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Paper 1
1. Gradient of line PQ
= QR
––––PR
Answer: A
2. Gradient of line EF
= FG
––––EG
= tan q
Answer: C
3. Gradient of the straight line passing through (1, −2) and (3, 6)
= 6 − (−2)––––––––3 – 1
= 8—2
= 4
Answer: C
4. Gradient of the straight line passing through (1, 5) and (−3, 4)
= 4 – 5–––––––3 – 1
= –1—–– 4
= 1—4
Answer: A
5. Gradient of line PQ
= 6 – 2–––––––7 – (–3)
= 4–––10
= 2—5
Answer: C
6. Gradient of line EF
= 5 – 2–––––––5 – 4
= 3––––9
= − 1—3
Answer: D
7. The straight line with negative gradient is JK.
Answer: C
8. The x-intercept = −4
Answer: C
9. The y-intercept = 7
Answer: B
10. Gradient of line PQ
= − y-intercept
––––––––––x-intercept
= − 10––––2
= 5
Answer: B
11. Gradient of line EF
= − y-intercept
––––––––––x-intercept
= − 5—6
Answer: C
CHAPTER
14 The Straight LineCHAPTER
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Mathematics SPM Chapter 14
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12. P(0, 1) and Q(2, 5)Gradient of line PQ, m = 5 – 1–––––
2 – 0
= 4—2
= 2The y-intercept, c = 1
The equation of line PQ isy = mx + cy = 2x + 1
Answer: A
13. M(0, −2) and N(4, 2)Gradient of line MN, m = 2 – (–2)–––––––
4 – 0 = 4—
4 = 1The y-intercept, c = −2
The equation of line MN isy = mx + cy = 1x − 2y = x − 2
Answer: C
14. m = 4 and c = 3
The equation of line PQ isy = mx + cy = 4x + 3
Answer: D
15. P(0, 4), Q(4, 0)Gradient of line PQ, m = 4 – 0–––––
0 – 4 = −1The y-intercept, c = 4
The equation of line PQ isy = mx + cy = −x + 4
Answer: B
16. y = 3x − 2 ...................1 The equation of a straight line is y = mx + c ..................2
Compare equations 1 and 2.m = gradient = 3c = y-intercept = −2
Answer: A
17. y = 2x − 1⇒ gradient, m = 2 the y-intercept, c = −1
Answer: C
18. y = − 1—2
x + 5 ..........................1
Substitute (6, 2) into equation 1.
2 = − 1—2
(6) + 5
2 = −3 + 52 = 2
Hence, (6, 2) lies on the straight line y = − 1—2
x + 5.
Answer: B
Paper 2
1.
θ
Given tan q = 2hence, q = 63.43°
2. (a) Gradient of the straight line passing through (1, 3) and (2, 5)
= 5 – 3–––––2 – 1
= 2—1
= 2
(b) Gradient of the straight line passing through (0, 4) and (4, 0)
= 0 – 4–––––4 – 0
= – 4–––4
= −1
(c) Gradient of the straight line passing through (−2, −5) and (3, 1)
= 1 – (–5)–––––––3 – (–2)
= 6—5
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(d) Gradient of the straight line passing through (−4, 0) and (3, −2)
= –2 – 0––––––––3 – (– 4)
= − 2—7
3. Gradient of the straight line passing through (1, −2) and (6, h)
= h – (–2)––––––––6 – 1
= h + 2–––––5
Given the gradient = 4—5
hence, h + 2–––––5
= 4—5
h + 2 = 4 h = 2
4. (a) x-intercept = −4 y-intercept = 6
(b) Gradient of line PQ
= − y-intercept
––––––––––x-intercept
= − 6–––––(– 4)
= 3—2
5.
x
y
B(0, 4)
A(–3, 0)3
4
D
O
(a) Gradient of line AB = − y-intercept
––––––––––x-intercept
4—3
= − 4—t
4t = −4(3)
t = – 12–––4
= −3
(b) AB = OA2 + OB2
= 32 + 42
= 5 units BD = AB = 5 units
Hence, the coordinates of D = (−5, 4)
6.
x
y
K(12, 8)J(0, 8)
L(4, 0)4
O
(a) The coordinates of J = (0, 8)
(b) Gradient of line LK
= 8 – 0––––––12 – 4
= 8—8
= 1
7.
