14[a math cd]

1 © Penerbitan Pelangi Sdn. Bhd.

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


= 6 – 2–––––––7 – (–3)

= 4–––10

= 2—5

Answer: C


= 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


= − y-intercept

––––––––––x-intercept

= − 10––––2

= 5

Answer: B


= − y-intercept


= − 5—6

Answer: C

CHAPTER

14 The Straight LineCHAPTER

2

Mathematics SPM Chapter 14

© Penerbitan Pelangi Sdn. Bhd.

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

3



(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


= − 6–––––(– 4)

= 3—2

5.

x

y

B(0, 4)

A(–3, 0)3

4

D

O

(a) Gradient of line AB = − y-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

4



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


= − 4—2

= −2

5



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


= − 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


= 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

6



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


– 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

7



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


− 1—3

= − (– 4)


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


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

8



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

9



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



2 = − 8


x-intercept = − 8––2

= −4

Answer: A

2. Gradient of line EF = − y-intercept


− 3––4

= – 6


x-intercept = 6 × 4–––––3

= 8

Answer: C



− 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

10



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


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

11



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


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

)

12



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


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

13



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


14



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


(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


= − 6–––– 4

= 6—4

= 3—2

m = 3—2

and P(− 4, 12)


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

15



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)


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


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.

14[a math cd]

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