BUKU TEKS MATHEMATICS FORM 3 (DLP)

Chapter 9 Straight Lines

Determine the equation of a straight line which passes through a point and is parallel to a
given straight line

%\ QRZ \RX ZRXOG NQRZ WKDW LI WZR VWUDLJKW OLQHV DUH SDUDOOHO WKHQ WKH JUDGLHQWV RI ERWK OLQHV DUH
equal.

Example 14

The diagram below shows the straight line AB with equation y = –2x + 6. Determine the equation of
D VWUDLJKW OLQH SDUDOOHO WR AB DQG SDVVHV WKURXJK SRLQW P(5, 4).

A

P(5, 4)

y = –2x + 6

B

Solution:

The equation of the straight line AB is y = –2x + 6, thus the gradient of AB is –2.

7KH VWUDLJKW OLQH ZKLFK SDVVHV WKURXJK SRLQW P LV SDUDOOHO WR AB, thus the gradient P of that line is –2.

Substitute the values of P, x and y into \ P[ F to determine the value of F

4 = (–2)(5) + F Given P(5, 4), thus
4 = –10 + F [ = 5 and \ = 4.
F = 4 + 10

F = 14

7KXV WKH HTXDWLRQ RI WKH VWUDLJKW OLQH SDUDOOHO WR AB DQG SDVVHV WKURXJK WKH SRLQW P is y = –2x + 14.

Example 15

'HWHUPLQH WKH HTXDWLRQ RI D VWUDLJKW OLQH SDUDOOHO WR WKH VWUDLJKW OLQH 2x y = 12 DQG SDVVHV WKURXJK
SRLQW G(6, 8).

Solution:

Given the equation of straight line 2x y = 12.

Thus y = –2x + 12

y = – —2 x + 4
G rad ient of the s traight line = – — 2 .

7KH VWUDLJKW OLQH ZKLFK SDVVHV WKURXJK SRLQW G LV SDUDOOHO WR WKH VWUDLJKW OLQH 2x y = 12.

Hen ce, the gradient of the straight line is – — 2 .

Substitute the values of P [ and y into \ P[ F and determine the value of F.
Thus, 8 = (– —2 )(6) + F
Given Q(6, 8), thus

8 = –4 + F [ = 6 and \ = 8. CHAPTER 9

F=8+4

F = 12

+HQFH WKH HTXDWLRQ RI WKH VWUDLJKW OLQH SDUDOOHO WR 2x y = 12 DQG SDVVHV WKURXJK SRLQW G

i s y = – — 2 x + 12.

241

MIND TEST 9.1g

1. Determine the equation of a straight line that is parallel to the given straight line and passes
through point P.

(a) y = 3x + 9, P(2, 7) (b) y = –2x + 7, P(–3, 4)

(c) 3x + 2y = 4, P(2, 6) (d) —2x + —3y = 1, P(–12, 9)

2. The diagram on the right shows a straight line PQ. Given that y

the equation of the straight line PQ is y = —31 x + 2 and O is the 4 A(2, 4) Q

origin, determine the equation of a straight line parallel to PQ 2

and passes through P

(a) point A(2, 4) O 24 x
(b) point B(4, –2)
(c) the origin –2 B(4, –2)

How do you determine the point of intersection of two LEARNING
straight lines? STANDARD

The point of intersection of two straight lines can be determined by the Determine the point
following methods: of intersection of two
straight lines.
1. Drawing both straight line graphs on the same Cartesian plane and
determine the point of intersection from the graphs. REMINDER

2. Solving simultaneous equations using The calculator should
only be used for checking
(a) substitution method (b) elimination method answers.

Example 16

Determine the point of intersection of the straight lines 2x + y = 5 and x + 2y = 1.

Graphical method y
(a) 2x + y = 5
234 8
y = –2x + 5 1 –1 –3 7
6
x –1 0 1 5
4
y753 3
2
CHAPTER 9 (b) x + 2y = 1 2345 1 Point of intersection
2y = –x + 1 – —21 –1 – —32 –2 = (3, –1)
y = – —12 x + —12 –3 –2 ––11O
–2 x
x –1 0 1 –3 123456
y 1 —21 0
x + 2y = 1

2x + y = 5

242

Chapter 9 Straight Lines

From the graph, it is found that the point of intersection of the straight lines 2x + y = 5 and
x + 2y = 1 is (3, –1).

Substitution method Elimination method

2x + y = 5 ---------------1- 2x + y = 5 ---------------1-
x + 2y = 1 ---------------2-
From 1 , y = 5 – 2x-- 3 x + 2y = 1 ---------------2-

Substitute y = 5 – 2x in 2 , 1 × 2 4x + 2y = 10 -----------3- (minus)
x + 2y = 1 ------------2-
x + 2(5 – 2x) = 1
x + 10 – 4x = 1 3x = 9
x – 4x = 1 – 10 x =3
–3x = –9
x=3 Substitute x = 3 in 1 ,

Substitute x = 3 in 3 , 2(3) + y = 5
y = 5 – 2(3) 6+y =5
y=5–6 y =5–6
y = –1 y = –1

Thus, the point of intersection is (3, –1).

Thus, the point of intersection is (3, –1).

Brainstorming 5 In pairs

Aim: To determine the coordinates of the intersection of two straight lines.

Materials: Dynamic software

Steps:
1. Start with New Sketch and click Graph next click Show Grid.

2. Click Graph again and select Plot New Function (Diagram 1).

3. Use Plot New Function to plot the intersection of the two straight lines.

4. Example: y = x + 3 and y = –x + 5.

5. Use Arrow Tool to select both straight line graphs. Click Construct and select Intersection.

6. Click Measure and select Coordinates. The intersection point A (1.00, 4.00) will be
displayed (Diagram 2).

7. Repeat steps 1 to 6 for intersection of the other two straight lines. CHAPTER 9
(a) y = x + 2 and y = 2x + 4 (Diagram 3)
(b) y = 4 and y = 3x – 2 (Diagram 4)

243

Diagram 1

Diagram 2



Diagram 3 Diagram 4

Discussion:

What can you conclude from the results above?

From Brainstorming 5, it is found that:

(a) The point of intersection of two straight lines can be determined by plotting both straight
lines on the Cartesian plane.

(b) Two straight lines that are not parallel intersect at only one point.

MIND TEST 9.1h

1. Determine the point of intersection of the following pairs of straight lines using the substitution
method.

(a) x = 3, 2x + y = 10 (b) y = 4, 3x – 2y = 7

(c) x + y = 5, 2x – y = 4 (d) 2x + y = 3, 3x – 2y = 8

CHAPTER 9 2. Determine the point of intersection of the following pairs of straight lines using the elimination
method.

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

(c) x – y = –5, 2x + 3y = –10 (d) 2x – 3y = 5, 3x + 2y = 14

244

Chapter 9 Straight Lines

How do you solve problems involving straight lines? LEARNING
STANDARD
Example 17
Solve problems involving
straight lines.

The diagram on the right shows a parallelogram PQRS. y
1
Given that the gradient of SR is 2 and y-intercept of straight %Q(0, h)

line PS is – 4, determine P(– 4, 6)%

(a) the value of h x

(b) equation of straight line PS O %R

(c) x-intercept for straight line PS %

Solution: T

S%

Understanding the problem Planning a strategy

Ɣ PQRS is a parallelogram. Ɣ 7KH YDOXH RI h can be determined by using the gradient
1
Ɣ *UDGLHQW RI PQ that is, gradient of PQ = gradient of SR = 2 .