Q(–3, 6)
y
R(0, 10)
10
xOP
(a) The y-intercept of line QR = 10
(b) Gradient of line QR
= 10 – 6––––––––0 – (–3)
= 4—3
8. y
xO
18
9
K(18, 0)
J
L
H
(a) OL = 1—2
OK
= 1—2
(18)
= 9
The coordinates of J = (18, 18)
(b) J(18, 18) and L(0, 9) Gradient of line JL
= 9 – 18––––––0 – 18
= 1—2
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9. y
x0 2G
E(6, h)
F
H
(a) E(6, h) and G(2, 0)
Gradient of line EG = h – 0–––––6 – 2
2 = h—4
h = 2 × 4 = 8(b) Let H = (0, k)
Gradient of line GH = k – 0–––––0 – 2
7 = − k—2
k = 7 × (−2) = −14
Hence, the y-intercept of line GH = −14
10. (a) y
x0
E(0, 16)
F(h, 6)
G(t, 0)
Let G = (t, 0)
Gradient of EG = 16 – 0––––––0 – t
−2 = − 16–––t
t = 16–––2
= 8
Hence, the x-intercept of line EG = 8
(b) Gradient of line EF = 16 – 6––––––0 – h
−2 = 10––––h
h = 10–––2
= 5
11. A(2, 3) and m = 4
The equation of line AB is y = mx + c3 = 4(2) + c3 = 8 + c c = 3 − 8 = −5
Hence, the equation of line AB is y = 4x − 5.
12. P(4, −2) and m = − 3—2
The equation of line PQ is y = mx + c
−2 = − 3—2
(4) + c
−2 = −6 + c −2 + 6 = c c = 4
Hence, the equation of line PQ is y = − 3—2
x + 4.
13. The equation of line QR is y = 4.
14. Q(5, 0) and m = −2
The equation of line PQ is y = mx + c 0 = −2(5) + c 0 = −10 + c c = 10
Hence, the equation of line PQ is y = −2x + 10.
15. x − 2y = 12 .................1 (a) Substitute x = 0 into equation 1. 0 − 2y = 12 2y = −12 y = −6
Hence, the y-intercept = −6
(b) Substitute y = 0 into equation 1. x − 2(0) = 12 x = 12
Hence, the x-intercept = 12
16. Gradient of line AB
= − y-intercept
––––––––––x-intercept
= − 4—2
= −2
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For the line CD, m = −2 and c = 6.
The equation of line CD isy = mx + cy = −2x + 6
17. Gradient of line QR= Gradient of line OP
= 2 – 0–––––4 – 0
= 2—4
= 1—2
18. Gradient of line DO= Gradient of line AB
= − y-intercept
––––––––––x-intercept
= − 3—4
19. Equation of line PQ:
y = − 1—2
x + 3
Gradient of line RS= Gradient of line PQ= − 1—
2
20. Equation of line AB:3x − 2y = 8 3x = 2y + 8 2y = 3x − 8
y = 3—2
x − 8—2
y = 3—2
x − 4
Gradient of line CD= Gradient of line AB
= 3—2
21. Gradient of line AB= Gradient of line OC
= 5 – 0–––––2 – 0
= 5—2
22. Gradient of line QR= Gradient of line PS
= 10 – 4––––––4 – 0
= 6—4
= 3—2
23. Gradient of line BC= Gradient of line OA= −2
For the line BC, m = −2 and c = 7.The equation of BC isy = mx + cy = −2x + 7
Paper 1
1. Substitute y = 0 into 3x + 2y – 12 = 0. 3x + 2(0) – 12 = 0 3x = 12
x = 123
= 4Hence, x-intercept = 4Answer: C
2. Gradient of line PQ
= 7 – 1–––––––2 – 3
= – 6—5
Answer: B
3. 3y + kx = 24 3y = –kx + 24
y = – k3
x + 8
OF : OE = 4 : 3
OFOE
= 43
Gradient = – 43
– k3
= – 43
k = 4Answer: B
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4. 2y = 8 – x2y = –x + 8
y = – 12
x + 4
Gradient = – 12
and y-intercept = 4
4
y
0 8x
Answer: A
5. Gradient = – y-interceptx-intercept
= – 34
Hence, the y-intercept = 3 and the x-intercept = 4.
