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CHAPTER Loci in Two Dimensions Mathematics Form 3 Chapter 8 Loci in Two Dimensions (c) Q

(c)
8
P
Lokus dalam Dua Dimensi


S Q


1. (a) The path traced by a moving wheel (c)
of a bicycle on a horizontal plane is a
straight line. R
(b) The path traced by a moving swing is
an arc of a circle.
P
(c) The locus of a spinning ball is a sphere. 4. (a)
P
(d) The locus of a rolling rod is a cylinder. O
5. (a)
2. (a) Q


S



3. (a) Q R
O
Q P



R
(b)
(b)
P S
(b) R
(b)
Q
P


Q P
P R
S
O


Q
S













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Mathematics Form 3 Chapter 8 Loci in Two Dimensions Mathematics Form 3 Chapter 8 Loci in Two Dimensions
(c) (c) S B R (c) PT3 Standard Practice 8
A
S
Section A
R

B E 1. Answer: C
Q
P Q 2. Locus of X is a circle of radius 6 cm with
P
Q centre N. From the rectangle KLMN, locus
X is arc KQR.
P R
Answer: D
C D
3. Answer: D
C D A
4. Answer: C

6. (a) D P C 5. Locus X is arc STU and locus Y is line OB.
Hence, the point of intersection is at A.
7. (a) Answer: A
A
Q Q
R 6. Locus X is line BD and locus Y is arc ADE at
centre B. Hence the point of intersection is
R S at D.

Answer: D

Section B

B D C
A Q B P 1. (a) An ant moves on a
straight horizontal
wire.
(b) Seekor semut bergerak
Q (b) pada wayar mendatar
A B yang lurus.
(b) The tip of a fan blade
that is spinning.
A B Hujung bilah kipas yang
sedang berpusing.
Z
(c) The tip of hour hand
in a clock from 4
P R Y X o’clock to 8 o’clock.
Hujung jarum jam dari
pukul 4 ke pukul 8.
D E C (d) A lift moving in tall
building.
D C Sebuah lif bergerak
dalam bangunan tinggi.




S



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Mathematics Form 3 Chapter 8 Loci in Two Dimensions Mathematics Form 3 Chapter 8 Loci in Two Dimensions
(c) (c) S B R (c) PT3 Standard Practice 8
A
S
Section A
R

B E 1. Answer: C
Q
P Q 2. Locus of X is a circle of radius 6 cm with
P
Q centre N. From the rectangle KLMN, locus
X is arc KQR.
P R
Answer: D
C D
3. Answer: D
C D A
4. Answer: C

6. (a) D P C 5. Locus X is arc STU and locus Y is line OB.
Hence, the point of intersection is at A.
7. (a) Answer: A
A
Q Q
R 6. Locus X is line BD and locus Y is arc ADE at
centre B. Hence the point of intersection is
R S at D.

Answer: D

Section B

B D C
A Q B P 1. (a) An ant moves on a
straight horizontal
wire.
(b) Seekor semut bergerak
Q (b) pada wayar mendatar
A B yang lurus.
(b) The tip of a fan blade
that is spinning.
A B Hujung bilah kipas yang
sedang berpusing.
Z
(c) The tip of hour hand
in a clock from 4
P R Y X o’clock to 8 o’clock.
Hujung jarum jam dari
pukul 4 ke pukul 8.
D E C (d) A lift moving in tall
building.
D C Sebuah lif bergerak
dalam bangunan tinggi.




S



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Mathematics Form 3 Chapter 8 Loci in Two Dimensions Mathematics Form 3 Chapter 8 Loci in Two Dimensions
2. (b)
A moving point that its P Q HOTS Challenge
distance from the line KL and
NM is always constant. Line
Suatu titik bergerak yang jaraknya POR
sentiasa sama dari garis KL dan
NM. U R
Locus L Pool
A moving point that its Kolam
distance is the same as LM Arc
from point L. KVM T S
Suatu titik bergerak yang jaraknya Locus M
adalah sama dengan LM dari titik L. 13 m

A moving point that its
distance from the line KN and (c) Tree
NM is always constant. Line Pokok
Suatu garis bergerak yang jaraknya LON 7.6 m
sentiasa sama dari garis KN dan
NM.

A moving point that is
equidistant from the points P Line 5 cm
and R. QOS S T X
Suatu titik bergerak yang berjarak
sama dari titik P dan R.


Stool
Section C Bangku

3. (a) (i) The locus of the screw is a parallel
line to the horizontal line. Y

(ii) The locus of the apple is a vertical
line going down to the ground.
The possible locations of the treasure are marked X or Y.
(iii) The locus of the cow is a circle and
the pole as its centre.

(iv) The locus of the ship is a
perpendicular line between Johor
and Singapore.




















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Mathematics Form 3 Chapter 8 Loci in Two Dimensions Mathematics Form 3 Chapter 8 Loci in Two Dimensions
2. (b)
A moving point that its P Q HOTS Challenge
distance from the line KL and
NM is always constant. Line
Suatu titik bergerak yang jaraknya POR
sentiasa sama dari garis KL dan
NM. U R
Locus L Pool
A moving point that its Kolam
distance is the same as LM Arc
from point L. KVM T S
Suatu titik bergerak yang jaraknya Locus M
adalah sama dengan LM dari titik L. 13 m

A moving point that its
distance from the line KN and (c) Tree
NM is always constant. Line Pokok
Suatu garis bergerak yang jaraknya LON 7.6 m
sentiasa sama dari garis KN dan
NM.

A moving point that is
equidistant from the points P Line 5 cm
and R. QOS S T X
Suatu titik bergerak yang berjarak
sama dari titik P dan R.


Stool
Section C Bangku

3. (a) (i) The locus of the screw is a parallel
line to the horizontal line. Y

(ii) The locus of the apple is a vertical
line going down to the ground.
The possible locations of the treasure are marked X or Y.
(iii) The locus of the cow is a circle and
the pole as its centre.

(iv) The locus of the ship is a
perpendicular line between Johor
and Singapore.




