= gradient of SR Ɣ 7KH y-intercept of the straight line PS is – 4. Thus, the

= 1 coordinates of T are (0, – 4).
2
Ɣ 7KH x-intercept of the straight line PS can be determined
Ɣ y-intercept of PS is – 4.
by substituting y = 0 into equation PS.

Implementing the strategy

(a) Gradient PQ = Gradient SR = 1 (b) Straight line PS passes through point
—0 h–—–(–—64–) = —21 2 T (0, – 4)

h—4–—6 = —12 Gradient PS = —0––—4(–—– 46–) = —–41–0 = – —25
y-intercept of straight line PS is – 4
h–6 =2 Thus, equation of straight line
h =2+6 PS is y = – —52 x – 4.
h = 8.
Making a conclusion
(c) Equation of straight line PS is y = – —25 x – 4
When y = 0 (a) The value of h is 8.
0 = – —25 x – 4
—25 x = – 4 (b) The equation of the straight line
x = – —58 5
x-intercept of the straight line PS is – —58 . PS is y = – 2 x – 4. CHAPTER 9

(c) x-intercept of the straight line PS
8
is – 5 .

245

Example 18 TIPS

Given straight lines y = – 1 x + 3 and 2x – y = 4 intersect at point A, A straight line can be
3 drawn if its x-intercept
determine the coordinates of point A using the graphical method. and y-intercept are
known.

Solution: 1
3
For the straight line y = – x + 3,

(a) when x = 0, (b) when y = 0,

y=– 1 (0) + 3 0=– 1 (x) + 3 y
3 3 2x – y = 4

y=3 1 x = 3 3
3
y-intercept = 3 x=9 2 A y = – —31 x + 3
1

x-intercept = 9 –1O 1 2 3 4 5 6 7 8 9 x
–2
For the straight line 2x – y = 4, –3
–4
(a) when x = 0, (b) when y = 0,
From the graph, it is found that the
2(0) – y = 4 2x – (0) = 4 coordinates of A are (3, 2).
–y = 4 2x = 4
y = –4 x=2

y-intercept = – 4 x-intercept = 2

MIND TEST 9.1i y G H(8, 2)
F K
1. The diagram on the right shows a parallelogram x
FGHK. Given that O is the origin and point K is located O
on the x-axis, the equation of straight line FG is
2y = x + 20, determine
(a) the gradient of straight line FG
(b) the y-intercept of straight line HK
(c) the equation of straight line HK

y

2. In the diagram on the right, O is the origin and PQRS P(–3, 8)
is a trapezium where PS and QR are parallel. The
CHAPTER 9 straight line RS is parallel to the y-axis, and points Q QO Sx
and S are on the x-axis. Determine R(5, –10)

(a) the coordinates of S

(b) the equation of straight line QR

(c) the x-intercept of straight line QR

246

Chapter 9 Straight Lines

Dynamic Challenge

Test Yourself

1. Given that 2x + 5y = 30 is an equation of a straight line, determine

(a) the x-intercept (b) the y-intercept (c) the gradient

2. State the equation of the straight line for each of the following diagrams.

(a) y (b) y

–6 O x x
O
–8

3. Determine the equation of a straight line that has a gradient of 3 and passes through point
R(– 4, 6).

4. Determine the equation of a straight line that passes through point P(–1, –2) and point Q(3, 14).

5. Determine the equation of a straight line that passes through point M(–3, 5) and is parallel to
the straight line 6x + 2y = 18.

6. Determine the point of intersection of the straight lines y = –8 and y = – 4x + 12.

Skills Enhancement y 2x – y = 5
P
1. The diagram on the right shows two straight lines x – 3y = –5
intersecting at point P. Given that O is the origin, x
determine the coordinates of P.

2. In the diagram on the right, GH, HK and KL are O x CHAPTER 9
straight lines. Point H lies on the x-axis, GH is y
parallel to KL, and HK is parallel to the y-axis.
K
Given the equation of GH is 2x + y = 6, G

(a) state the equation of straight line HK OH

(b) determine the equation of straight line KL L(10, – 4)
and then state the x-intercept of KL

247

3. The diagram on the right shows a parallelogram y E(7, 21) F
OEFG. Given O is the origin, determine O G(6, –12) x

(a) the equation of straight line OG

(b) the equation of straight line EF

(c) the x-intercept of straight line EF

4. The diagram on the right shows a trapezium y D(9, 10)
ABCD drawn on the Cartesian plane. Given AB A(4, 6) C
is parallel to DC, determine
O B(12, 2)
(a) the equation of straight line AB x

(b) the equation of straight line CD KEDAI MAKAN y

(c) if the straight lines AB and CD intersect. Restaurant Clinic
State the reasons for your answer. (2, 4)

Self Mastery (–1, 0) O (5, 0) x

1. The diagram on the right shows a parallelogram Kamal’s School
drawn on a Cartesian plane and it represents house
the locations of Kamal’s house, the school, the Town Q
clinic and the restaurant. Given that the scale is y (6, 7)
1 unit = 1 km,
Town P x
(a) calculate the distance, in km, between (–9, 4)
CHAPTER 9 Kamal’s house and the school
O
(b) determine the coordinates of the restaurant
(c) calculate the distance, in km, between Town R
(–3, – 4)
Kamal’s house and the restaurant
(d) determine the equation of the straight line

connecting the school and the clinic

2. The diagram on the right shows the positions of
town P, town Q and town R drawn on a Cartesian
plane. Given that the scale is 1 unit = 2 km,

(a) calculate the distance in km, between town R
and the origin O

(b) determine the equation of straight line
connecting town P and town Q

(c) calculate the nearest distance, in km, between
town P and town R

(d) calculate the time taken, in minutes, by Encik
Mazlan to arrive at town Q if he drives from
town R to town Q using the shortest route at
an average speed of 50 km h–1

248

Chapter 9 Straight Lines

3. The original height of a plant F is 9 cm. Its height is y cm after x days and is represented by the
3
equation y = 16 x + 9. Plant G has the same growth rate as plant F. Plant G reaches a height of

15 cm after 8 days. Determine the equation to represent the height of plant G. Then, state the

original height, in cm, of plant G.

y

4. JK is a straight road that passes through the J Town F
midpoint between town E and town F. (3, 3)
(a) The equation for the straight road JK is Town E
y = –2x + k, where k is a constant. Determine (–7, –1) x
the value of k. O

(b) Another straight road GH with the equation K
y = 2x + 17 ZLOO EH FRQVWUXFWHG $ WUD൶F
light will be installed at the junction of both
roads JK and GH. Determine the coordinates
RI WKH WUD൶F OLJKW.

TIPS

Solution by scale drawing
is not accepted.

P ROJ EC T

Title: Gradient and speed.

Material: Toy car, board, brick, long ruler and stopwatch.

Steps:
1. Place a board over two bricks as in the diagram below.

Toy car

Board }Brick height

P
%
}
Horizontal distance

2. 0HDVXUH WKH KRUL]RQWDO GLVWDQFH ¿[HG DQG WKH KHLJKW RI WKH FDU IURP WKH JURXQG &DOFXODWH CHAPTER 9
the gradient of the board and record it.

3. Release the toy car. Record the time, in seconds, for the toy car to reach point P.

4. Add the bricks one by one. Repeat steps 2 and 3.

5. What can you conclude about the relationship between the gradient of the board and the
speed of the car?

249

CONCEPT MAP
Straight Lines

Equation of a straight line Rewriting the equation of Point of
y = mx + c a straight line intersection of two
m = gradient
y = mx + c in the form of straight lines
c = y-intercept
ax + by = c and —ax + —by = 1
and vice versa.