3
y
0 4x
Answer: A
6. 3y + 6—5
x = 1
3y = − 6—5
x + 1
y = − 2—5
x + 1—3
Hence, the gradient = − 2—5
Answer: D
7. y
xP R
Q
O–4
–4
The y-intercept of QR = –4
Gradient of line QR = – y-interceptx-intercept
3 = – –4x-intercept
x-intercept = – –43
= 43
Answer: D
8. Gradient = – y-intercept
––––––––––x-intercept
– 23
= – y-intercept
––––––––––6
y-intercept = 23
× 6
= 4Answer: A
9. Midpoint of EF = –4 + 62
, 2 + 82
2 = (1, 5)
Gradient of GM = –2 – 54 – 1
= − 73
Answer: B
10. Substitute (0, 4) into y = 2x + c.4 = –2(0) + cc = 4The equation of the straight line is y = –2x + 4.When the straight line intersects x-axis, y = 0.0 = –2x + 4x = 2Hence, the point of intersection is (2, 0).Answer: C
11. 3x – 5y = 15 5y = 3x – 15
y = 35
x – 3
Hence, the y-intercept = –3
Answer: C
12. y
E
F x12
13 units5
O
OE2 = EF 2 – OF 2 OE = 132 – 122
= 25 = 5 unitsThe y-intercept = 5
Gradient of line EF = – y-interceptx-intercept
= – 512
Answer: B
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13.
x
E(1, –4)
F
y
0
m = 3 and E(1, −4)The equation of line EF is y = mx + c −4 = 3(1) + c − 4 − 3 = c c = −7
The equation of line EF is y = 3x − 7.
Substitute (4, 5) into y = 3x − 7.5 = 3(4) − 75 = 12 − 75 = 5
Hence, F = (4, 5)
Answer: C
14. Gradient of line PQ = − y-intercept
––––––––––x-intercept
− 1—3
= − (– 4)
––––––––––x-intercept
x-intercept = 4 × (−3) = −12
Answer: D
Paper 2
1. (a) Equation of line EF: 2y + x = 3 2y = −x + 3
y = − 1—2
x + 3—2
The y-intercept = 3—2
Hence, the equation of line FG is y = 3—2
.
(b) Gradient of line GH = Gradient of line EF
= − 1—2
m = − 1—2
and H(−2, 9)
The equation of line GH is y = mx + c
9 = − 1—2
(−2) + c
9 = 1 + c c = 9 − 1 = 8
Hence, the equation of line GH is y = − 1—2
x + 8.
Substitute y = 0 into y = − 1—2
x + 8.
0 = − 1—2
x + 8
1—2
x = 8 x = 2 × 8 = 16
Hence, the x-intercept = 16
2. (a) Equation of line CD: 4y = x − 7
y = 1—4
x − 7—4
Gradient of line AB = Gradient of line CD = 1—
4
m = 1—4
and B(4, 6)
The equation of line AB is y = mx + c
6 = 1—4
(4) + c
6 = 1 + c c = 6 − 1 = 5
Hence, the equation of line AB is y = 1—4
x + 5.
(b) Substitute y = 0 into y = 1—4
x + 5.