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CHAPTER Straight Lines Mathematics Form 3 Chapter 9 Straight Lines 9. (a) LHS: 3(2) = 6

y
x
9
(b) = – + 1

5
10
RHS: 2(0) + 6 = 6

Conclusion: Yes
5x
Garis Lurus
y = – + 5
10
1 (b) LHS: 4(–2) = –8
y = – x + 5
2 RHS: 3(–1) + 1 = –2
Conclusion: No
1. (a) m = undefined –7 –7 y x
(c) Gradient m = = (c) – = – + 1
(b) m < 0 –7 11 – 5 6 4 3 (c) LHS: 0
(c) m = 0 y = 6 x + c, y = 4x – 4 RHS: 5(–3) – 1 = –16
3
substitute (5, 0) Conclusion: No
y
2. (a) y = –5x + 2 0 = –7 (5) + c (d) – = 4x + 1
9
6
(b) y = 5 35 9 y = –4x – 9 10. (a) L 1 : 3y = –6x + 7
7
(c) y = 3 c = 6 y = –2x + , gradient m 1 = –2
1 3
7
(d) y = x y = – x + 35
1
2 6 6 7. (a) y = x – 9 L 2 : –4y = 8x – 5
3 2 6 5
(e) y = x + y = –2x + , gradient m 2 = –2
4 5 6y = x – 54 4
5. (a) 3y = –5x + 2
–5 2 x – 6y – 54 = 0 m 1 = m 2 , therefore L 1 is parallel to L 2 .
3. (a) m = –5, c = –6 y = 3 x + 3
1
(b) m = , c = 7 x y 1
2 (b) 6y = 2x +3 (b) 2 + = 1 (b) L 1 : 2y + 1 = – x + 6
8
(c) m = 0, c = 8 2 3 x y 2

y = x +  + = 1 × 8 1 5 1
6 6 2 8 y = – x + , gradient m 1 = –
4
4
2
3 – 5 1 1 1 4x + y = 8
4. (a) Gradient m = –4 – 0 = 2 y = x + 2 L 2 : 3y = 8x – 3
3
1 4x + y – 8 = 0
y = x + c, 8 8
2 y = x – 1, gradient m 2 =
(c) 4y = –x – 1 3 3
substitute (0, 5) x 1
y = – – 8. (a) 5y = 2x + 10
5 = 0 + c 4 4 m 1 ≠ m 2 , therefore L 1 is not parallel to
–2x + 5y = 10
c = 5 2x 5y 10 L 2 .
1 (d) –8y = –4x +12 – + =
y = x + 5 10 10 10
2 –4 12 x
y
y = x – – + = 1 (c) L 1 : 3y + 2x = –6
–8 8 5 2
2
2
–2 – 1 1 3 y = – x – 2, gradient m 1 = –
(b) Gradient m = y = x –
10 – 9 2 2 (b) 6y – 2x – 8 = 0 3 3
–3 (–2x + 6y = 8) ÷ 8 L 2 : 6 – 5y = 8x – 1
=
1 y x –x 3y 8 7 8
= –3 6. (a) = – + 1 4 + 4 = 1 y = – x + , gradient m 2 = – 5
8
4
5
5
8
y = –3x + c, y = – x + 8 1
substitute (9, 1) 4 (c) y = x – 6 m 1 ≠ m 2 , therefore L 1 is not parallel to
2
y = –2x + 8

1 = –3(9) + c  1 x – y = 6 ÷ 6 L 2 .
c = 28 2
1
y
y = –3x + 28 x – = 1
12 6
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CHAPTER Straight Lines Mathematics Form 3 Chapter 9 Straight Lines 9. (a) LHS: 3(2) = 6

y
x
9
(b) = – + 1

5
10
RHS: 2(0) + 6 = 6

Conclusion: Yes
5x
Garis Lurus
y = – + 5
10
1 (b) LHS: 4(–2) = –8
y = – x + 5
2 RHS: 3(–1) + 1 = –2
Conclusion: No
1. (a) m = undefined –7 –7 y x
(c) Gradient m = = (c) – = – + 1
(b) m < 0 –7 11 – 5 6 4 3 (c) LHS: 0
(c) m = 0 y = 6 x + c, y = 4x – 4 RHS: 5(–3) – 1 = –16
3
substitute (5, 0) Conclusion: No
y
2. (a) y = –5x + 2 0 = –7 (5) + c (d) – = 4x + 1
9
6
(b) y = 5 35 9 y = –4x – 9 10. (a) L 1 : 3y = –6x + 7
7
(c) y = 3 c = 6 y = –2x + , gradient m 1 = –2
1 3
7
(d) y = x y = – x + 35
1
2 6 6 7. (a) y = x – 9 L 2 : –4y = 8x – 5
3 2 6 5
(e) y = x + y = –2x + , gradient m 2 = –2
4 5 6y = x – 54 4
5. (a) 3y = –5x + 2
–5 2 x – 6y – 54 = 0 m 1 = m 2 , therefore L 1 is parallel to L 2 .
3. (a) m = –5, c = –6 y = 3 x + 3
1
(b) m = , c = 7 x y 1
2 (b) 6y = 2x +3 (b) 2 + = 1 (b) L 1 : 2y + 1 = – x + 6
8
(c) m = 0, c = 8 2 3 x y 2

y = x +  + = 1 × 8 1 5 1
6 6 2 8 y = – x + , gradient m 1 = –
4
4
2
3 – 5 1 1 1 4x + y = 8
4. (a) Gradient m = –4 – 0 = 2 y = x + 2 L 2 : 3y = 8x – 3
3
1 4x + y – 8 = 0
y = x + c, 8 8
2 y = x – 1, gradient m 2 =
(c) 4y = –x – 1 3 3
substitute (0, 5) x 1
y = – – 8. (a) 5y = 2x + 10
5 = 0 + c 4 4 m 1 ≠ m 2 , therefore L 1 is not parallel to
–2x + 5y = 10
c = 5 2x 5y 10 L 2 .
1 (d) –8y = –4x +12 – + =
y = x + 5 10 10 10
2 –4 12 x
y
y = x – – + = 1 (c) L 1 : 3y + 2x = –6
–8 8 5 2
2
2
–2 – 1 1 3 y = – x – 2, gradient m 1 = –
(b) Gradient m = y = x –
10 – 9 2 2 (b) 6y – 2x – 8 = 0 3 3
–3 (–2x + 6y = 8) ÷ 8 L 2 : 6 – 5y = 8x – 1
=
1 y x –x 3y 8 7 8
= –3 6. (a) = – + 1 4 + 4 = 1 y = – x + , gradient m 2 = – 5
8
4
5
5
8
y = –3x + c, y = – x + 8 1
substitute (9, 1) 4 (c) y = x – 6 m 1 ≠ m 2 , therefore L 1 is not parallel to
2
y = –2x + 8