Equation of straight line parallel to the x-axis Gradients of parallel lines
y = k; k = constant are equal

Equation of straight line parallel to the y-axis
x = h; h = constant

SELF-REFLECT
At the end of this chapter, I am able to:

1. Determine the gradient and y-intercept when the equation of the straight line in
the form of y = mx + c is given.

2. Determine the gradient and y-intercept when the equation of the straight line in
the form of ax + by = c is given.

3. Determine the gradient and y-intercept when the equation of the straight line in
—ax —y
the form of + b = 1 is given.

4. Determine whether a point lies on a given straight line.

5. Determine whether two straight lines are parallel.

6. Determine the equation of a straight line.

CHAPTER 9 7. Determine the point of intersection of two straight lines.

8. Solve problems involving straight lines.

250

Chapter 9 Straight Lines

EXPLORING MATHEMATICS

The area under a straight line can be determined if y y=x
enough information is given. 6
For example, the area under the graph of the straight Area under the graph
line y = x for the range 0 x 6 in the diagram on O of the straight line
the right can be determined as follows: y=x
6x
The area under the graph = 1 × base × height
2

= 1 × 6 units × 6 units
2

= 18 unit2

Worksheet 2. y=x+2 3. y = 2x + 1

1. y y

y

y=4

O 6 x 6 x 4x
O O

4. 5. 6.

yy y

y = –x + 6 y = – —21 x + 6

x + 2y = 12

O x 14 x O 8x
O

Steps:

1. Get into groups. CHAPTER 9
2. Calculate the area under each graph of straight line provided.
3. 3UHVHQW \RXU JURXS¶V ¿QGLQJV.
4. Propose at least two other ways to determine the area under the graph of a straight line.

251

Answers

CHAPTER 1 Indices MIND TEST 1.2c

MIND TEST 1.1a 1. (a) 4 (b) 72 (c) m4n5
(d) 3xy3 (e) m (f) –5h
1. Base Index 2. (a) 66
3 (b) (0.5)7 2. (a) 8 8 ÷ 84 ÷ 83 = 8 3. 8
5 7
–4 ( )(c) —21 4 (b) m4n 6 ÷ m 2 n5 = m2n
—12 10
m 6 (d) (–m)5 (c) —m1—0n—4m=7—nm—2 —n2 = m5n 5
n 0
0.2 9 ( )(e) 1—32 3 (d) 27x3y6 = xy 2 = 3x 2 y5
– —3 ( )(f) – —1n 6 ——9—x2—y3 —–
4
7 MIND TEST 1.2d
x 20
2—1 1. (a) 1210 (b) 320 (c) 76 (d) (– 4)21
3 2 (e) k24 (f) g26
8 (g) (–m)12 (h) (–c)21
1 2. (a) True (b) False
(c) False (d) False

3. (a) (–3) = (–3) = (–3) MIND TEST 1.2e

(b) 2.5 = 2.5 = 2.5 = 2.5 1. (a) 22 × 38 (b) 119 × 915 (c) 136 ÷ 712
(d) 515 × 320 (e) m15n20 p10 (f) 16w8x12
(c) 2 = 2 = 2 = 2 = —2 (g) —72—9a—30 (h) —8a—15
— — — — b24 (c) 6—464
33 333 27b12 (e) x4y4
2. (a) 112 × 44 (h) b2d6
(d) (–2—41 ) = (–2—14 ) = (–2—41 ) (b) 33 × 62 (c) de

(e) k = k = k = k = k = k (d) (– 4)6 × (–5)4
(f) (–p) = (–p) = (–p) = (–p) = (–p) = (–p) = (–p)
(f) h10k6 (g) m11n15
1 1 1 1 = —m1 1 1 = —m1
(g) —m = —m = —m = —m = —m = —m 3. (a) 6mn8 (b) 10x8y3

(h) (3n) = (3n) = (3n) = (3n) = (3n) MIND TEST 1.2f

MIND TEST 1.1b 1. (a) 5—13 (b) 8—14 (c) x—18 (d) —y11–6

1. (a) 34 (b) 56 ( )(c) —4 3 (e) a4 (f) 202 (g) n—34 (h) – —n56–
5
(d) (0.2)5 (e) (– 4)7 (i) 7—2m–5 (k) —5 12 (l) – —7 14
( )(f) – —1 2 2 3
MIND TEST 1.1c 4 ( ) ( )(j) – 8–m3—4

(c) 15.625 (n) —3y– 4
(f) —1—– 2x
1. (a) 6 561 (b) –1 024 ( ) ( )(m) —yx 10 (o) (2x)5
(d) –32.768 (e) —2—43— 1 296
(g) 2—7 2. (a) 5–4 (b) 8–3 (c) m–7 (d) n–9
9 32 768
(h) –121—9 1 111
(e) —10–––2 (f) —(–4–)—–3 (g) —m–—12 (h) —n–—16
27
( ) ( )(i) —7 –9
4
MIND TEST 1.2a (j) —xy –10

1. (a) 37 (b) (– 0.4)8 ( )(c) —4 9 3. (a) —1 (b) —24– (c) 26 × 52
7 4 314
(f) —n1–2
( )(d) –1—2 10 (e) – 6m9 (d) 3—m1—3n–7 (e) 8—m1–8 (f) m—6–n
5 (h) y12 5 18
(g) –15x7
MIND TEST 1.2g
MIND TEST 1.2b (b) (0.4)3 × (1.2)9 1. (a) 125–31 (b) 2 187–71 (c) (–1 024)–51 (d) n—110
1. (a) 55 × 95 (d) – —3 k6p11 2. (a) ¥4 (b) 5¥32 (c) 3¥–729 (d) 15¥n
3. (a) 7 (b) – 6 (c) 8 (d) –8
(c) 4x6y7 2

252

252

MIND TEST 1.2h w–73 ( ) ( )x–25 —1816 –34 —hk –23 3. (a) 3 (b) 0 (c) –8
a–mn 729–56 121–32 (d) –5 (e) 5 (f) 2
(g) 2 (h) –1 (i) 1

81—16 3 –14 —hk 2 –13 Self Mastery

(am)–1n (7295)–61 (1213)–12 (w3)–71 [( ) ] [( ) ](x2)–15 1. (a) 1 000 (b) 500 000 (c) 50

81—61 –41 3 —hk –13 2 2. (a) —3 (b) —3 (c) 15
(a–n1)m (729–16)5 (121–12)3 (w–17)3 [( ) ] [( ) ](x–15)2 4 2

3. (a) –1, 6 (b) 1, –7 (c) –1, 4

( ) ( )n¥ am 6¥ 5 ¥ 3 7¥ w 3 5¥ x 2 4 —1816 3 3 —hk 2 4. (a) x = —1 , y = 2 (b) x = 1, y = – —2
( ) ( )(n¥ a )m (6¥ 5 ¥ 3 (7¥ w )3 (5¥ x )2 4 —1816 3 3 —hk 2 6 3

5. 12$C 6. RM27 130 7. RM61 462.77

MIND TEST 1.2i CHAPTER 2 Standard Form

1. (a) 9 (b) 4 (c) 4 (d) 8 MIND TEST 2.1a (d) 4 s.f.
(h) 6 s.f.
(e) 256 (f) 16 (g) 216 (h) 343 1. (a) 2 s.f. (b) 5 s.f. (c) 5 s.f.
(e) 2 s.f. (f) 5 s.f. (g) 4 s.f.