0 = 1—4
x + 5
− 1—4
x = 5
x = −4 × 5 = −20 Hence, the x-intercept = −20
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3. (a) Equation of line EF: 2y = px + 18
y = p2
x + 9
Equation of line HG: y = –2x + 7
Gradient of line EF = Gradient of line HG p
2 = – 2
p = –4
(b) Substitute y = 0 into y = –2x + 9. 0 = –2x + 9 2x = 9
x = 92
4. (a) Gradient of line QR = Gradient of line OP
= 5 – 0–4 – 0
= – 54
Using Q(2, –8) and y = mx + c
–8 = − 54
(2) + c
–8 = – 52
+ c
c = – 112
Hence, the equation of line QR is y = − 54
x – 112
.
(b) Substitute y = 0 into y = − 54
x – 112
.
0 = − 54
x – 112
54
x = – 112
x = – 225
Hence, the x-intercept of line QR is – 225
.
5. (a) Gradient of line GH = Gradient of line EF = – 3
2 m = – 3
2 and G(–4, –2)
The equation of line GH is y = mx + c
–2 = – 32
(–4) + c
c = –8
Hence, the equation of line GH is y = – 32
x – 8.
(b) Substitute y = 0 into y = – 32
x – 8.
0 = – 32
x – 8
32
x = – 8
x = –8 × 23
= – 163
Hence, the x-intercept = – 163
6.
OB (–2, 0)D x
yA(–2, 9)
C(3, –6)
(a) Gradient of line AD = Gradient of line BC
= 0 – (– 6)–––––––––2 – 3
= − 6—5
m = − 6––5
and A(−2, 9)
The equation of line AD is y = mx + c
9 = − 6––5
(−2) + c
9 = 12–––5
+ c
c = 9 − 12–––5
= 33–––5
Hence, the equation of line AD is
y = − 6––5
x + 33–––5
.
(b) Substitute y = 0 into y = − 6––5
x + 33–––5
.
0 = − 6––5
x + 33–––5
6––5
x = 33–––5
6x = 33
x = 33–––6
= 11–––2
Hence, the x-intercept = 11–––2
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7. (a) Gradient of line GH = Gradient of line EF
= 6 – 0–––––0 – 3
= 6––––3
= −2
m = −2 and G(−4, 5) The equation of line GH is y = mx + c 5 = −2(−4) + c 5 = 8 + c c = 5 − 8 = −3
Hence, the equation of line GH is y = −2x − 3.
(b) Substitute y = 0 into y = −2x − 3. 0 = −2x − 3 2x = −3
x = − 3—2
Hence, the x-intercept = − 3—2
Paper 1
1. Gradient of line PQ = − y-intercept
––––––––––x-intercept
2 = − 8
––––––––––x-intercept
x-intercept = − 8––2
= −4
Answer: A
2. Gradient of line EF = − y-intercept
––––––––––x-intercept
− 3––4
= – 6
––––––––––x-intercept
x-intercept = 6 × 4–––––3
= 8
Answer: C
3. Gradient of line PQ = − y-intercept
––––––––––x-intercept
− 5—2
= − y-intercept
––––––––––4
y-intercept = 4 × 5–––––2
= 10
Answer: A
4. P(0, 12), Q(−4, 0), R(−3, h) Gradient of line RQ = Gradient of line PQ
h – 0–––––––––3 – (–4)
= 12 – 0––––––––0 – (–4)
h––1
= 12–––4
h = 3
Answer: D
5. Q(2, 0), R(6, k)
Gradient of line PQR = k – 0–––––6 – 2
3 = k––4
k = 3 × 4 = 12
Answer: A
6. Let T = (0, k)
Gradient of line TV = k – 0–—––0 – 3
− 4—3
= − k—3
k = 4TV 2 = OV 2 + OT 2 = 32 + 42
TV = 25 = 5 units WV = TV = 5 unitsHence, the coordinates of point W = (3, 5)
Answer: C
7. Substitute x = 0 into 2x − 3y = 12.0 − 3y = 12 3y = −12
y = − 12–––3
= −4Hence, the y-intercept = −4
Answer: B
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8. 3x + 4y = −16 4y = −3x − 16
y = − 3––4
x − 16–––4
y = − 3––4
x − 4
Hence, the gradient = − 3––4
Answer: B
9. Substitute y = 0 into y = 2––3
x − 4.