1 = –3(9) + c  1 x – y = 6 ÷ 6 L 2 .
c = 28 2
1
y
y = –3x + 28 x – = 1
12 6
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Mathematics Form 3 Chapter 9 Straight Lines Mathematics Form 3 Chapter 9 Straight Lines
11. (a) 2y = 6 – 3x 12. (a) y = 2x – 4 (b) y = –x – 2 From the graph,
–3
y = x +3 x y x y Point of intersection = (2, 0)
2 y
3 0 –4 0 –2
m = –
2 2 0 –2 0
new equation y = –x + 2 4
3 y = 6 – 2x y = 2x – 2
y = – x + c
2 x y x y 2
1
2
pada titik (5, 0) 0 6 0 –2 y = –x – 1
3
0 = – (5) + c 3 0
2 1 0 –4 –2 0 2 4 x
15
c = From the graph,
2 From the graph, –2
Point of intersection
3 15
Maka, y = – x + Point of intersection
2 2 = (2.5, 1)
= (0, –2)
y = 6 – 2x
(b) 6x – 3y = 8 y y
6 8 13. (a) y = 2x – 1 ................(1)
y = x – 6 y = 3 – x ..................(2)
3 3 4 y = 2x – 2
8
= 2x – (1) – (2)
3 4 y = 2x – 4 3x – 4 = 0
m = 2 y = –x – 2 2 1
x = 1
new equation 2 3
1
y = 2x + c –4 –2 0 2 4 x Substitute x = 1 into (2)
(2.5, 1)
3
at the point (8, –1) x 1 2
–4 –2 0 2 4 –2 y = 3 – 1 = 1
–1 = 2(8) + c 3 3
c = –17 Therefore the point of intersection is
–2

1
Therefore, y = 2x – 17 = 1 , 1 2 
3
3
1
–4 (c) y = x – 1
(c) 2y = 3x + 10 2 (b) 2y = 3x – 2 ................(1)
3
y = x + 5 x y y = 2x + 1 ................(2)
2
0 –1
3 (2) × 2
m =
2 2 0 2y = 4x + 2 ................(3)
new equation y = –x + 2 (1) – (3)
3 0 = –x – 4
y = x + c
2 x y x = –4
at the point(–2, –1) 0 2 Substitute x = –4 into (2)
3 2 0
–1 = (–2) + c y = 2(–4) + 1
2 y = –7
c = 2
Therefore the point of intersection is
3
Therefore, y = x + 2 = (–4, –7)
2



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Mathematics Form 3 Chapter 9 Straight Lines Mathematics Form 3 Chapter 9 Straight Lines
11. (a) 2y = 6 – 3x 12. (a) y = 2x – 4 (b) y = –x – 2 From the graph,
–3
y = x +3 x y x y Point of intersection = (2, 0)
2 y
3 0 –4 0 –2
m = –
2 2 0 –2 0
new equation y = –x + 2 4
3 y = 6 – 2x y = 2x – 2
y = – x + c
2 x y x y 2
1
2
pada titik (5, 0) 0 6 0 –2 y = –x – 1
3
0 = – (5) + c 3 0
2 1 0 –4 –2 0 2 4 x
15
c = From the graph,
2 From the graph, –2
Point of intersection
3 15
Maka, y = – x + Point of intersection
2 2 = (2.5, 1)
= (0, –2)
y = 6 – 2x
(b) 6x – 3y = 8 y y
6 8 13. (a) y = 2x – 1 ................(1)
y = x – 6 y = 3 – x ..................(2)
3 3 4 y = 2x – 2
8
= 2x – (1) – (2)
3 4 y = 2x – 4 3x – 4 = 0
m = 2 y = –x – 2 2 1
x = 1
new equation 2 3
1
y = 2x + c –4 –2 0 2 4 x Substitute x = 1 into (2)
(2.5, 1)
3
at the point (8, –1) x 1 2
–4 –2 0 2 4 –2 y = 3 – 1 = 1
–1 = 2(8) + c 3 3
c = –17 Therefore the point of intersection is
–2

1
Therefore, y = 2x – 17 = 1 , 1 2 
3
3
1
–4 (c) y = x – 1
(c) 2y = 3x + 10 2 (b) 2y = 3x – 2 ................(1)
3
y = x + 5 x y y = 2x + 1 ................(2)
2
0 –1
3 (2) × 2
m =
2 2 0 2y = 4x + 2 ................(3)
new equation y = –x + 2 (1) – (3)
3 0 = –x – 4
y = x + c
2 x y x = –4
at the point(–2, –1) 0 2 Substitute x = –4 into (2)
3 2 0
–1 = (–2) + c y = 2(–4) + 1
2 y = –7
c = 2
Therefore the point of intersection is
3
Therefore, y = x + 2 = (–4, –7)
2



