Page 165)

1. (a) y = 5x2 + 3x

y =5+ 3
x2 x

Through comparison,
y 1
Y= x2 , X = x , m = 3, c = 5

(b) y = p!wx + q
!wx

y!wx = px + q

Through comparison,
Y = y!wx, X = x, m = p, c = q

(c) y = axb

log10 y = log10 a + b log10 x

Through comparison,
Y = log10 y, X = log10 x, m = b, c = log10 a
(d) x = mxy + ny

x = mx + n
y

Through comparison,

Y = x , X = x, m = m, c = n
y
(e) ypx = q

log10 y = –log10 px + log10 q
Through comparison,

Y = log10 y, X = x, m = –log10 p, c = log10 q
(f) y(b – x) = ax

yb – yx = ax
ax
b – x = y

x = – ax + b
y a

TMherloaulugihpceormbapnadrisnogna,n,
x – 1a , b
Y= y , X = x, m = c = a

2. (a) y = ax3 + bx2
y
(b) x2 = ax + b

x 0.5 1.0 1.5 2.0 2.5 3.0
5.00
y 1.24 2.05 2.75 3.50 4.21
x2

9

–xy–2
5.0 Graph of –xy–2 against x
4.5

4.0

3.5

3.0

2.5

2.0

1.5

1.0

0.5

0 0.5 1 1.5 2.0 2.5 3.0 x

(c) a = 5.0 – 1.24
3.0 – 0.5
= 1.504

b = 0.5

3. (a) y = ab + x

log10 y = (b + x) log10 a
= b log10a + x log10a

(b)

x1 2 3 4 5
1.05 1.35 1.66
log10y 0.45 0.75
x
log10 y
1.8 Graph of log10 y against x 10
1.6

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0 12345

(c) log10a = 0.75 – 1.66
2–5
91
= 300

a = 2.0106

b log10a = 0.16

( ) b 39010 = 0.16
b = 0.5275

Self Practice 6.6 (Page 167)

1. (a) y = pqx
log10y = log10p + xlog10q

x 246 8 10 16
1.2 1.4 2.0
log10y 0.6 0.8 1.0
110 140
log10y Graph of log10y against x 6.00 1.40
2.0

1.5

1.0

0.5

0 2 4 6 8 10 12 14 16 x

(b) (i) log10 p = c
= 0.4

p = 2.51189

(ii) log10 q = 2.0 – 1.4
16 – 10
= 0.1

q = 1.25893

(c) When x = 5 hours,
log10 y = 0.9
y = 7.94328

2. (a) xy – yb = a
yb = xy – a

y= 1 xy − a
b b

xy 10 28 60 90
13.25 8.80
y 20.60 18.00

11

y
Graph of y against xy

22
20
18
16
14
12

10
8

6
4
2
0 20 40 60 80 100 120 140 160 xy

(b) 1 = 1.4 – 20.6
b 140 – 10
1
b = −0.14769

b = −6.77094
a
b = 22.2

a = 22.2 × −6.77094

= −150.314868

(c) Alternative method
xy – yb = a

y(x – b) = a

(x –a b) = 1
y

1 = 1 x – b
y a a

x 0.485 100
15 5
y 103 7

m = 5 – 5
7 103
100 – 0.485
= 0.0066899

Y = 0.0066899x + c

5 = 0.0066899(0.485) + c
103c = 0.0452991

Gradient, m = 0.0066899

Y-intercept = 0.0452991

12

Intensive Practice 6.3 (Page 168 –169)
1. Diagram (a), p = 10

Diagram (b), y =

p = 10(2) 30 40 50 60
= 20 121 159 199 225

2. (a)(b)
p 10 20
t2 40 81

t2

220 Graph of t2 against p

200
180
160

140
120

100

80
60

40

20

0 10 20 30 40 50 60 p

t2 = 240

t = 15.5

(c) !wp = t
k
1
p = k2 t2

t2 = k2p

k2 = 121 – 81
30 – 20
k2 = 4

k=2

3. (a) 2N2H – aH = b

2N 2H = b + aH
b a
N 2H = 2 + 2 H

MTehlraoluighpecrobmanpdairnigsoan, Y = N 2H, X = H, m = a , c = b .
2 2

13

H 20 40 60 80 100
N2H 30 54 78 103 127

N2H
140 Graph of N2H against H
120

100

80

60
40
20

0 20 40 60 80 100 H

(b) a = 127 – 307
2 100 – 20
= 1.2125

a = 2.425

2b = 6
b = 12

(c) N 2H = 18

N2 (10) = 18

N2 = 1.8

N = 1.3416

(d) N 2H = 1.213H + 6

(1.1183)2H = 1.213H + 6
H ≈ 159.6
º H = 160, Number of workers = 160 = 20 people
b8
4. (a) L = A(3) T
1
log10L = log10A + T log103b 1
T
Through comparison, Y = log10L, X = , m = log10 3b,c= log10 A

