26

Chapter 7 Geometrical constructionsNote for the teacher:

This chapter deals mostly with constructions. Students are required to use the relevant mathematical equipment to make accurate constructions of various two–dimensional shapes. In the student’s book where diagrams have been shown for students to copy, the answer will read “construction”. Where students have been asked, in words, to construct a shape, a diagram of the shape has been shown. Please note that students must draw the shapes accurately and to scale.

Exercise 7.11. Construction

2. (a) P

Q R

4 cm

4 cm

4 cm

(b) A

B C

8 cm

7 cm

6 cm

(c) E

F G5 cm

9 cm

9 cm

(d)

4 cm

8 cm

5 cm

D

E F

Exercise 7.21. Construction

2. (a) A

B C26°

31°

123°

6.5 cm

(b) N

M O70° 60°

50°

8 cm

(c) P

Q R54° 54°

72°

7.5 cm

Exercise 7.31. Construction

2. (a) X

Y Z

6.9 cm

4 cm

8 cm

60°

(b) X

Y Z5.4 cm

10.6 cm

7.5

cm

(c) X

Y Z4.5 cm

7 cm

7 cm

70° 70°

(d) Z

X Y8 cm

6 cm

3.5 cm

25°

3. ∆XYZ XY YZ XZ Angle

(a) 4 cm 8 cm 6.9 cm XYZ = 60º

(b) 7.5 cm 5.4 cm 10.6 cm XYZ = 110º

(c) 7 cm 4.5 cm 7 cm XZY = 70º

(d) 8 cm 3.5 cm 6 cm YXZ = 25º

Activity 7.11.

Triangle EFG Triangle TPR Equal or not equal

EF = 4.8 cm TP = 4.8 cm EF = TP

FG = 5 cm PR = 5 cm FG = PR

EG = 2.6 cm TR = 2.6 cm EG = TR

F = 30º P = 30º F = P = 30º

G = 70º R = 70º G = R = 70º

E = 80º T = 80º E = T = 80º

2. Corresponding sides and angles are equal.

3. Yes, since corresponding sides and angles are equal.

Activity 7.21. Answers will vary. Students’ triangle pairs should be

congruent and satisfy the minimum requirements of congruence.

2. All pairs of triangles should be congruent (identical in every way).

3. No, knowing three angles are correspondingly equal is not enough to prove congruence. For example, if I start with a base line equal to 7 cm, then construct the triangle using the angles given, this does not mean that my partner will also start with a base line of 7 cm. Therefore, if three angles are correspondingly equal, it does not follow that the three sides will be correspondingly equal. Triangles are only congruent if all corresponding angles are equal AND all corresponding sides are equal.

Exercise 7.41. (a) SSS (b) SAS (c) SAA (d) SAA (e) RHS (f) RHS (g) SSS (h) SAS (i) SAA (j) ASA (k) RHS

2. (a) (b)

27

(c) (d)

(e)

Note that diagrams may vary, but should show the essential elements of congruence.

Exercise 7.51.

Triangle RST Triangle PQR Equal or not equal

ST = 4 cm QR = 4 cm ST = QR = 4 cm

RS = 5.7 cm PQ = 4.6 cm RS ≠ PQ

RT = 4 cm PR = 2.4 cm RT ≠ PR

R = 45º P = 60º R ≠ P

S = 45º Q = 30º S ≠ Q

T = 90º R = 90º T = R

No, the triangles are not congruent. Not all corresponding sides are equal and not all corresponding angles are equal.

2. (a) K

L M2 cm

3.5

cm

4 cm

30°

60°

(b) TU = 3.5 cm; TV = 4 cm; UV = 2 cm; TUV = 90º; TVU = 60º; UTV = 30 º

3. MN = MO = 4 cm; NO = 5 cm; MNO = NOM = 50º; NMO = 80º

4. (a) a = 30º; x = 10 units; p = 125º;r = 25º; q = 30º

(b) q = 43º; x = p = 47º; m = 4 units (c) x = 3 units; y = 4 units; z = 5 units;

n = 37º; j = 53º (d) e = 58º; f = 33º; a = d = 89º

Exercise 7.61. (a) SSS (b) Not similar (c) SAS (d) SSS (e) SSS (f) AAA

2. Yes, their sides are in proportion. Scale factor is 5.

3. No, this is not correct. The example diagram shows how two triangles with two pairs of corresponding angles equal, may not be similar. Using the ratio 3 __

7

as the ratio to compare with, the triangles have one other pair of corresponding sides with ratio 2.8 ___

6.9 .

Using a calculator, it is clear that the sides are not in proportion.

6.9

cm

7 cm50° 60° 50° 60°

3 cm

2.8

cm

4. (a) Yes, the triangles are similar by AAA. At point C there are two vertically opposite angles which are equal. This means that angles A and E are also equal.