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Mathematics Form 3 Chapter 9 Straight Lines Mathematics Form 3 Chapter 9 Straight Lines
1 7 2 – (–2) 4 Substitute x = 0, y = 4
(c) y = x – 3 ..................(1) substitute x = into (3) (b) Gradient CD = =
2 2 4 – (–1) 5 2
4y = 6 – x ...................(2) 7 y – 0 4 4 = (0) + c
y = 8   – 26 Gradient AB = = 3
(1) × 4 2 3 – (–3) 5 c = 4
y = 2 y 4 2
4y = 2x – 12 ...............(3) = y = x + 4
6 5 3
2
4
(2) – (3) Therefore, the point of intersection is y = 4 4 m QR = – = –
0 = –3x + 18 =  7 , 2  4 5 6 3
x = 6 2 y = 5 x + c Equation of QR is :
2
Substitute x = 6 into (1) 1 16 At the point (–3, 0 ) y = – x + c
3
1 (c) y = x – ....................(1) substitute x = –3 and y = 0 2
y = (6) – 3 = 0 5 5 at the point (0, 4), c = 4, y = – x + 4
2 4 3
–3x + 7y = –8 ..................(2) 0 = (–3) + c
5
Therefore, the point of intersection is 12 (e) R(–3, 3)
Substitute (1) into (2) c =
= (6 , 0) 5 m RP = 0
–3x + 7  1 x – 16  = –8 Therefore equation of AB is Equation of RP is y = 3
14. (a) 7x – 4y = –7 ..........(1) 5 5
5x + y = 22 ............(2) 7 112 y = 4 x + 12 (f) Q(4, 0)
–3x + x – = –8 5 5
From (2) 5 5 and the equation of AB in general form m PQ = 8 = –2
y = 22 – 5x ............(3) 8 72 is 0 – 4
– x = 5 m RS = m PQ = –2
5
Substitute (3) into (1)  y = 4 x + 12  × 5
7x – 4(22 – 5x) = –7 x = –9 5 5 Equaton of RS:
7x – 88 + 20x = –7 substitute x = –9 into (2) 5y = 4x + 12 y = –2x + c
27x = 81 4x – 5y + 12 = 0 Substitute x = –3, y = 8
x = 3 –3(–9) + 7y = –8
y = –5 6 – 1 8 = –2(–3) + c
Substitute x = 3 into (3) (c) Gradient m PQ = = 5 c = 8 – 6 = 2
y = 22 – 5(3) Therefore the point of intersection is m OR = m PQ = 5 7 – 6
y = 7 = (–9, –5) y = –2x + 2
Equation of OR : y = 5x + c
Therefore, the point of intersection is
= (3, 7) 15. (a) y = –2x + 5, Substitute x = 0, y = 0
c = 0
Gradient PQ = –2
(b) 4x – 7y = 0 .......................(1) y = 5x
Equation PQ, y = –2x + c
8x – y – 26 = 0 .................(2)
Substitute x = 2 and y = –1,
From (2) –1 = –2(2) + c (d) m PR = undefined
y = 8x – 26 ........................(3) c = 3 Equation of PR is x = 6
substitute (3) into (1) Therefore the equation is y = –2x + 3 Q (0,4)
4x – 7(8x – 26) = 0 m PQ = 8 – 4 = 2
4x – 56x +182 = 0 6 3 2
52x = 182 Equation of PQ is y = x + c
3
7
x =
2










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Mathematics Form 3 Chapter 9 Straight Lines Mathematics Form 3 Chapter 9 Straight Lines
1 7 2 – (–2) 4 Substitute x = 0, y = 4
(c) y = x – 3 ..................(1) substitute x = into (3) (b) Gradient CD = =
2 2 4 – (–1) 5 2
4y = 6 – x ...................(2) 7 y – 0 4 4 = (0) + c
y = 8   – 26 Gradient AB = = 3
(1) × 4 2 3 – (–3) 5 c = 4
y = 2 y 4 2
4y = 2x – 12 ...............(3) = y = x + 4
6 5 3
2
4
(2) – (3) Therefore, the point of intersection is y = 4 4 m QR = – = –
0 = –3x + 18 =  7 , 2  4 5 6 3
x = 6 2 y = 5 x + c Equation of QR is :
2
Substitute x = 6 into (1) 1 16 At the point (–3, 0 ) y = – x + c
3
1 (c) y = x – ....................(1) substitute x = –3 and y = 0 2
y = (6) – 3 = 0 5 5 at the point (0, 4), c = 4, y = – x + 4
2 4 3
–3x + 7y = –8 ..................(2) 0 = (–3) + c
5
Therefore, the point of intersection is 12 (e) R(–3, 3)
Substitute (1) into (2) c =
= (6 , 0) 5 m RP = 0
–3x + 7  1 x – 16  = –8 Therefore equation of AB is Equation of RP is y = 3
14. (a) 7x – 4y = –7 ..........(1) 5 5
5x + y = 22 ............(2) 7 112 y = 4 x + 12 (f) Q(4, 0)
–3x + x – = –8 5 5
From (2) 5 5 and the equation of AB in general form m PQ = 8 = –2
y = 22 – 5x ............(3) 8 72 is 0 – 4
– x = 5 m RS = m PQ = –2
5
Substitute (3) into (1)  y = 4 x + 12  × 5
7x – 4(22 – 5x) = –7 x = –9 5 5 Equaton of RS:
7x – 88 + 20x = –7 substitute x = –9 into (2) 5y = 4x + 12 y = –2x + c
27x = 81 4x – 5y + 12 = 0 Substitute x = –3, y = 8
x = 3 –3(–9) + 7y = –8
y = –5 6 – 1 8 = –2(–3) + c
Substitute x = 3 into (3) (c) Gradient m PQ = = 5 c = 8 – 6 = 2
y = 22 – 5(3) Therefore the point of intersection is m OR = m PQ = 5 7 – 6
y = 7 = (–9, –5) y = –2x + 2
Equation of OR : y = 5x + c
Therefore, the point of intersection is
= (3, 7) 15. (a) y = –2x + 5, Substitute x = 0, y = 0
c = 0
Gradient PQ = –2
(b) 4x – 7y = 0 .......................(1) y = 5x
Equation PQ, y = –2x + c
8x – y – 26 = 0 .................(2)
Substitute x = 2 and y = –1,
From (2) –1 = –2(2) + c (d) m PR = undefined
y = 8x – 26 ........................(3) c = 3 Equation of PR is x = 6
substitute (3) into (1) Therefore the equation is y = –2x + 3 Q (0,4)
4x – 7(8x – 26) = 0 m PQ = 8 – 4 = 2
4x – 56x +182 = 0 6 3 2
52x = 182 Equation of PQ is y = x + c
3
7
x =
2










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Mathematics Form 3 Chapter 9 Straight Lines Mathematics Form 3 Chapter 9 Straight Lines

PT3 Standard Practice 9 Section B (b) Equation of line PQ 2. (a) (i) k located at line y = 3
y = – 5 x + 5 ..............(1) Therefore k(a, 3)
Section A 1. Diagram Equation 3 y = 2x – 4
Rajah Persamaan Equation of line RS
6 – (–4) 3 = 2(a) – 4
1. Gradient of KL = (a) y y = x + 3 ....................(2)
8 – (–2) a = 3.5
= 1 K 2 y = – + 2 (1) = (2), ∴ k(3.5, 3)
2