(i) 7 (j) 1 331 (k) 169 (l) 1 000 MIND TEST 2.1b

2. (a) 2 ¥6 561 1, 3 4, 9 2, 81 1, 243—54 , 27—34 1. (a) 47 200 47 000 50 000
(b) 25—23 , 125 1, 625—34 , 2 ¥15 625 1, 3 125—53 , 5 3 (b) 5 260 5 300 5 000
(c) 306 310 300
MIND TEST 1.2j (b) mn6 (c) 1—30zx–2 (d) 20.7 21 20
(b) 648 (c) 86 400 (e) 8.60 8.6 9
1. (a) —c7 (e) 81 (f) —12–5 (f) 5.90 5.9 6
de 4. 48 (g) 0.694 0.69 0.7
8 (h) 0.0918 0.092 0.09
2. (a) —1—– (i) 0.00571 0.0057 0.006
2 401 (b) 2.83 (c) 11.1
2. (a) 12.02 (e) 6.61 (f) 13
(d) —7 (d) 24 (h) 36.0
54 (g) 20

3. 3 456

Dynamic Challenge MIND TEST 2.2a

Test Yourself 1. (a) 3.5 = 101 (b) 4.81 = 102

1. (a) True (b) False (25) (c) False (1) (c) 5.075 = 103 (d) 9.725 = 101

(d) False (32x15) (e) True ( )(f) False —a24 (e) 3.1243 = 103 (f) 9.0 = 10–1
( )(i) False —62—15 m–
(g) 2.3 = 10–1 (h) 3.75 = 10–2
53( 3 )
(g) False [(5¥32)2] (h) True 2. (a) 2.5 (b) 37.5 (c) 423
2. 5 4 × 55
(d) 5 070 (e) 91 000 (f) 0.62

(g) 0.0729 (h) 0.001034 (i) 0.0008504

3. (a) 1.05 = 106 metres (b) 2.16 = 1011 bytes

512 ÷ 5 3 (¥25) 9 (c) 7.5 = 1011 litres (d) 9.5 = 10–5 metres

(e) 1.23 = 10–7 metres (f) 8.9 = 10–17 metres

(—51 ) –9 ( 3 ¥125) 9 MIND TEST 2.2b

(5 6 )–23 —56—5×2—5 5– 1. (a) 5.97 = 104 (b) 3.93 = 106
—51––9 (c) 1.021 = 108 (d) 1.574 = 105
( )—51––3 3 (e) 5.46 = 108 (f) 8.59 = 104
(g) 5.77 = 104 (h) 1.08 = 10–3
( )3. —3 –2 (i) 6.09 = 10–5 (j) 9.91 = 10–3
20 as —31––4 as 5 as 72 × 5–3 as (5–1 × ¥25)3 (k) 7.68 = 10–4 (l) 8.685 = 10–6

1 34 (—35 )2 —7523 1 MIND TEST 2.2c

Skills Enhancement 1. (a) 1.48 = 108 (b) 3.75 = 10–8
(c) 2.52 = 108 (d) 2.12 = 103
1. (a) —n7– (b) —x4—y7 (c) xy2 (e) 4.5 = 10–3 (f) 6.4 = 103
m 2 (c) 1 (d) 2 (g) 2.95 = 103 (h) 8.6 = 108
3. 63 4. 103 micrometres
2. (a) —4– (b) 2—5 (e) 7 (f) 1 2. 3.126 = 103
125 7

253

253

MIND TEST 2.2d 5. (a) Real estate
(b) Risk potential = Low
1. 2.02 = 105 m3 Return = High
2. (a) 9.17 = 107 km (b) 4.44 = 109 km (c) 4.35 = 109 km Liquidity = Low
(c) Encik Osman’s action is wise because our
Dynamic Challenge country focuses on the tourism sector. Therefore,
it is appropriate to set up the homestay. Besides,
Test Yourself the investment in the homestay has low risk.

1. (a) 24 000 (b) 54 300 (c) 9 000 (d) 300 000 MIND TEST 3.1e

(e) 5 000 (f) 5.00 (g) 0.28 (h) 40 1. Purchase of shares every month or periodically but
not at a lump sum.
(i) 420 (j) 10 (k) 1.04 (l) 502
2. (a) Investor 2. This is because the purchase of 2
2. (a) 3.48 = 108 (b) 5.75 = 104 (c) 5.11 = 104 shares on a regular basis allows him to purchase
many units of shares and the average cost per
(d) 2.96 = 109 (e) 8.84 = 10–2 (f) 3.31=10–4 unit can be reduced.

(g) 9.77 = 10–8 (h) 5.43 = 104 (b) RM1.80. 13 268 units of shares
F Ɣ The average cost per unit share can be reduced
3. (a) –2, 0.025, 0.025, 1.35, 1.375 Ɣ Reduce the risk of loss

(b) –3, 0.0034, 5.74, 0.0034, 5.7434 MIND TEST 3.1f

(c) –3, 0.0042, 1.75, 0.0042, 1.7458 1. (a) Mr Rasamanie – Real estates (Low Risk)
Encik Nik Izwan – Savings (Low Risk)
(d) –3, 0.0043, 3.7, 0.0043, 3.657 Real estates (Low Risk)
Shares (High Risk)
4. (a) 1.2 = 104 (b) RM214 5. 97 people
(b) Encik Nik Izwan. This is because if there is a loss
Skills Enhancement in one of the investments, it can be covered by
other investments.
1. (a) 5.57 = 102 m2 (b) RM10 824
(c) Economic factor and political factor of the
2. (a) (i) 70.9 km h–1 (ii) 47.1 km h–1 (iii) 68.4 km h–1 location of the real estate.

Self Mastery 2. 25.74%

1. (a) Mercury = 7.48 = 107 km2 MIND TEST 3.2a

Neptune = 7.62 = 109 km2 1. Personal loans are short term loans for consumer use.
2. Ɣ Prepare your budget
Jupiter = 6.14 = 1010 km2 Ɣ Plan your expenses
3. Credit card – He is not required to pay interest if his
(b) 6.133 = 1010 km2
debts are settled in interest-free period as compared
2. (a) 4.37 g (b) 4.99 g to loan.

CHAPTER 3 Consumer Mathematics: Dynamic Challenge
Savings and Investments, Credit and Debt
Test Yourself
MIND TEST 3.1a
1. Savings is the balance after deducting mandatory
1. Ɣ For a well-planned life in the future expenditure from salary.
Ɣ As an additional income
Ɣ For emergency use 2. Ɣ High interest rate.
2. Ɣ Open a Fixed Deposit Account Ɣ 6DYLQJV SHULRG LV VXEMHFWHG WR D VSHFL¿HG WLPH.
Ɣ This is because the money will not be used for a 3. RM8 640

given period Skills Enhancement
Ɣ Higher interest rates are also offered
3. Cheques are commonly used by businessmen/ 1. Increase the number of shares purchased and the
average cost per unit will be lower as compared to if
businesswomen for payments in large amounts while the units are purchased all at once.
most people only make daily payments in small
amounts. 2. Purchase of land lots, houses, factories and so on.
3. (a) Dividend (b) Capital gain (c) Bonus share
MIND TEST 3.1b 3. RM106.17 4. (a) Lee Chong needs to have the knowledge to assess
1. RM610.10 2. RM1 159.70
and select shares while Mokhtar’s investment is
MIND TEST 3.1c assisted by a professional company.
(b) Lee Chong’s risk is higher compared to
1. Return on investment is the value of return of the Mokhtar’s.
investment. 5. RM300 6. (a) RM360 (b) 3 000 units (c) 9 000 units
7. RM1 000, 3%, 3 years 8. RM634.12
2. (a) RM2 000
(b) RM24 000 + RM230 000 = RM254 000

3. RM320

MIND TEST 3.1d

1. The higher the risk, the higher the return.
2. Bank Negara Malaysia guarantees deposits in the

bank.
3. It can be cashed immediately.
4. Real estate price usually increases but rarely falls.