(b) Yes, the triangles are similar by AAA. At point C there are two vertically opposite angles which are equal. Angles B and D must be equal as they are alternate angles. Angles A and E must be equal as they are alternate angles.

5. Challenge: False. Substitute any values for x, y and z and see if

a contradiction arises. For example, if x = 1,y = 2 and z = 3 in the first triangle, then in the second triangle, corresponding sides will be x + 1 = 2, y + 1 = 3 and z + 1 = 4. To find whether or not the sides are in proportion we look at x ____ x + 1

, y ____ y + 1

and z ____ z + 1 , which should be equal ratios.

But 1 __ 2 ≠ 2 __

3 ≠ 3 __

4 , therefore the triangles are not similar

since their sides are not in proportion.

Exercise 7.71. (a) x = 12 units; y = 4 units (b) x = 75º; y = 60º; z = 45º (c) x = 4.5 units (d) x = 24 units; y = 6 units

2. 12 units

3. Challenge: 2.7 m

Activity 7.31. G

H

J

I

4 cm

4 cm

2 cm

2 cm

2. Students discuss their diagrams.

Exercise 7.81. 4 cm

4 cm

4 cm

4 cm

2. 6 cm

6 cm

6 cm

6 cm

28

Exercise 7.111. Congruent by either SASAS or ASASA.

2. Congruent by either SASAS or ASASA.

3. Not congruent.4. Congruent by SASAS.

5. Not congruent. 6. Not congruent.

7. Congruent by either SASAS or ASASA.

8. Congruent by either SASAS or ASASA.

9. Congruent by either SASAS or ASASA.

10. Congruent by either SASAS or ASASA.

Exercise 7.121. (a) x = 125º; y = 55º (b) x = 45º; y = 35º (c) x = 75º; y = 100 º (d) x = 95º; y = z = 108º (e) x = 8.5 units; y = 2.8 units; z = 117º

Revision exercise1. A

B C132° 25°

4 cm

2. D

E F

5 cm

5 cm

45°

A = 23º

3. (a) Not congruent. (b) RHS (c) Not congruent. (d) SAA

4. Three pairs of corresponding angles are equal; three pairs of corresponding sides are proportional; two pairs of corresponding sides are in the same proportion and one included angle is equal.

5. A

B C

D4 cm

4 cm

2 cm2 cm

55°

55°125°

125°

6. Congruent by SASAS.

7. A =76º; B = 110º; C = 64º; D = 110º

8. QT = RS = 6 cm; QR = TS = 2 cm

3. A

B

D

C

4 cm

4 cm

2 cm

2 cm

4. P

Q

S

R

55 mm

55 mm

35 m

m

35 mm

Exercise 7.91. A

B C

D

60° 120°

120° 60°

4 cm 2. P

Q R

S

3 cm3

cm

6 cm

6 cm45° 135°

45°135°

3. X

Y Z

W3.5 cm

3.5 cm

3.5 cm

3.5 cm

40°

40°140°

140°

4. x = 144º, so 1 __ 4 x = 36 º

Activity 7.41. E Y

R T55° 75°

5 cm

2. U

V W

X

76°

84°

105°

95°

4.5 cm

4.1 cm

3.5

cm

2 cm

UX = 4.1 cm; U = 84º; X = 105º

Exercise 7.101. D

E G

F

122° 122°

94°

5 cm5 cm

2. W

X Z

Y

90°

100° 100°

70°

3.1

cm3.1 cm

3.9 cm

3.9

cm

XY = 3.9 cm and Y = 70º

3. J

M

K L

4 cm

5 cm72°

95°

124°

4. S V

T U83° 83°

97° 97°

8 cm

9 cm

4 cm

4 cm

M = 69º S = V = 97º SV = 8 cm

5. Construction

29

(d) Each side is 4.2 cm long. (e) Each interior angle is 90 º. (f) A square.

4. Challenge: Students follow the instructions and construct a

matching diagram.

Exercise 8.21. (a) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

9

10

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

A

C' CB

(b) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

9

10

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

P S

Q R

P' S'

Q' R'

(c) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

9

10

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

T

U W

V

T'

U' W'

V'

Chapter 8 PolygonsActivity 8.1(a)

72° 3 cm

(b)

3 cm51.4°

Exercise 8.11. (a)

2.5 cm 45°

(b)

40°

2.5 cm

(c)

2.5 cm 90°

(d)

36°

2.5 cm

2. (a) A pentagon with 5 sides 3 cm in length. Shape not drawn within a circle, but rather each interior angle of 108º has been measured.

(b) An octagon with 8 sides 2 cm in length. Shape not drawn within a circle, but rather each interior angle of 135º has been measured.