Equation of KL x 5  – 5 x + 5 = x + 3 × 3
y = x + c O 5 L 3 (ii) x = –4
at L(8, 6) (b) –5x + 15 = 3x + 9 (iii) m ML = m KN
6 = 8 + c y x = 3 = 2
c = –2 K(8, 4) 4
1 3 y = 2x + c
∴ y = x – 2 y = x Substitute x = into (2)
x 2 4 Substitute x = –4, y = 0,
Answer: D L O y = 3 + 3
4 0 = 2(–4) + c
2. 4x + 3y = 15 y = 15 c = 8
3y = –4x + 15 2. Equation Point 4 ∴ y = 2x + 8
4 Persamaan Titik 3 15
y = – x + 5 ∴ k  ,  (iv) Coordinates P = (p, 0)
3 4 4
(a) y = 3x – 7 (6, 11) ✓ Substitute x = p, y = 0 into y = 2x – 4,
4
∴ m = – , c = 5 (c) Substitute x = 2, y = 8, 0 = 2p – 4
3
(b) –5x + y = 2 (–3, –13) ✓ y = k x + 5m p = 2
Answer: B 3 Length of MP = 2 – (–4)
1
3. x-intercept, y = 0 (c) 3y = x + 8 (2, –3) ✗ 8 = 2k + 5m = 6 units
2 3
2x – 3y = 6 24 = 2k + 15m ...................(1) Area of KLMP = (3 × 6)
1
2x – 3(0) = 6 (d) 1 y + x = 1 (–10, 6) ✗ Substitute x = –5, y = –13 = 18 units 2
2x = 6 3 5 k
x = 3 y = x + 5m (b) 2x + 3y – 4 = 0
Section C 3 3y = –2x + 4
∴ x-intercept = 3 –13 = –5k + 5m
2
1. (a) (i) Rhombus has equal length of sides. 3 y = – x + 4
Answer: B 3 3
OK = KL = LM = MO –39 = –5k + 15m ...............(2) 2
2
4. m KL = –1, c = –5 OK = (–5) + 12 2 (1) – (2) gradient, m = – 3
= 13 units 24 – (–39) = 2k – (–5k)
∴ equation of the straight line KL is (c) 5x + 2y = 8
y = –x – 5 ∴ OM = 13 units 63 = 7k
M(–13, 0) k = 9 Divide both sides by 8,
Answer: C Substitute k = 9 into (1) 5x 2y 8
(ii) m KO = 12 – 0 8 + 8 = 8
2 –5 – (0) 24 = 2(9) + 15m
5. y – x = 5 x y
3 = – 12 15m = 6 8 + = 1
4
3y – 2x = 15 5 2
2x – 3y + 15 = 0 y = mx + c m = 5 5
12 Therefore, x-intercept = 8
Answer: C y = – x + c ∴ m = 2 , k = 9 5
5 5 y-intercept = 4
Substitute x = 0, y = 0,
c = 0
Equation of LM

y = – 12 x
5
(iii) y = 12
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Mathematics Form 3 Chapter 9 Straight Lines Mathematics Form 3 Chapter 9 Straight Lines

PT3 Standard Practice 9 Section B (b) Equation of line PQ 2. (a) (i) k located at line y = 3
y = – 5 x + 5 ..............(1) Therefore k(a, 3)
Section A 1. Diagram Equation 3 y = 2x – 4
Rajah Persamaan Equation of line RS
6 – (–4) 3 = 2(a) – 4
1. Gradient of KL = (a) y y = x + 3 ....................(2)
8 – (–2) a = 3.5
= 1 K 2 y = – + 2 (1) = (2), ∴ k(3.5, 3)
2

Equation of KL x 5  – 5 x + 5 = x + 3 × 3
y = x + c O 5 L 3 (ii) x = –4
at L(8, 6) (b) –5x + 15 = 3x + 9 (iii) m ML = m KN
6 = 8 + c y x = 3 = 2
c = –2 K(8, 4) 4
1 3 y = 2x + c
∴ y = x – 2 y = x Substitute x = into (2)
x 2 4 Substitute x = –4, y = 0,
Answer: D L O y = 3 + 3
4 0 = 2(–4) + c
2. 4x + 3y = 15 y = 15 c = 8
3y = –4x + 15 2. Equation Point 4 ∴ y = 2x + 8
4 Persamaan Titik 3 15
y = – x + 5 ∴ k  ,  (iv) Coordinates P = (p, 0)
3 4 4
(a) y = 3x – 7 (6, 11) ✓ Substitute x = p, y = 0 into y = 2x – 4,
4
∴ m = – , c = 5 (c) Substitute x = 2, y = 8, 0 = 2p – 4
3
(b) –5x + y = 2 (–3, –13) ✓ y = k x + 5m p = 2
Answer: B 3 Length of MP = 2 – (–4)
1
3. x-intercept, y = 0 (c) 3y = x + 8 (2, –3) ✗ 8 = 2k + 5m = 6 units
2 3
2x – 3y = 6 24 = 2k + 15m ...................(1) Area of KLMP = (3 × 6)
1
2x – 3(0) = 6 (d) 1 y + x = 1 (–10, 6) ✗ Substitute x = –5, y = –13 = 18 units 2
2x = 6 3 5 k
x = 3 y = x + 5m (b) 2x + 3y – 4 = 0
Section C 3 3y = –2x + 4
∴ x-intercept = 3 –13 = –5k + 5m
2
1. (a) (i) Rhombus has equal length of sides. 3 y = – x + 4
Answer: B 3 3
OK = KL = LM = MO –39 = –5k + 15m ...............(2) 2
2
4. m KL = –1, c = –5 OK = (–5) + 12 2 (1) – (2) gradient, m = – 3
= 13 units 24 – (–39) = 2k – (–5k)
∴ equation of the straight line KL is (c) 5x + 2y = 8
y = –x – 5 ∴ OM = 13 units 63 = 7k
M(–13, 0) k = 9 Divide both sides by 8,
Answer: C Substitute k = 9 into (1) 5x 2y 8
(ii) m KO = 12 – 0 8 + 8 = 8
2 –5 – (0) 24 = 2(9) + 15m
5. y – x = 5 x y
3 = – 12 15m = 6 8 + = 1
4
3y – 2x = 15 5 2
2x – 3y + 15 = 0 y = mx + c m = 5 5
12 Therefore, x-intercept = 8
Answer: C y = – x + c ∴ m = 2 , k = 9 5
5 5 y-intercept = 4
Substitute x = 0, y = 0,
c = 0
Equation of LM

y = – 12 x
5
(iii) y = 12
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Mathematics Form 3 Chapter 9 Straight Lines