0 = 2––3
x − 4
4 = 2––3
x
4 × 3 = 2x
x = 12–––2
= 6
Hence, the x-intercept = 6
Answer: D
10. E(0, 3), F(4, 5)Gradient of line EF
= 5 – 3–––––4 – 0
= 2—4
= 1—2
Answer: A
11. Substitute point (3, 1) into 2y − kx + 7 = 0. 2(1) − k(3) + 7 = 0 2 − 3k + 7 = 0 9 = 3k
k = 9—3
= 3
Answer: C
12. Substitute x = 0 into 4x − 5y = 20. 0 − 5y = 20 5y = −20
y = − 20–––5
= −4
Hence, the y-intercept = −4
Answer: C
13. Substitute y = 0 into 2x + 5y − 7 = 0. 2x + 0 − 7 = 0 2x − 7 = 0 2x = 7
x = 7—2
Answer: C
14. 2y − 4—5
x = 1
2y = 4—5
x + 1
y = 4–––––5 × 2
x + 1—2
y = 2—5
x + 1—2
Hence, the gradient = 2—5
Answer: B
Paper 2
1.
R(2, 4)
P(–6, –4) S(0, –4)
Q
y
x0
(a) Gradient of line PQ = Gradient of line SR
= 4 – (–4)–––––––2 – 0
= 8—2
= 4
m = 4 and P(−6, −4) The equation of line PQ is y = mx + c −4 = 4(−6) + c −4 = −24 + c −4 + 24 = c c = 20 Hence, the equation of line PQ is y = 4x + 20.
(b) Substitute y = 0 into y = 4x + 20. 0 = 4x + 20 −20 = 4x
x = – 20–––4
= −5 Hence, the x-intercept = −5
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2.
F(3, –4)
H(–5, 6) G(3, 6)
0
E(0, –4)
y
x
(a) The equation of line EF is y = −4.
(b) The coordinates of G are (3, 6).
(c) Gradient of line HE
= 6 – (–4)––––––––5 – 0
= 10––––5
= −2
m = −2 and c = −4 The equation of line HE is y = mx + c y = −2x − 4
Substitute y = 0 into y = −2x − 4. 0 = −2x − 4 2x = −4 x = −2
Hence, the x-intercept = −2
3. (a) Gradient of line BC = Gradient of line OA
= 1 – 0–––––4 – 0
= 1—4
m = 1—4
and C(1, 2)
The equation of line BC is y = mx + c
2 = 1—4
(1) + c
2 − 1—4
= c
c = 7—4
Hence, the equation of line BC is
y = 1—4
x + 7—4
.
(b) Substitute y = 0 into y = 1—4
x + 7—4
.
0 = 1—4
x + 7—4
− 7—4
= 1—4
x
x = −7
Hence, the x-intercept = −7
4. y
xO
P(0, 2)
Q(3, 0)
R(3, k)
(a) Equation of line PQ: 2x + 3y = 6 ............................1
Substitute x = 0 into equation 1. 2(0) + 3y = 6
y = 6—3
= 2
Hence, P = (0, 2)
Substitute y = 0 into equation 1. 2x + 3(0) = 6 2x = 6
x = 6—2
= 2 Hence, Q = (3, 0)
The equation of line QR is x = 3.
(b) Gradient of PQ
= 2 – 0–––––0 – 3
= − 2—3
Let R = (3, k)
Gradient of line PR = k – 2–––––3 – 0
1—2
= k – 2–––––3
k − 2 = 3—2
k = 3—2
+ 2
= 7—2
m = − 2—3
and R(3, 7—2
)
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The equation of the straight line parallel to line PQ and passing through point R is
y = mx + c
7—2
= − 2—3
(3) + c
7—2
= −2 + c
c = 7—2
+ 2
= 11–––2
Hence, the equation is y = − 2—3
x + 11–––2
.