254

254

Self Mastery CHAPTER 5 Trigonometric Ratios
MIND TEST 5.1a
1. RM3 750 2. 8.85%

3. RM7 000 4. RM400

5. RM233.33 6. RM52.87 Angle Hypotenuse Opposite side Adjacent side

7. (a) Masnah Rasam’s view is not recommended QPR PR QR PQ
PRQ PR PQ QR
because she has to pay interest. MNK KN KM MN
MKN KN MN KM
(b) RM320, 8% FEG EG FG EF
EGF EG EF FG
(c) Cash, because no interest needs to be paid. BAE AE BE AB
AEB AE AB BE
8. RM15 000 BCD CD BD BC
BDC CD BC BD
9. 4%

10. RM900

CHAPTER 4 Scale Drawings
MIND TEST 4.1a
1. Diagram 1, Diagram 2, Diagram 4

MIND TEST 4.1b (d) 1 : —23 MIND TEST 5.1b

1. (a) 1 : —21 (b) 1 : 3 (c) 1 : —12 ¨DEF cos x = —DDFE– tan x = —DEFE–
2. Length = 6 cm Width = 2 cm 1. sin x = —DEFF– cos y = —EF– tan y = —DE–
3. 10 km
4. 6 cm sin y = —DE– DF EF
DF
MIND TEST 4.1c (ii) 1 : 2 cos x = —LM– tan x = —KL–
2. (b) (i) 1 : —1 ¨KLM KM LM
sin x = —KL–
2 cos y = K—KML– tan y = —KLML–
KM
MIND TEST 4.1d sin y = K—LMM– cos x = —PS– tan x = —QS–
¨PQR PQ PS
1. 1 944 cm2 2. 34.8 cm sin x = —QS–
3. 560 m2 4. 20 cm cos y = —RS– tan y = —QS–
5. (a) 7 200 m2 (b) 2 hours 24 minutes PQ QR RS
sin y = —QS–
Dynamic Challenge
QR

Test Yourself MIND TEST 5.1c

1. 1 : —1 1. Trigonometric ratios of angle x and angle y are the
5

2. (a) I and III (b) I = 1 : 2 same. This is because all lengths of side are reduced

III = 1 : —1 by the same rate.
2
2. (a) (i) 1—3485– (ii) 22—98 (iii) 71—09
(c) (i) I = 1.5 cm2 (ii) I = 1 : 4
(iv) —1 (v) —7 (vi) —4
III = 24 cm2 III = 1 : —1 2 8 7

4 (b) No
The ratio of area is not proportional to the

scale of the scale drawings.

3. (a) 17.0 cm (b) 203.5 m2 MIND TEST 5.1d

Skills Enhancement 1. (a) sin e = —1359 cos e = 11—23 tan e = 1—356

1. 540 km h–1 (b) sin e = —24 cos e = —7 tan e = —24
25 25 7
2. 50 cm × 50 cm tile. RM633.20 can be saved.

3. (a) 2 829 m2 (b) 4 : 13 (c) sin e = —15 cos e = —8 tan e = —15
17 17 8
(c) 1 971 m2 (d) RM3 960

Self Mastery (d) sin e = —5 cos e = 1—2 tan e = —5
13 13 12
1. (a) 48 m2 (b) 8 : 1 (c) 1 440 m3
2. (a) 8 400 m2 (e) sin e = —15 cos e = —8 tan e = —15
17 17 8
(b) 1 : 500. The most relevant value for scale.
(f) sin e = 0.6 cos e = 0.8 tan e = 0.75
(c) (i) 60 pieces (ii) RM31 500

255

255

2. (a) —1– (b) —1– (c) —¥ 3—9 (d) —4¥–2– CHAPTER 6 Angles and Tangents for Circles
¥3 ¥2 89
MIND TEST 6.1a

3. (a) 3 m (b) 21 m (c) 25 mm 1. (a) 35$ (b) 25$ (c) 30$ (d) 35$
2. (a) 40$ (b) 35$ (c) 70$ (d) 105$
4. (a) 10 cm (b) 15 cm (c) 30 mm 3. (a) 40$ (b) 30$ (c) 10$ (d) 80$
4. (a) 24$ (b) 25$
5. (a) 18 cm (b) 20 cm (c) 9 mm

6. (a) 15 cm (b) 20 cm

7. 51.61 cm

MIND TEST 5.1e MIND TEST 6.1b (c) 3.6 cm (d) 10.4 cm
(c) 40$
1. (a) 2 (b) 3.5 (c) 2.5 (d) 0.5 1. (a) 40$ (b) 30$
(e) –0.5 (f) 3 (g) —5¥—3 (h) —9¥—3 2. (a) 70$ (b) 30$
3. (a) 22$ (b) 114$
(i) 10 (j) 9 2 2

MIND TEST 6.1c

MIND TEST 5.1f 1. (a) 40$ (b) 80$ (c) 50$
2. (a) 50$ (b) 65$ (c) 50$
1. (a) 37$48' (b) 74$36' (c) 58$6' 3. (a) 110$ (b) 55$ (c) 125$
(d) 60$12' (e) 41$30' (f) 16$54' 4. (a) 124$ (b) 34$ (c) 54$
(g) 5$24' (h) 72$18'
(b) 47.7$ (c) 18.2$ MIND TEST 6.1d (c) 10 cm (d) 55$
2. (a) 65.9$ (e) 70.1$ (f) 36.6$ (b) 5 cm (b) ORQ and OQR
(d) 69.4$ (h) 20.3$ 1. (a) 45$
(g) 35.5$ 2. (a) 40$ (b) 10 cm
3. (a) 40$

MIND TEST 5.1g MIND TEST 6.1e
(b) 25$
1. (a) 0.6947 (b) 0.2840 (c) 2.6746 1. (a) 55$ (c) 27.5$ (d) 30$
(d) 0.7815 (e) 0.8630 (f) 1.5051 2. 216$ 3. 90$

MIND TEST 5.1h (c) 24$ MIND TEST 6.1f
(f) 75.3$
1. (a) 12.2$ (b) 54$ (i) 35.8$ 1. (a) 110$ (b) 10.3 cm 2. 176$ 3. 132$
(d) 65.8$ (e) 14.4$ (l) 84.3$
(g) 55.9$ (h) 8.7$ MIND TEST 6.2a
(j) 78.3$ (k) 45.3$ 3. 173.9 m
1. (a) (i) No – vertex P does not lie on the
MIND TEST 5.1i circumference

1. 2.15 m 2. 83.2 m (ii) Yes – DEFG
4. (a) 13 cm (b) 67.4$ (iii) Yes – KNPQ and KLMN

Dynamic Challenge (iv) Yes – ABDE
(b) (i) None
Test Yourself (b) —15 (c) —8
17 17 (ii) D and F, DEF and DGF
1. (a) 28$ 4, (iii) KQP and KNP, NPQ and NKQ,
(c) 22.6$
2. (a) 39 cm KLM and KNM, LMN and LKN
(iv) BAE and BDE, ABD and AED
3. (a) 27 cm
4. (a) 6 (b) —152 MIND TEST 6.2b