(c) A decagon with 10 sides 1.5 cm in length. Shape not drawn within a circle, but rather each interior angle of 144º has been measured.

(d) A nonagon with 9 sides 1 cm in length. Shape not drawn within a circle, but rather each interior angle of 140º has been measured.

(e) A heptagon with 7 sides 25 mm in length. Shape not drawn within a circle, but rather each interior angle of 128.6º has been measured.

3. (a) – (c)

A B

C

D

E

F

G

H7 cm

30

(d) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

9

10

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

D

E

G

F

D'G'

F' E'

(e)

M

LK

J

K'

J'

L'

M'

y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

9

10

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

2. (a) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

A F

EB

C D

A'

B'

C'D'

E'

F'

(b) A(–5,3), B(–4,1), C(–5,–1), D(–7,–1), E(–8,1), F(–7,3)

3. (a) A(–3,–3), B(–3,0), C(–1,0) (b) A(3,0), B(2,2), C(4,2) (c) A(–2,–2), B(–4,–4), C(1,–4) (d) A(0,–6), B(3,–3), C(0,–3)

4.

A B

CD

D''

C'' C''' D'''

B''' A'''

B' C'

D'A'

A'V B'V

C'VD'V

(b)

(d)

(a)

(c)

y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

Activity 8.21. Original triangle: A(–7,6), B(–7,1), C(–3,1)

Translated triangle: A(–3,3), B(–3,–2), C(1,–2)

2. If we add the top number of the vector, i.e. 4, to the x–coordinate of every point, and if we add the bottom number of the vector, i.e. –3, to the y–coordinate of every point, we get the new points of the translated triangle.

3. Yes, we could have used the method in (2) to find the new coordinates, without drawing a diagram.

4. (a) Translated vector: P(–4, 4), Q(–8,0), R(0,0) (b) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

P

Q R

P'

Q' R'

(c) Yes.

Exercise 8.31. In each case, the parallelogram is drawn. The

coordinates are as follows: (a) A(–1,3), B(–2,1), C(4,1), D(5,3) (b) A(–5,–2), B(–6,–4), C(0,–4), D(1,–2) (c) A(–2,–3), B(–3,–5), C(3,–5), D(4, –3) (d) A(–7,3), B(–8,1), C(–2,1), D(–1,3) (e) A(0,1), B(–1,–1), C(5, –1), D(6,1)

31

2. y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

J

M

L

K

J''

M''

L''

K''

J'V

M'V

L'V

K'V

J'''

M'''

L'''

K'''

J'

M'

L'

K' (a)(b)

(c)

(d)

3. (a) False. The vector should be ( 5 –8

) . (b) False. The translated decagon will be congruent

(exactly the same in size and shape) to the original decagon.

4. (a) ( 9 ) 0 (b) ( –2

–8 ) (c) ( 6

–4 )

(d) ( 7 –10

) (e) ( –1 –3

) Exercise 8.41. (a) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

P

Q R

Q' P'

(b) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

A D

B CC'

B' C'

D'

E

(c) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

K N

L M

ON'

M'

L'

K'

2. (a) – (b) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

A C

B

C' A'

A'

C'

1

2

3

3. (a) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

E

F G

H

H'

E'

G'

G''

H'' E''

G''' H'''

E'''

(b) The picture does not look like a windmill. He should use point G as the centre of rotation and not point F.

4. y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

S V

T UT'

S'V'

32

Activity 8.3y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

R

S TT'

R'

Exercise 8.51. y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

F

G H

I

F'

G'H'

I'

2. y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

F

G H

I

F'

G'H'

I'

3. (a) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

O'

N'

M'

O

NM

C

centre (1,1)scale factor 2

(b) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

O

N M

N''(–1,–1) M''( , )

(–1,–3)

32

12

O''(–1, )12

centre (–1,–3)scale factor 1

2

(c) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

O

NM

N'''

O'''(–3,12)

M'''(12,6)

centre (0,0)scale factor 3

(d) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

O

N

MO'V(-1, )13

4

M'V( , )114

74

centre (-1,1)scalefactor 3

4

4. (a) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

O(3,3)A

BC

D

A'(–7,3)

B'(–9,–7)

C'(-3,-7)

D'(1,1)

centre (3,3)scale factor 2

33

Revision exercise1.

72° 3 cm 40°

2.5 cm

2.

A

B

CD

E

A'B'

C'

D'

E'

y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

3. y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

P'

Q'

R'

S'

4. (a) ( 5 –3

) (b) H(1,–1), I(0,–3), J(4,–3), K(6,–1)

(b) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

A

BC

DA'( ,3)4

3D'( , )8

383

C'(2, )43

B'(1, )53

5. Centre (–6,7) and Scale factor 2 1 __ 2

Activity 8.4Investigation:

Students will enjoy rearranging the tangram shapes to make the candle and the rabbit. Encourage them to use words such as flip, rotate, translate as they move the tangram pieces. If this task is too difficult for your group, show them the answer diagrams and let them simply match the shapes to form the pictures.