5. (a) The equation of line CD is y = 3.
(b) Gradient of line BC = Gradient of line AD
= 3 – 0––––––––0 – (–1)
= 3—1
= 3
m = 3 and B(4, 0) The equation of line BC is y = mx + c 0 = 3(4) + c c = −12 Hence, the equation of line BC is y = 3x − 12.
(c) The y-intercept of line BC is −12.
6. y
xO
P
L
N(6, 10)
M(6, 0)
(a) m = 1—2
and N(6, 10)
The equation of line PN is y = mx + c
10 = 1—2
(6) + c
10 = 3 + c 10 − 3 = c c = 7
Hence, the equation of line PN is y = 1—
2x + 7.
(b) Let L = (0, p)
Gradient of line LM = − y-intercept
––––––––––x-intercept
1—2
= − p—6
−p = 6—2
p = −3
Hence, the coordinates of L are (0, −3).
7. y
xO
Q(4, –10)
P(0, –4)
R(4, 0)
S
(a) Gradient of line RS = Gradient of line PQ
= –4 – (–10)–––––––––0 – 4
= 6––––4
= − 3—2
m = − 3—2
and R(4, 0)
The equation of line RS is y = mx + c
0 = − 3—2
(4) + c
0 = −6 + c c = 6
Hence, the equation of line RS is y = − 3—2
x + 6.
(b) The y-intercept = 6
8. (a) Gradient of line SRT = Gradient of line PQ
= 3 – 1––––––––4 – 4
= 2––––8
= − 1—4
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m = − 1––4
and R(2, 9)
The equation of line SRT is y = mx + c
9 = − 1––4
(2) + c
9 = − 1––2
+ c
c = 19–––2
Hence, the equation of line SRT is
y = − 1––4
x + 19–––2
.
(b) Substitute y = 0 into y = − 1––4
x + 19–––2
.
0 = − 1––4
x + 19–––2
1––4
x = 19–––2
x = 4 × 19–––2
= 38 Hence, the x-intercept = 38
9. y
xO
A(6, 8)B(9, 6)
D(p, q)
C(6, 0)
(a) Gradient of line OA
= 8 – 0–––––6 – 0
= 4—3
(b) m = 4—3
and C(6, 0)
Using y = mx + c
0 = 4—3
(6) + c
0 = 8 + c c = −8 Hence, the equation of the straight line parallel
to OA and passes through C is y = 4—3
x − 8.
(c) Let D = (p, q) Given B is the midpoint,
then 6 + p–––––
2 = 9 and 8 + q–––––
2 = 6
6 + p = 2 × 9 8 + q = 6 × 2 p = 18 − 6 q = 12 − 8 = 12 = 4
Hence, D = (12, 4)
10. (a) Gradient of MN = 2
6 – k–––––2 – 0
= 2
6 − k = 2 × 2 6 − k = 4 6 − 4 = k k = 2
(b) m = 2 and T(0, 9) The equation of line ST is y = mx + c y = 2x + 9
(c) Substitute y = 0 into y = 2x + 9. 0 = 2x + 9 −9 = 2x
x = − 9––2
Hence, the x-intercept = − 9––2
11. y
x
D(k, 21)
6
15A
C F
E
B
O
(a) Line ABC: 3x + y = 6 y = −3x + 6 Gradient of line ABC = −3 Gradient of line DEF = −3
m = −3 and c = 15 The equation of line DEF is y = mx + c y = −3x + 15
(b) Substitute D(k, 21) into y = −3x + 15. 21 = −3k + 15 3k = 15 − 21
k = − 6––3
= −2
(c) Substitute y = 0 into y = −3x + 15. 0 = −3x + 15 3x = 15
x = 15–––3
= 5
Hence, the x-intercept = 5
Page 14
14
Mathematics SPM Chapter 14
© Penerbitan Pelangi Sdn. Bhd.