(b) 3399$$ 48, 1. (a) 30$ (b) 20$ (c) 120$
(b) 2. 50$ 3. 40$
4. (a) 125$ (b) 117.5$

Skills Enhancement E F ¥ 6 MIND TEST 6.2c
(b) 35 cm (c) 45$
1. D ¥ 1. Exterior angle = a
2. (a) 12 cm Corresponding opposite interior angle = d
3. 8.66 m Exterior angle = e
Corresponding opposite interior angle = b
Self Mastery
2.
1. (a) —7 (b) 15.56 cm (c) 26$ 45,
12 șq

2. (a) 90$ (b) (3b0)$ 63$ 26(c, ) 10.4 m (d) 1 : 2
3. D ¥ 5 cm
p
(c) Not true. The actual ratio is 3 : 5 Į

256

256

MIND TEST 6.2d MIND TEST 7.1b
1. 97$ 2. 38$
3. 79$ 4. 99$ 5. 108$ 1. (a) (i) (ii) (iii)

N M E/D H/I J C

MIND TEST 6.3a VC 4 cm

1. (a) (i) RS and ST – touching the circle at only one K 4 cm L 4 cm F/A G/L K 2 cm B
point. V
(b) (i) (ii) (iii)
(ii) X and Y.
(iii) PQ – passes through 2 points on the circle. C F/E 2 cm G/H 1 cm
(iv) A and B.
(b) (i) BC and BD – touching the circle at only one K/J

point. 5 cm 5 cm 4 cm L/I
(ii) H and E.
(iii) BF – passes through 2 points on the circle. K/N 4 cm L/M A 4 cm B A/D 6 cm B/C
(iv) F and G.
MIND TEST 7.1c
MIND TEST 6.3b D F/E
1. Diagram 1 A E 3 cm
1. 34$ (a) (i) (a) (ii)
2. (a) 120$ (b) 60$
(c) 30$ 3. 114$ 4 cm

MIND TEST 6.3c B/A 2 cm C/D

1. (a) 60$ (b) 30$ (c) 8.66 cm (d) 10 cm B 1 cm F C

2. (a) 40$ (b) 3.575 cm (c) 7.667 cm (b) As viewed from Z:

MIND TEST 6.3d Changed – the lengths of sides AE, ED, BF

and FC.

1. (a) y = z (b) x = b (c) x = y Unchanged – the lengths of sides EF, AB, DC,
f = e
x = a y = a z = a AD, BC and all sizes of angles.

4. 44$ As viewed from X:

2. 27$ 3. 52$ No changes in length of sides and size of

angles.

MIND TEST 6.3e Diagram 2 S/T 1 cm U

1. 50$ 2. x = 26°34', y = 31°43' (a) (i) T/Q 1 cm U R (a) (ii)

3. (a) 130$

(b) (i) 12.87 cm (ii) 8.578 cm (iii) 23.66 cm S 3 cm

4. (a) 4 cm (b) 3.87 cm (c) 11.61 cm

P

MIND TEST 6.4a (b) As viewed from Z: P/Q 2 cm R

1. (a) 8.49 cm (b) 38.21 cm2 Changed – the lengths of sides SP and UR.
2. (a) 35$ (b) 55$
(c) 11.31 cm Unchanged – the lengths of sides ST, TU, PQ,

QR and all sizes of angles.

Dynamic Challenge As viewed from X:

Test Yourself Changed – the lengths of sides SP, SU and PR.

1. x = 40$, y = 150$ 2. 100$ Unchanged – the lengths of sides TQ, QR,
3. x = 30$, y = 60$ 4. 230$
5. x + y = 180$ 6. 86$ TU, UR and all sizes of angles.

MIND TEST 7.2a

Skills Enhancement 1. (a), (b), (c) Side elevation Front elevation

S/R V/W R/W 3 cm S/V

1. 30$ 2. 130$ 3. 114$ 4. 60$

Self Mastery (b) 80$ 2. 64.8 cm2 2 cm
(b) 3.6 cm (c) 30 cm2
1. (a) 61$ (b) 13 cm
3. (a) 36°52'
4. (a) 5 cm Q/P 4 cm U/T P/T 1 cm Q/U

45°
W TU V

CHAPTER 7 Plans and Elevations

MIND TEST 7.1a (d) Yes

1. (a) Yes (b) Yes (c) No RPQ S
2. (a) Correct (b) Wrong Plan 1 cm

257

257

2. (a), (b), (c) Side elevation Front elevation 1 cm (b) (i) 75 cm3 (ii) 1 : 1
F/E G/H N/M O/P S/T V/U 2. (a) (i), (ii), (iii)
V/S/O/N/G/F U/T/P/M/H/E
R/K/J 1 cm J/I K/L R/Q 3 cm Elevation as
Q/L/I G/F viewed from Y
Elevation as
2 cm
F viewed from X G

B/A 3 cm C/D A/D 6 cm B/C

45° 4 cm 4 cm
E/D H/I M/L P T/Q U/C
B/D 2 cm C
C/B/A 3 cm D/E A/E 6 cm D
45°
E

F/A G/J N/K O S/R V/B
Plan

3. (a), (b), (c) Side elevation 3 cm
IJ
J/I

F/A 6 cm G/B 2 cm C

Front elevation 4 cm 4 cm Plan

D/C 4 cm D/E
2 cm 2 cm
E/H C/H A/F/K (b) CD CG DG
A/B Plan 3.6 cm 2 cm 3 cm
C/B F/G 2 cm K/L B/G/L 5 cm Elevation as viewed from X 2 cm 4.5 cm 4 cm
Plan I/H/G L 45° Elevation as viewed from Y 3 cm 4 cm 5 cm

5 cm (c) Original object
CD = 3.61 cm, CG = 4.47 cm, DG = 5 cm.
D/A 4 cm J/E/F 2 cm K
(d) Elevation as viewed from X = BCG, BGC
MIND TEST 7.2b Elevation as viewed from Y = AEF, AFE
Plan = BCD, BDC.

1. Plan 2. Plan Dynamic Challenge

K Test Yourself
J
1. (a) True (b) True (c) False (d) True
G 2. Three vertical cylinders with diameters 1 cm, 2 cm
H 5 cmÎ
E G Î and 3 cm. The height of all cylinders is 4 cm. The
three cylinders are arranged symmetrically from all
I F 2 cm Î directions.
D

CJ

L 4 cm L FE D
HI

AK 8 cm ÎSide Skills Enhancement
3 cm A 4 cm B 6 cm C
B 2 cm 10 cm elevation 1. (a) (i), (ii), (iii)
Front Î
Î Side Elevation as viewed from Y
elevation C/D E/F
elevation

Front elevation Elevation as viewed from X
D/F 3 cm C/E

MIND TEST 7.2c

1. (a) (i), (ii), (iii) Elevation as Elevation as 4 cm 4 cm
viewed from Y 1 cm viewed from X

N/G/F M/H/E F/E G/H N/M

B/A 5 cm H/G A/G 5 cm B/H
E/H
5 cm 2 cm 6 cm 45° F
G
K/J B/C
B/A L/I J/I K/L M/C

5 cm C/D A/D

45° E/D H/I L 5 cm

5 cm

A 2 cm D 3 cm C/B

2 cm 2 cm Plan

F/A G/J K N/B (b) AD = 4.5 cm, ADC = 116°

1 cm Plan

258

258

2. (a) (i), (ii), (iii) CHAPTER 8 Loci in Two Dimensions

Elevation as viewed from Y Elevation as viewed from X MIND TEST 8.1a
G 2 cm H G/H

1. (a) A straight line parallel to the inclined plane

3 cm 3 cm (b) A curve
4 cm E/F J
(c) A vertical straight line
6 cm D/F
C/D C/E/J (d) A straight line parallel to slide
2 cm K/I A/I 2 cm
45° H/I B/K 2. (a) A vertical straight line (c) A curve
B/A
J/K (b) A horizontal straight line
2 cm
4 cm E MIND TEST 8.1b
1. (a) T (b) T
(c) T (d) T

G/F

4 cm
S S SS

MIND TEST 8.2a

D/A 4 cm C/B 1. (a) Locus of X is a circle centred at P with radius

Plan 3 cm.