T 10 30 50 71 91 10
log10 L 8.8 10 11.2 12.6 13.4 14.2

14

log10L

16
14 Graph of log10L against –T1–

12
10

8

6

4

2

0 10 20 30 40 50 60 70 80 90 100 –T1–

(b) (i) log10A = 8.2
A = 1.585 × 108

(ii) log10 3b = 14.2 – 8.8
100 – 10
= 0.06

3b = 1.14815

b= log10 1.14815
log10 3

= 0.1258

(c) Y = 0.06X + 8.2

2 1.5 = 0.06X + 8.2

X = 221.7°C

5. (a) 1 0.07 0.05 0.04 0.02 0.01
u

1 0.03 0.05 0.06 0.08 0.09
v

15

–1v– Graph of –1v– against –1u–
0.10

0.09

0.08

0.07

0.06

0.05

0.04

0.03

0.02

0.01

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 –u1–

(b) (i) m = 0.09 – 0.03
0.01 – 0.07
m = –1

1 = –  1 + 1
v u 10
1 –10 + u
v = 10u

10u = –10v + uv

10v – uv = –10u

v(10 – u) = –10u
–10u
v= 10 – u

(ii) W1vhe=n0.1u1 = 0, 1 = 0.1
v

1 = 0.1
f

f = 10

Mastery Practice (Page 171 – 173)

1. (a) y = 3x + 4
x2
yx2 = 3x3 + 4

yx = 3 + 4
x3

16

(b) y = px3 + qx2
xxyy23 ==
px + q
p+ q
x
p q
(c) y= x + p x

xy = p + q x2
p
xy = p q
x2 + p

(d) y = pk!wx
log10 y = log10 p + !wx log10 k

(e) y = pkx – 1

log10 y = log10 p + (x – 1)log10 k

(f) y = kx2
p

log10 y = x2 log10 k – log10 p

2. (a) y = px2 + qx
y
x = px + q

(b) p= 5–3
1–9
= −0.25

5 = −0.25(1) + q

q = 4.75

x2
3. (a) y = pq 4
1
log10 y = log10 p + x2 log10 q 4
1
= 5–4
log10 q 4 6–4

1 = 0.5
4
log10 q = 0.5
log10 q = 2
q = 102

= 100

c = 4 – 0.5(4)

log10 p = 2
p = 102

p = 100

4. y = 5x – 3x2
y
x = 5 – 3x

k = 5 – 3(2)

= −1
5–3
h= 3

= 2
3
17

5. (a)
log2 y

Qʹ(3, 5)

Pʹ(1, 2)
0x

When x = 1, log2y = log24
=2

Thus, P(1, 2)
When x = 3, log2y = log232

=5
Thus, Q(3, 5)

(b) y = abx
log2 y = log2 abx

log2 y = log2 a + x log2 b

log2 b = 5–2
3–1
= 1.5

b = 21.5

= 2.828

log2 a = 2 – 1.5
= 0.5

a = 20.5

= 1.414

6. (a) x2y = 8x + c

19 = 8(2) + c

c = 3
x2y = 8x + 3

y = 8x + 3
x2
(b) When x = 9.4
8(9.4) + 3
y = (9.4)2

y = 0.8850

18

7.
m
Graph of m against V

3

2

1

0 1234 V

8. (a)

y
35 Graph of y against x

30

25

20

15

10

5

0 x
10 20 30 40 50 60

(b) m= 35.0 – 20.0
60 – 20
3
= 8

c = 25
2
EPqeurastaimonaaonf sgtararisghlut rluinse, y = mx + c
3 25
y= 8 x + 2

19

9. (a)

T°C Graph of T against t
40

30

20

10

t(s)
0 2 4 6 8 10

(b) 30.0

(c) (i) When t = 0, T = 28°C
(ii) After t = 9, T = 33°C
(iii) When T = 30.5°C, t = 5 s

10. (a) y = stx

log10 y = log10 s + x log10 t

x 1.5 3.0 4.5 6.0 7.5 9.0
log10 y 0.40 0.51 0.64 0.76 0.89 1.00

log10 y Graph of log10 y against x
1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 x
123456789