12. y
xO
P(–4, 12)
Q
N(0, 6)
M(–4, 0)
(a) Substitute y = 0 into 3x + 2y = 12. 3x + 0 = 12 3x = 12
x = 12–––3
= 4
Hence, the x-intercept = 4
(b) The equation of line PM is x = − 4.
(c) Line NP: 3x + 2y = 12 2y = −3x + 12
y = − 3—2
x + 6
Hence, N = (0, 6)
Gradient of line PQ = Gradient of line MN
= − y-intercept
––––––––––x-intercept
= − 6–––– 4
= 6—4
= 3—2
m = 3—2
and P(− 4, 12)
The equation of line PQ is y = mx + c
12 = 3—2
(− 4) + c
12 = −6 + c c = 12 + 6 = 18
Hence, the equation of line PQ is y = 3—2
x + 18.
13. y
x0
P(0, 7)
Q(2, 1)
R(8, 3)
S
(a) Gradient of line PQ
= 7 – 1–––––0 – 2
= 6––––2
= −3
m = −3 and c = 7 The equation of line PQ is y = mx + c y = −3x + 7
(b) m = −3 and R(8, 3) The equation of line RS is y = mx + c 3 = −3(8) + c 3 = −24 + c c = 3 + 24 = 27 Hence, the y-intercept = 27
14. (a) Line TUV: 3y + 4x = 24 3y = −4x + 24
y = − 4––3
x + 8
Hence, the y-intercept = 8
(b) y
xO
U(3, 4)
P T
V(0, 8)
Gradient of line PV = Gradient of line OU
= 4 – 0–––––3 – 0
= 4—3
m = 4—3
and c = 8
The equation of PV is y = mx + c
y = 4—3
x + 8
Page 15
15
Mathematics SPM Chapter 14
© Penerbitan Pelangi Sdn. Bhd.
15. (a) Line QR: y = 3x − 9 .......................... 1
Substitute R(2, w) into equation 1. w = 3(2) − 9 = 6 − 9 = −3(b) (i) Gradient of line PQ = Gradient of line SR
= 2—5
m = 2—5
and Q(5, 6)
The equation of line PQ is y = mx + c
6 = 2—5
(5) + c
6 = 2 + c c = 6 − 2 = 4
Hence, the equation of line PQ is
y = 2—5
x + 4.
(ii) Substitute y = 0 into y = 2—5
x + 4.
0 = 2—5
x + 4
2—5
x = −4
x = – 4 × 5–––––––2
= −10
Hence, the x-intercept = −10
16. (a) Gradient of line EF
= 3 – (–2)–––––––– 4 – 1
= 5––––5
= −1
(b) m = −1 and H(2, 9) The equation of line HG is y = mx + c 9 = −1(2) + c 9 = −2 + c c = 9 + 2 = 11
Hence, the equation of line HG is y = −x + 11. Substitute y = 0 into y = −x + 11. 0 = −x + 11 x = 11 Hence, the x-intercept = 11
17. (a) Substitute y = 0 into 4x − y + 3 = 0. 4x − 0 + 3 = 0 4x = −3 x = − 3—
4
Hence, the x-intercept = − 3—4
(b) Line UV: 4x − y + 3 = 0 y = 4x + 3
Gradient of line PQR = Gradient of line UV = 4
m = 4 and Q(2, 0) The equation of line PQR is y = mx + c 0 = 4(2) + c c = −8 Hence, the equation of line PQR is y = 4x − 8.
18. (a) Perimeter, y = 2(10 – 2x) + 2(x) y = 20 – 4x + 2x y = 20 – 2x
(b)
20
10x
O
y
(c) The x-intercept and the y-intercept do not represent the sides of the rectangle.
When x = 0, the length of the rectangle does not exit. When y = 0, the breadth of the rectangle does not exit.