(b) 60 cm3 (b) Locus of Y is a circle centred at Q with radius
3. 462.5 cm3
4. 477.75 cm3 5. 96 cm3 4 cm.

2. (a) HF (b) AC (c) EG

Self Mastery (d) BD (e) AD and BC

1. (a) (i), (ii), (iii) 3. (a)

Elevation as viewed from Y Elevation as viewed from X 1.5 cm
CƔ ƔD locus of T
G/F 5 cm H/E F/E 6 cm G/H
1 cm I/L 4 cm K/L 3 cm 1 cm
J/K J/I 1.5 cm

3 cm (b) The locus of point T is a pair of straight lines

of 6 cm parallel to the straight line CD with a

perpendicular distance of 1.5 cm.

B/A C/D A/D 3 cm B/C 4. (a) (b) (c) locus of Y

45° E/D L/C H/I Q CƔ PR
PƔ

R ƔD

locus of Y locus of Y Q

5 cm 5.

lane 7

F/A 3 cm K/B 3 cm G/J lane 4
Plan
(b) 45 cm3 lane 1
(c) RM264
2. (a) E F (b) 44—2 cm3 MIND TEST 8.2b ƔB
7 1. A Ɣ
locus of W

HG
5 cm

locus of V

D C
2 cm 3 cm

A 4 cm B DƔ ƔC

2. Point G 259

259

3. locus of G 2.
locus of X M
y
RN
4
locus of Y
3
QP
2 locus of F

1

0 1234 x 3. R
locus of Y
locus of X

MIND TEST 8.2c QP
1. (a), (b) P
Q Skills Enhancement Q R
MO
locus of Y 1. (a) ABS locus of Y
(b)
locus of X
P
N

locus of Y

S 6 cm R

(c) The intersection between the locus of X and A BS
locus of Y is arc OP.

2. P 6 m Q locus of Z

7m DT
8m C

2. M N
locus of X
locus of Y

S R PO
(b)
3. Self Mastery
(a)

1. (a) N (b) L (c) I (d) II (e) VI

Jalan Bahagia 2. (a) I (b) IV (c) III

ƔP 3. (a) locus of X – moving constantly 1 cm from O.

locus of Y – equidistant from P and R.

(b) locus of X – moving constantly 1 cm from O.

locus of Y – equidistant from Q and S.

CHAPTER 9 Straight Lines

Dynamic Challenge MIND TEST 9.1a

Test Yourself A S 1. (a) gradient = 3 (b) gradient = 2
ƔB y-intercept = 5 y-intercept = –7
1.
locus of X (c) gradient = –1 (d) gradient = 4
y-intercept = 4 y-intercept = 3
locus of Y
(e) gradient = – —1 (f) gradient = —1
C 3 2

y-intercept = 6 y-intercept = – —45
(b) h = 4, k = –3
2. (a) h = –2, k = 4

260

260

MIND TEST 9.1b xy MIND TEST 9.1g (b) y = –2x – 2
xy (b) — + — = 1 (d) y = – —3 x – 9
1. (a) y = 3x + 1
1. (a) — – — = 1 4 14 (c) y = – —3 x + 9 2
86 y = – —7 x + 14 2 (b) y = —1 x – 1—0
y = —3 x – 6
4 2 2. (a) y = —1 x + —10 33
(d) – 2—x + —y = 1 33
(c) —x – —y = 1
35 93 (c) y = —1 x
y = —5 x – 5 y = —2 x + 3 3
3
3 MIND TEST 9.1h
2. (a) 3x + 4y = 12
y = – —3 x + 3 (b) –6x + 3y = 18 1. (a) (3, 4) (b) (5, 4) (c) (3, 2) (d) (2, –1)
4 2. (a) (–2, 3) (b) (0, 4) (c) (–5, 0) (d) (4, 1)
y = 2x + 6
(c) 9x + y = 6 MIND TEST 9.1i
(d) 8x – 3y = 12
y = –9x + 6 y = —8 x – 4 1. (a) —1 (b) –2 (c) y = —1 x – 2
3 2 2
3. (a) –2x + y = 6 3x – y = 12
xy xy 2. (a) (5, 0) (b) y = –x – 5
(c) x-intercept = –5
–—+—=1 (b) — – — = 1
36 4 12 Dynamic Challenge (b) y-intercept = 6

(c) x + y = 5 (d) 2x + y = – 4 Test Yourself (b) y = –8
xy xy 4. y = 4x + 2
—+—=1 1. (a) x-intercept = 15 6. (5, –8)
55 –—–—=1 (c) gradient = – —25
24
2. (a) x = –6
MIND TEST 9.1c (d) No 3. y = 3x + 18
(d) No 5. y = –3x – 4
1. (a) No (b) Yes (c) Yes (d) No
2. (a) Yes (b) No (c) Yes Skills Enhancement
3. (a) Yes (b) No (c) Yes
4. (a) h = 2 (b) k = –2 (c) n = 3 1. (4, 3)
2. (a) x = 3
MIND TEST 9.1d (b) Not parallel
(d) Not parallel (b) y = –2x + 16, x-intercept = 8
1. (a) Parallel (b) k = 6 3. (a) y = –2x
(c) Parallel (d) —3
(b) y = –2x + 35
2. (a) k = 3 2 (c) x-intercept = 3—25
(c) k = 1—5 4. (a) y = – —1 x + 8
8
2
3. h = – —5 , k = 3 (b) y = – —1 x + 2—9 or x + 2y = 29
2
22

(c) No, because the two straight lines are parallel.

MIND TEST 9.1e (b) y = –3x – 14 Self Mastery (b) (– 4, 4)
(d) y = – —1 x – 4 (d) 4x + 3y = 20
1. (a) y = 2x + 1 1. (a) 6 km (b) y = —x + —29
(c) y = —2 x – 3 2 (c) 5 km
3 55
2. (a) 10 km (d) 34.11 minutes
MIND TEST 9.1f (b) y = 4x + 8
1. (a) y = – —1 x + 2 (c) 20 km (b) (–5, 7)
(d) y = —23 x + —23 3. y = —3 x + —27, 13.5 cm
3 (f) y = – x – 2
(c) y = 4x – 13 16 2
(e) y = x + 4 4. (a) k = –3

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261

? Glossary

Accuracy Gradient Side elevation
The degree of approximation of the
measured value to the actual value. The ratio of the vertical distance to the The orthogonal projection of an object to