(b) (i) log10 s = 0.28
s = 1.90546

20

(ii) log10 t = 1 – 0.4
9 – 1.5
= 0.08

t = 1.20226

(iii) y = 4, log10 y = 0.60
MTahkuas, x = 4

11. (a) 2y – p = q
x
2xy – px = q

xy = p x + q
2 2

x12 3 4 5 6
9.3 10.8 13 15
xy 5 7

xy
16 Graph of xy against x

14

12
10

8

6
4
2
0 123456 x

(b) (i) p = 15 – 5
2 6–1
p = 4

(ii) q =3
2
q=6

(iii) When x = 3.5, xy = 10

y= 10
3.5
= 2.8571

(c) xy = 2x + 3

x( 50) = 2x + 3

48x = 3

x = 0.0625

21

CHAPTER 7 COORDINATE GEOMETRY

Inquiry 1 (Page 176)

3. KTheedupdouskitaionntiotifkpPoimntePmbdaivhiadgeisttehmebleinrensgeggmareisntAABBkienptaodtawdoupaabratshawgiitahnrdateinogman: nni.sbah m : n.

4. (a) 2 pbarhtasgian
(b) 8 bpaahrtasgian
(c) 10 pbarhtasgian

(d) AP = 2 AB
10

= 1 AB
5

PB = 8 AB
10

= 4 AB
5

(e) AP : PB = 2 : 8

=1:4
(f) KPoesdiutidounkPandPivimdeesmtbhaehlaingei tseemgmbeernetnAgBgawritshArBatidoemng:an.nisbah m : n.

5. Yes. The length of AP is equal to the length of PB.
The position of point P is in the middle (midpoint) of line segment AB when the ratio m : n
is equal for each parts.

6. Yes. The position of point P change according to the changes in the value of the ratio m : n.

Self Practice 7.1 (Page 177)

1. (a) Point P divides the line segment AB in the ratio 1 : 2.

Point Q divides the line segment AB in the ratio 1 : 1.

Point R divides the line segment AB in the ratio 11 : 1.

(b) S B
A

2. (a) m = 2, n = 5

(b) P divides rope AB in the ratio 2 : 5.
(c) P(6, 0)

Self Practice 7.2 (Page 180)

1. (a) x = 2(3) + 3(–7) , y = 2(7) + 3(2)
5 5
–15 20
= 5 = 5

= –3 =4

P(–3, 4)

1

(b) x = 2(– 4) + 1(2) , y = 2(–1) + 1(5)
3 3
–6 3
= 3 = 3

= –2 =1

P(–2, 1)

(c) x= 3(7) + 2(–3) , y = 3(–3) + 2(2)
5 5
15 –5
= 5 = 5

= 3 = –1

P(3, –1)

2. p= 3(2h) + 2(2p) t= 3(h) + 2(3t)
5 5
6h + 4p 3h + 6t
= 5 = 5

5p = 6h + 4p 5t = 3h + 6t
p
h= 6 –t = 3h
p
3( 6 ) = –t

p = –2t

3. (a) x = 1(–2) + 3(6) , y = 1(–5) + 3(7)
4 4
16 16
= 4 = 4

= 4 = 4

C(4, 4)

(b) x = 1(–2) + 1(6) , y = 1(–5) + 1(7)
2 2
4 2
= 2 = 2

=2 =1

D(2, 1)

4. (a) n – 5m = –1
m+n

n – 5m = –m – n

2n = 4m

m = 2
n 4

m = 1
n 2

AP : PB = 1 : 2

2k + 10 = 2
3

2k + 10 = 6

2k = – 4

k = –2

2

(b) 2n + 6m =4
m+n

2n + 6m = 4m + 4n

2m = 2n

m = 1
n 1

AP : PB = 1 : 1

1+k =3
2
1+k=6

k=5
3n + 8m
(c) m+n =4

3n + 8m = 4m + 4n

4m = n

m = 1
n 4

AP : PB = 1 : 4

4k + 2 =6
5
4k + 2 = 30

4k = 28

k=7
–3n + 2m
(d) m+n = –1

–3n + 2m = –m – n

3m = 2n

m = 2
n 3

AP : PB = 2 : 3

3(–2) + 2(8) = k
5
5k = 10

k = 2

Self Practice 7.3 (Page 182)

1. x= 4 + 2(40) , y = 6 + 2(45)
3 3
84 96
= 3 = 3

= 28 = 32

º(28, 32)

The coordinate of the ball when it touches the surface of the field is (28, 32).

2. The first rest house divides highway AB in the ratio 1 : 2.

x= –233(–4)=3+ 1(5) , y = 2(5) + 1(2)
= 12 3
3

= –1 =4

º(–1, 4)

3

The second rest house divides highway AB in the ratio 2 : 1.

x= 631(–4=)3+ 2(5) , y = 1(5) + 2(2)
= 93
3

= 2 = 3

º(2, 3)

The coordinate of both rest house is (–1, 4) and (2, 3).