HOTS Challenge PT3 Model Paper



Types A B
Packaging
Section A 5. 4.3 × 10 + 0.00174
–5
Packet (x) 10 15 = 4.3 × 10 + 1.74 × 10 –3
–5
Boxes (y) 8 20 1. = (0.043 + 1.74) × 10 –3
–3
740 1 350 = 1.783 × 10
Answer: C
10x + 8y = 740 ....................(1)
15x + 20y =1 350 ................(2)
1 + 2 2 + 2 –3 3

17
10

From (1) 6.   =  
10x + 8y = 740 17 10
x = (740 – 8y) ÷ 10 3 1
17
x = 74 – 0.8y =    2
10
Substitute (3) into (2) 3
17
15(74 – 0.8y) + 20y = 1 350 =   2
1 110 – 12y + 20y = 1 350 10
8y = 240 Answer: C
y = 30 3 + 2 4 + 2
x = 74 – 0.8 × 30 7. x + 90° + 0.5x + 69° + 2x + 68° = 360°
x = 50 = 5 3.5x + 227° = 360°
∴ 5n + 2, n = 1, 2, 3, 4, … 3.5x = 133°
x = 38°
Answer: A
Answer: A
2.
8 8, 16, 64 8. Answer: B
1, 2, 8
Answer: B x + x y + y
9. Midpoint =  1 2 2 , 1 2 2 

3. Scale S =  –6 + (–2) , –9 + 3 
= length of drawing : length of object 2 2
= 0.16 cm : 5 m = (–4, –3)
= 0.16 : 500 Answer: A
= 1 : 3 125

Answer: C 10. Coordinates of image
= (–1 + 7, –5 + –5)
= (6, –10)
4. Answer: D
Answer: C









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Mathematics Form 3 Chapter 9 Straight Lines

HOTS Challenge PT3 Model Paper



Types A B
Packaging
Section A 5. 4.3 × 10 + 0.00174
–5
Packet (x) 10 15 = 4.3 × 10 + 1.74 × 10 –3
–5
Boxes (y) 8 20 1. = (0.043 + 1.74) × 10 –3
–3
740 1 350 = 1.783 × 10
Answer: C
10x + 8y = 740 ....................(1)
15x + 20y =1 350 ................(2)
1 + 2 2 + 2 –3 3

17
10

From (1) 6.   =  
10x + 8y = 740 17 10
x = (740 – 8y) ÷ 10 3 1
17
x = 74 – 0.8y =    2
10
Substitute (3) into (2) 3
17
15(74 – 0.8y) + 20y = 1 350 =   2
1 110 – 12y + 20y = 1 350 10
8y = 240 Answer: C
y = 30 3 + 2 4 + 2
x = 74 – 0.8 × 30 7. x + 90° + 0.5x + 69° + 2x + 68° = 360°
x = 50 = 5 3.5x + 227° = 360°
∴ 5n + 2, n = 1, 2, 3, 4, … 3.5x = 133°
x = 38°
Answer: A
Answer: A
2.
8 8, 16, 64 8. Answer: B
1, 2, 8
Answer: B x + x y + y
9. Midpoint =  1 2 2 , 1 2 2 

3. Scale S =  –6 + (–2) , –9 + 3 
= length of drawing : length of object 2 2
= 0.16 cm : 5 m = (–4, –3)
= 0.16 : 500 Answer: A
= 1 : 3 125

Answer: C 10. Coordinates of image
= (–1 + 7, –5 + –5)
= (6, –10)
4. Answer: D
Answer: C









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Mathematics Form 3 PT3 Model Paper Mathematics Form 3 PT3 Model Paper
11. Total time = 2345 – 2015 16. Mean Section B 0.5 cm
= 3.5 hours (12 × 5) + (15 × 3) + (18 × 7) + (21 × 2) 0.5 cm
250 + (24 × 6) 2
Average speed = = 21. (i) K = –2.4 = –2
3.5 5 + 3 + 7 + 2 + 6 5
4
= 71.43 km/h 417 L = –1.8 = –1
    = 5
Answer: B 23
= 18 .13 (ii) KL – 1.203
= –2.4 (–1.8) – 1.203
Answer: A
12. (Cos 60° + sin 45°)(Cos 60° – sin 45°) = 4.32 – 1.203

2
2
 1 +   1 –    17. Percentage of students weight more than = 3.117
2
2
2
2
= – 1 60 kg. 22. (a) (i) Line KM / Garis KM
4 = 25 + 5 × 100% (ii) Line AC / Garis AC
Answer: D 15 + 30 + 10 + 15 + 25 + 5 (b) 6 – 3y  4y – 8
= 30%
6 + 8  4y + 3y
13. Price after discount Answer: D 14  7y
25. (i) ✓
80
P = RM60 ×  = RM48 y  2
100 18. The interest Vadivoo has to pay (ii) ✗
5.85
75 = RM55 000 ×  × 2
Q = RM78 ×  = RM58.50 100 23. (a) Centre (b) (i) FALSE
100 = RM6 435 (b) Chord (ii) TRUE
95 (c) Major sector
R = RM53 ×  = RM50.35 Answer: B
100 (d) Circumference Section C
60
S = RM94 ×  = RM56.40 19. Answer: A 1 1
100 24. (a) Scale = 2 cm : 10 cm 26. (a) (i)  =
3
64
4
= 1 : 5
∴ The cheapest type of diaper is diaper P 2 2
 