Arc horizontal distance. a vertical plane as viewed from the side.
A curved line connecting any two points
on the circumference of a circle. Grid 6LJQL¿FDQW ¿JXUH
A set of straight lines that cross each 7KH GLJLWV LQ D QXPEHU VSHFL¿HG H[DFWO\
Bisector other and are usually in the form of to a degree of accuracy required.
A line that divides an angle into two squares or equilateral triangles.
angles of the same size. Simultaneous equation
Hypotenuse Two equations with two variables that
Centre of a circle The opposite side of the right angle in a can be solved simultaneously.
The given point from which all points right-angled triangle.
on the circumference are of the same Sinus
distance. Index The ratio of the length of the opposite
A number that states the power. side to the hypotenuse of an angle in a
Chord Generally, an with n is the index for a. right-angled triangle. Its abbreviation
A line segment connecting any two is sin.
points on a curve. Linear function
A function of the form , where a and b Standard form
Circumference DUH FRQVWDQWV DQG D  7KH JUDSK RI The standard method to write a real
A closed curve which is the boundary of linear function is of straight line form. QXPEHU 7KURXJK VFLHQWL¿F QRWDWLRQ
a circular shape. all real numbers are written in the form
Locus A = n, with 1 р A Ͻ DQG n is an
Cyclic quadrilateral The path formed by a set of points in a integer.
A quadrilateral inscribed in a plane or three-dimensional space that
circle where all vertices lie on the meets one or more conditions. Substitution method
circumference of the circle. A method for solving simultaneous
Normal to a plane equations by substituting one of the
Common tangent A line that is perpendicular or at right variables.
A straight line that touches two circles, angle to the corresponding plane.
each at one point only. Subtend
Origin To include an angle at the circumference
Corresponding angles The point of intersection of the or centre of the circle, opposite a
A pair of angles formed when a line cuts horizontal axis and vertical axis. The particular arc.
two parallel lines. These two angles are FRRUGLQDWHV RI RULJLQ DUH
equal. Tangent
Orthogonal projection The ratio of the length of the opposite
Cosine The image formed on a plane as a side to the adjacent side of an angle in
The ratio of the lengths of the adjacent result of the line projection of an object a right-angled triangle. Its abbreviation
side to the hypotenuse of an angle in a perpendicular to the plane. is tan.
right-angled triangle. Its abbreviation
is cos. Parallel line Tangent to circle

Degree Straight lines that are on the same plane The straight line touches the circle at
A unit of measurement for angles. The
degree symbol is shown as °. and do not intersect each other. The only one point without cutting it.

Diameter perpendicular distance between the Three-dimensional shape
A straight line connecting two points parallel lines are always the same. Shapes that have length, width, height
on the circumference of the circle and
passes through the centre of the circle. Perpendicular bisector and volume.

Elevation A line perpendicular to a line segment
A vertical sketch of an object viewed and divides the segment into two equal Trigonometric ratio
from a certain side. A ratio describes the relationship between
parts.
Elimination method the sides in a right-angled triangle.
A method for solving simultaneous Plan
equations by eliminating one of the
variables. A sketch of an object on a horizontal Trigonometry

Factor plane and is viewed from above. A mathematical branch related to the
Numbers, terms or algebraic expressions
that divides exactly the given number, Plane relationship between the sides and angles
term or algebraic expression. $ ÀDW VXUIDFH LQ DOO GLUHFWLRQV DQG LV of a triangle and their applications.

Front elevation two-dimensional. Two-dimensional shape
The orthogonal projection of an object
to a vertical plane as viewed from the Proportion Shapes that are of two measurements,
front and is a uniform cross section of A mathematical statement that indicates namely length and width.
the object.
the relationship between two quantities Uniform cross section
or values in the same ratio.
The cross section resulting from cutting

Scale a solid, which is of same size and shape
The ratio of the size of the drawing to as the base.

the size of the actual object. x-intercept

Scale drawing The point where a straight line or curve
A drawing that represents the actual cuts the x-axis.

object according to a certain scale. Scale y-intercept
drawing will be larger or smaller or of The point where a straight line or curve
equal size as the actual object. cuts the y-axis.

262

262

References

Channon, J. B., McLeish, A. Smith et. al. 1972. Malaysian General Mathematics Book
Four. Longman Malaysia Sdn. Bhd.

&KDSLQ 6X]DQQH + HW DO Middle Grades Maths Tools for Success Course 2.
Prentice-Hall, Inc.

&KDSLQ 6 + ,OOLQJZRUWK 0 /DQGDX 0 Middle Grades Maths Tools for Success
Course 2. New Jersey: Prentice Hall.

Curriculum Development Centre Ministry of Education Kuala Lumpur, 1973. Modern
Mathematics for Malaysia Form Four. Eastern Universities Press Sdn. Bhd.

Eliezer, E.J. and Idaikkadar, N.M., 1996. 0DWKHPDWLFV IRU 6FKRRO &HUWL¿FDWH 6WXGHQWV LQ
Malaysia. Dewan Bahasa dan Pustaka.

Istilah Matematik untuk Sekolah-sekolah Malaysia .XDOD /XPSXU 'HZDQ %DKDVD GDQ
Pustaka.

Kamus Dewan Edisi Keempat .XDOD /XPSXU 'HZDQ %DKDVD GDQ 3XVWDND
/LP 6 + +DVKLP 6 .RR 6 + &KRQJ * & Matematik Tingkatan 5. Darul Fikir.
5\DQ 0 Geometry for Dummies. Wiley Publishing Inc.
6SHVL¿NDVL .XULNXOXP 0DWHPDWLN 7LQJNDWDQ .XULNXOXP %HUVHSDGX 6HNRODK 0HQHQJDK ,

3XWUDMD\D %DKDJLDQ 3HPEDQJXQDQ .XULNXOXP .HPHQWHULDQ 3HODMDU 0DOD\VLD
7D\ & + 5LGGLQJWRQ 0 *ULHU 0 New Mathematics Counts Secondary

1RUPDO $FDGHPLF QG (GLWLRQ. Singapore: Marshall Cavendish Education.
7HK . 6 &RRL & . New Syllabus Mathematics. Singapore: Shinglee Publisher

Pte Ltd.

263

263

Index

Accuracy 32 Grid 88 Point of intersection 242
Acute angle 108 Horizontal distance 236 Repeated multiplication 2, 6
Adjacent side 108 Horizontal plane 170 Round off 35
Alternate segment 155 Hypotenuse 108 Scale drawing 88
Approximation 32 Inclined plane 170 Simultaneous equation 242
Axis 228 Image 171 Single number 37
Base 2 Index 2 Solid line 182
Centre of a circle 130 Index notation 2 Standard form 37
Chord 130 Interior angle 147 Straight line 226
Circumference 130 Linear function 226 Subtend 130
&RHI¿FLHQW 229 Locus 200 Substitution method 243
Common tangent 157 Major arc 130 Surface 170
Constant 111 Measurement 88 6LJQL¿FDQW ¿JXUH 33
Correspond 147 Metric system 39 Two-dimensional 201
Cosine 111 Minor arc 130 Scale 89
Cyclic quadrilateral 144 Minute 120 Second 120
Dashed line 183 Normal to a plane 170 Sine 111
Degree 120 Object 88, 171 Tangent 111
Degree of approximation 32 Opposite side 108 Tangent of circle 150
Diameter 130 Origin 235, 242, 246 Three-dimensional 203
Elevation 182 Orthogonal projection 171 Trigonometric ratio 111
Elimination method 243 Orthographic projection 183 Trigonometry 108
Estimation 32 Parallel 228, 236, 237 Uniform cross section 172
Exponent 2 Place value 34, 37 Vertical distance 236
Exterior angle 147 Plan 182 Vertical plane 170
Factor 6 Perpendicular bisector 211 Viewing direction 180
Fixed point 204 Proportional 136, 142 x-axis 226, 231
Gradient 226 Point of tangency 150 y-axis 226, 231

264

264

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