3. (a) HL : LK = 2 : 1

(b) x = 1(–3) + 2(6) , y = 1(–2) + 2(10) LK = ! (6 – 3)2 + (10 – 6)2
3 3 = ! 25
9 18 = 5 units
= 3 = 3

=3 =6

º(3, 6)

Intensive Practice 7.1 (Page 183)

1. x= 1(2) + 4(7) , y = 1(8) + 4(3)
5 5
30 20
= 5 = 5

=6 =4

º R(6, 4)

2. (a) 6 = 5(4) + 2(x) , 3= 5(5) + 2(y)
7 7

42 = 20 + 2x 21 = 25 + 2y

2x = 22 2y = – 4

x = 11 y = –2

º Q(11, –2) 5 + (–2)
2
( )(b) TitMikidtepnoginath PQ = 4 + 11 ,
2

= ( 15 , 3 )
2 2

3. 1= 3(–3) + 2(h) , 4 = 3(6) + 2(k)
5 5

5 = –9 + 2h 20 = 18 + 2k

2h = 14 2k = 2

h = 7 k=1

4. e= 16r + 9e
7
7e = 16r + 9e

2e = –16r

r = –  1 e…1
8

f= 4r + 12f
7

7f = 4r + 12f

5f = –4r

f = –  4 r…2
5
4

SGuabnsttiitkuatne 1 kientdoalam 2:

( ) f = – 45 –  1 e
8
1
= 10 e

e = 10f

5. (a) x = 1(1) + 2(7) , y = 1(4) + 2(–8)
3 3
15 –12
= 3 = 3

= 5 = – 4

º U(5, – 4)

( )(b) TitMikidtepnogianht QR = 7 + 9, –8 + 5
2 2

9n + 5m ( )= 8, –  3
m+n 2
(c) =6

9n + 5m = 6m + 6n

3n = m

m = 3
n 1

º RT : TS = 3 : 1
(d) PS = ! (5 – 1)2 + (1 – 4)2

= ! 25
= 5 units

6. (a) n + 5m = 2
m+n

n + 5m = 2m + 2n

3m = n

m = 1
n 3

ºm:n=1:3

(b) 3(–2) + 1(2) = k
4–6 + 2 = 4k

4k = – 4

k = –1

7. x= 1(3) + 4(13) , y = 1(11) + 4(1) x= 2(4) + 1(10) , y = 2(4) + 1(7)
5 5 3 3
55 15 18 15
= 5 = 5 = 3 = 3

= 11 = 3 = 6 = 5

º P1(11, 3) º P2(6, 5)

11 + 6 3+5
2 2
( )TitiMk itdepnogianht P1P2 = ,

= ( 17 , 4)
2
( )MakTah, utist,ikTthkeedpuodsuitkioan orufmHaahziHq'sazhioquisaelaihs17
2 , 4 .

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Inquiry 2 (Page 184)

ACTIVITY 1
4. The gradient of straight line L1 is equal to the gradient of straight line L2, m1 = m2.
5. Both angles formed are equal to each other, q1 = q2.

ACTIVITY 2
4. The product of the gradient of straight line L1 with the gradient of straight line L2 is –1, m1m2 = –1.
6. The product of tan q1 with tan q2 is –1, tan q1tan q2 = –1.

Self Practice 7.4 (Page 187)

1. (a) 2x + 3y = 9 4x + 6y = 0
3y = –2x + 9 6y = – 4x

y = –  2 x + 3 y = – 32 x
3

º m1 = – 23 º m2 = – 32

Since the pair of straight lines have the same gradient, thus the straight lines are

parallel.

(b) y = 3 x – 5 4y – 3x = 12
4
3 4y = 3x + 12
º m1 = 4
y= 3 x + 3
4
3
º m2 = 4

Since the pair straight lines have the same gradient, thus the straight lines are

parallel.

(c) x – 2y = 6 2x + y = 5

2y = x – 6 y = –2x + 5
1 º m2 = –2
y= 2 x – 3

º m1 = 1
2

Since the pair of straight lines have the product of their gradients as -1, thus the

straight lines are perpendicular.

(d) 2x + 3y = 9 2y = 3x + 10

3y = –2x + 9 y= 3 x + 5
2
y = –  2 x + 3 3
3 º m2 = 2

º m1 = – 23

Since the pair of straight lines have the product of their gradients as -1, thus the

straight lines are perpendicular.

2. (a) 2y = 10 – x y = 3px – 1
º m2 = 3p
y = –  1 x + 5
2

º m1 = – 12

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