20. K(0, y) J(–2, 6) (ii)  2 – 9 = 2 – 3 
1
Answer: A (b) AB = 36 × = 6 cm 3 3
Gradient JK = 3 6 7 2

y – 6 1 = – 
  = 3 BC = EF = 21 × = 3.5 cm 3
14. ∠UQP = ∠TSQ = 65° 0 – (–2) 1 6 49
y – 6 = 6 DE = 9 × = 1.5 cm =
∠QUP = 180° – 65° – 72° y = 12 6 9
1
= 43° AH = 42 × = 7 cm
∴ OK = 12 units 6
Answer: C x 2 –5 x –3 1
2
HE = AB × 1 (b) (i) 8y 2 × x = 8y 2 = 8x y
3 2
2
OL = 13 – 12 2 3 6
2 1 = 5 units 2 1 1
15. Probability of green dresses = 1 –  – = (36) × = 4 cm 2
5 5 ∴ L(5, 0) 3 6 (ii) 9 3k – 1 = 729 × 81
2 3 –2
= 3 6k – 2 × 3 = 3 × 3 2
–2
6
5 x-intercept of KL is 5
3 6k – 4 = 3 8
2 Answer: C
 × Total dresses = 10
5 ∴ 6k – 4 = 8
Total dresses = 10 ×  5 6k = 12
2 k = 2
    = 25 pieces
Answer: A
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Mathematics Form 3 PT3 Model Paper Mathematics Form 3 PT3 Model Paper
11. Total time = 2345 – 2015 16. Mean Section B 0.5 cm
= 3.5 hours (12 × 5) + (15 × 3) + (18 × 7) + (21 × 2) 0.5 cm
250 + (24 × 6) 2
Average speed = = 21. (i) K = –2.4 = –2
3.5 5 + 3 + 7 + 2 + 6 5
4
= 71.43 km/h 417 L = –1.8 = –1
    = 5
Answer: B 23
= 18 .13 (ii) KL – 1.203
= –2.4 (–1.8) – 1.203
Answer: A
12. (Cos 60° + sin 45°)(Cos 60° – sin 45°) = 4.32 – 1.203

2
2
 1 +   1 –    17. Percentage of students weight more than = 3.117
2
2
2
2
= – 1 60 kg. 22. (a) (i) Line KM / Garis KM
4 = 25 + 5 × 100% (ii) Line AC / Garis AC
Answer: D 15 + 30 + 10 + 15 + 25 + 5 (b) 6 – 3y  4y – 8
= 30%
6 + 8  4y + 3y
13. Price after discount Answer: D 14  7y
25. (i) ✓
80
P = RM60 ×  = RM48 y  2
100 18. The interest Vadivoo has to pay (ii) ✗
5.85
75 = RM55 000 ×  × 2
Q = RM78 ×  = RM58.50 100 23. (a) Centre (b) (i) FALSE
100 = RM6 435 (b) Chord (ii) TRUE
95 (c) Major sector
R = RM53 ×  = RM50.35 Answer: B
100 (d) Circumference Section C
60
S = RM94 ×  = RM56.40 19. Answer: A 1 1
100 24. (a) Scale = 2 cm : 10 cm 26. (a) (i)  =
3
64
4
= 1 : 5
∴ The cheapest type of diaper is diaper P 2 2
 
20. K(0, y) J(–2, 6) (ii)  2 – 9 = 2 – 3 
1
Answer: A (b) AB = 36 × = 6 cm 3 3
Gradient JK = 3 6 7 2

y – 6 1 = – 
 = 3 BC = EF = 21 × = 3.5 cm 3
14. ∠UQP = ∠TSQ = 65° 0 – (–2) 1 6 49
y – 6 = 6 DE = 9 × = 1.5 cm =
∠QUP = 180° – 65° – 72° y = 12 6 9
1
= 43° AH = 42 × = 7 cm
∴ OK = 12 units 6
Answer: C x 2 –5 x –3 1
2
HE = AB × 1 (b) (i) 8y 2 × x = 8y 2 = 8x y
3 2
2
OL = 13 – 12 2 3 6
2 1 = 5 units 2 1 1
15. Probability of green dresses = 1 –  – = (36) × = 4 cm 2
5 5 ∴ L(5, 0) 3 6 (ii) 9 3k – 1 = 729 × 81
2 3 –2
= 3 6k – 2 × 3 = 3 × 3 2
–2
6
5 x-intercept of KL is 5
3 6k – 4 = 3 8
2 Answer: C
 × Total dresses = 10
5 ∴ 6k – 4 = 8
Total dresses = 10 ×  5 6k = 12
2 k = 2
    = 25 pieces
Answer: A
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Mathematics Form 3 PT3 Model Paper Mathematics Form 3 PT3 Model Paper
7
(c) Amount to be paid by instalment c = (ii) New probability of green pen: n() = 60
2
= RM84.62 18 2 = 1 – = 3 17 – x + x + 8 + x + 2 + 23 – x + 6 + y = 60
= RM1 523.16 ∴ Equation of the straight line is 5 5 17 – 3 + 3 + 8 + 3 + 2 + 23 – 3 + 6 + y = 60
1
Amount of interest to be paid y = – x + 7 (c) 2 – p  2p – 4 56 + y = 60
= RM1 523.16 – RM1 360 2 2 2 + 4  2p + p y = 4
= RM163.16 3p  6
28. (a) x + y = 12 ......................(1)
4y – x = 18 .....................(2) p  2 30. (a) (i) (a) Transformation A is reflection
k + 5 + 6 + 6 + 8 + 12 + 12 2p – 4  26 – 3p at line y = 1.
27. (a) (i) + 12 = 8 From (1) 2p + 3p  26 + 4 (b) Transformation B is an
8 y = –x + 12 ...................(3) 5p  30
k + 61 = 64 enlargement with scale factor
k = 3 From (2) p  6 of 2 at centre (–4, 0).
(ii) mode = 12 x + 18
y = ......................(4) p 6
2
(iii) 3, 5, 6, 6, 8, 12, 12, 12 4 p 2 (ii) ∆STU = 2 × ∆JKL
6 + 8 (3) = (4) = 4 × 25 cm 2
median, m = 2 6
2 x + 18 2 p 6 = 100 cm 2
= 7 –x + 12 = 4
k – m = 3 – 7 –4x + 48 = x + 18 Therefore, the possible values of p are

= – 4 –5x = –30 3, 4 and 5. (b) (i) z = 1 x – 2
x = 6 3 y
(b) x – 2x – 3 = x(3 – x) Substitute x = 6 into (1) 8 x – 2
2
x – 2x – 3 = 3x – x 2 6 + y = 12 29. (a) sin x = 17 3z = 
2
y
2
2x – 5x – 3 = 0 x – 2
y = 6 AG = 8 9z =
2
y
2x 1 x ∴ coordinates T is (6, 6) BG 17 9yz = x – 2
2
x –3 –6x 16 8 2
= x = 9yz + 2
2x 2 –3 –5x 3 4 BG 17
(b) (i) P(Green pen) = 1 – = BG = 34 cm
2
(2x + 1)(x – 3) = 0 7 7 (ii) x = 9yz + 2
1 Total number of pens: DG = BG – BD
1
x = – , x = 3 = 34 – 29   2
2 4 x = 9(–1) + 2
× Total number of pens = 24 = 5 cm x = 1 3
7
1 1 7 5
(c) x + y = 5 Total number of pens = 24 × cos y =
4 2 4 13 31. (a) Volume of small cubes
1
1 y = – x + 5 = 42 y = cos –1 5
2 4 13 = Volume of sphere
1 Number of pink pens = 42 – 24 y = 67.4° 4 851
y = – x + 10 = 18 4 22
2   (21) 3
1 New probability of pink pen: (b) Price of juice Kavita has to pay = 3 7
∴ Gradient of the line, m = – 18 + 2 4 851
2 = = (150 × RM0.10) + (200 × RM0.08)
42 + 2 + 6 = 8 cm 3
Equation of straight line 2 + [(1 000 – 150 – 200) × RM0.03]
–1 = = RM15 + RM16 + RM19.50 3
8
y = x + c 5 Length of the side of small cube = 
2 = RM50.50 = 2 cm
Substitute x = 3, y = 2
1 (c) n(P  R  Q) = 16 Area of one of the faces of the small
2 = – (3) + c cube
2 x + 8 + x + 2 = 16
2x = 6 = 2 × 2 2
x = 3 = 4 cm
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Mathematics Form 3 PT3 Model Paper Mathematics Form 3 PT3 Model Paper
7
(c) Amount to be paid by instalment c = (ii) New probability of green pen: n() = 60
2
= RM84.62 18 2 = 1 – = 3 17 – x + x + 8 + x + 2 + 23 – x + 6 + y = 60
= RM1 523.16 ∴ Equation of the straight line is 5 5 17 – 3 + 3 + 8 + 3 + 2 + 23 – 3 + 6 + y = 60
1
Amount of interest to be paid y = – x + 7 (c) 2 – p  2p – 4 56 + y = 60
= RM1 523.16 – RM1 360 2 2 2 + 4  2p + p y = 4
= RM163.16 3p  6
28. (a) x + y = 12 ......................(1)
4y – x = 18 .....................(2) p  2 30. (a) (i) (a) Transformation A is reflection
k + 5 + 6 + 6 + 8 + 12 + 12 2p – 4  26 – 3p at line y = 1.
27. (a) (i) + 12 = 8 From (1) 2p + 3p  26 + 4 (b) Transformation B is an
8 y = –x + 12 ...................(3) 5p  30
k + 61 = 64 enlargement with scale factor
k = 3 From (2) p  6 of 2 at centre (–4, 0).
(ii) mode = 12 x + 18
y = ......................(4) p 6
2
(iii) 3, 5, 6, 6, 8, 12, 12, 12 4 p 2 (ii) ∆STU = 2 × ∆JKL
6 + 8 (3) = (4) = 4 × 25 cm 2
median, m = 2 6
2 x + 18 2 p 6 = 100 cm 2
= 7 –x + 12 = 4
k – m = 3 – 7 –4x + 48 = x + 18 Therefore, the possible values of p are

= – 4 –5x = –30 3, 4 and 5. (b) (i) z = 1 x – 2
x = 6 3 y
(b) x – 2x – 3 = x(3 – x) Substitute x = 6 into (1) 8 x – 2
2
x – 2x – 3 = 3x – x 2 6 + y = 12 29. (a) sin x = 17 3z = 
2
y
2
2x – 5x – 3 = 0 x – 2
y = 6 AG = 8 9z =
2
y
2x 1 x ∴ coordinates T is (6, 6) BG 17 9yz = x – 2
2
x –3 –6x 16 8 2
= x = 9yz + 2
2x 2 –3 –5x 3 4 BG 17
(b) (i) P(Green pen) = 1 – = BG = 34 cm
2
(2x + 1)(x – 3) = 0 7 7 (ii) x = 9yz + 2
1 Total number of pens: DG = BG – BD
1
x = – , x = 3 = 34 – 29   2
2 4 x = 9(–1) + 2
× Total number of pens = 24 = 5 cm x = 1 3
7
1 1 7 5
(c) x + y = 5 Total number of pens = 24 × cos y =
4 2 4 13 31. (a) Volume of small cubes
1
1 y = – x + 5 = 42 y = cos –1 5
2 4 13 = Volume of sphere
1 Number of pink pens = 42 – 24 y = 67.4° 4 851
y = – x + 10 = 18 4 22
2   (21) 3
1 New probability of pink pen: (b) Price of juice Kavita has to pay = 3 7
∴ Gradient of the line, m = – 18 + 2 4 851
2 = = (150 × RM0.10) + (200 × RM0.08)
42 + 2 + 6 = 8 cm 3
Equation of straight line 2 + [(1 000 – 150 – 200) × RM0.03]
–1 = = RM15 + RM16 + RM19.50 3
8
y = x + c 5 Length of the side of small cube = 
2 = RM50.50 = 2 cm
Substitute x = 3, y = 2
1 (c) n(P  R  Q) = 16 Area of one of the faces of the small
2 = – (3) + c cube
2 x + 8 + x + 2 = 16
2x = 6 = 2 × 2 2
x = 3 = 4 cm
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Mathematics Form 3 PT3 Model Paper
Total distance (ii) ∠FBC = 180° – 150°
(b) Average speed =
Time = 30°
[1.5 × (10 × 60) + 5 000 + x] 7 000
= ∴ x + 130° + 30° = 180°
10 + 32.5 + 17.5 60 x = 20°
x = 1 100m
∠DEB + x = 180°
∴ Ken walked for 1 100 m. ∠DEB = 180° – 20°
= 160°
10.24
(c) (i) tan ∠KJL = y = 360 – 160°
8 = 200°

∠KJL = tan –1 10.24 
8
∠KJL = 52°





























































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A32



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Mathematics Form 3 PT3 Model Paper
Total distance (ii) ∠FBC = 180° – 150°
(b) Average speed =
Time = 30°
[1.5 × (10 × 60) + 5 000 + x] 7 000
= ∴ x + 130° + 30° = 180°
10 + 32.5 + 17.5 60 x = 20°
x = 1 100m
∠DEB + x = 180°
∴ Ken walked for 1 100 m. ∠DEB = 180° – 20°
= 160°
10.24
(c) (i) tan ∠KJL = y = 360 – 160°
8 = 200°

∠KJL = tan –1 10.24 
8
∠KJL = 52°





























































61 © Penerbitan Pelangi Sdn. Bhd. © Penerbitan Pelangi Sdn. Bhd. 62

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BOOKLET ANS MATH F3.indd 32 03/01/2020 10:21 AM