34

5. y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

M P

N OQP'(–2,–4)

M'(–2,–9) N'(1,–9)

O'(1,–4)

6. (a) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

(1,–2)

X

Y

Z

X'

Y'

Z'

(b) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

X

Y

Z

X'(-1 , )14

14

Z'(–1 ,–3 )14

12Y'(–4 ,–2 )1

434

35

4. (a) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y =

2x +

4

y = –x + 5

y = x

– 3

(b) Graph (c) goes in the other direction, from the top left corner to the bottom right corner. This could be because in (a) and (b) the value of m is a positive number and in (c), the value of m is a negative number.

Exercise 9.21. Note: A range of answers is acceptable for the

“other point” column.

Graph x–intercept y–intercept Other point

(a) y = x + 4 (–4,0) (0,4) (1,5)

(b) y = –x – 6 (–6,0) (0,–6) (3,–9)

(c) y = 2x + 1 (– 1 __ 2 , 0) (0,1) (–1, –1)

(d) y = –4x + 7 (7/4,0) (0,7) (2, –1)

(e) y = –x – 5 (–5,0) (0,–5) (1,–6)

(f) y = –x + 4 (4,0) (0,4) (2,2)

2. (a) – (f)

y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y = x

+ 3

y = –x – 2

(–1,2)(0,3)

(3,0) (–2,0)

(–1,–1)(0,–2)

y =

2x –

3

(–1,–5)

(0,3)

( ,0)32

Chapter 9 Coordinate geometryActivity 9.11.

Equation x–value y–value Ordered pair

y = 2x + 5 1 7 (1,7)

0 5 (0,5)

–1 3 (–1,3)

–2 1 (–2,1)

2. y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y =

2x +

5

3. It is a straight line.

4. The graph cuts the y–axis at the point C(0,5).

Exercise 9.11. (a) m = 3, c = 5 (b) m = 2, c = –7 (c) m = –5, c = 2 (d) m = –2, c = 4 (e) m = 3, c = –1 (f) m = 3, c = 4

2. (a) (0,3) (b) (0,–6) (c) (0,–8) (d) (0,3) (e) (0,1) (f) (0,–3)

3. (a) y = x – 3

x–value –2 –1 0 1 2 3 4

y–value –5 –4 –3 –2 –1 0 1

Ordered pair

(–2,–5) (–1,–4) (0,–3) (1,–2) (2,–1) (3,0) (4,1)

(b) y = 2x + 4

x–value –4 –3 –2 0 1 2

y–value –4 –2 0 4 6 8

Ordered pair

(–4,–4) (–3,–2) (–2,0) (0,4) (1,6) (2,8)

(c) y = –x + 5

x–value –5 –4 –3 0 1 3

y–value 10 9 8 5 4 2

Ordered pair

(–5,10) (–4,9) (–3,8) (0,5) (1,4) (3,2)

36

(b) n

t0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

n = t +

3

n =

2t +

1

(c) Sewela (d) After 2 hours they both have planted 5 trees.

Activity 9.2Enter the formula in the formula bar, only once for the first y–value. Then position the cursor over the cell that has the worked out y–value in it. The mouse must hover in the bottom right-hand corner of the cell, until a + appears in the cursor’s place. Hold down the left mouse button and drag down, highlighting the y–value cells that you wish to have calculated for you. The computer will apply the formula used for the first y–value, to calculate the remaining values.

y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y = –x + 4

(–4,0)

(–1,–3)(0,–4)

y = x

– 5

y = x

+ 5

(0,5)

(–2,3)

(–5,0)

(0,–5)

(5,0)

(2,–3)

3. (a) Rate of planting: Sewela

n = 2t + 1

t (hours) 0 1 2 3

Number of trees planted (n) 1 3 5 7

Rate of planting: Msanan = t + 3

t (hours) 0 1 2 3

Number of trees planted (n) 3 4 5 6

Exercise 9.31 (a)

(b)

37

(c)

(d)

(e)

38

(f)

2. (a) – (b)

3. (a)

(b) (i) Hover over chart area and right click. Select: Format Chart Area. Select “Fill Effects”. Select: Two colour shading. Click on the drag down menus for each colour, to choose colours.

(ii) Hover over plot area and right click. Select: Format Plot Area. Select the light blue colour.

(iii) Select the whole chart area by clicking once on it. Right click to display options. Select: Format Chart Area. Click on the Font tab at the top. Select: Comic Sans MS.

(iv) Hover over the chart area and right click. Select: Format Chart Area. Check the box next to “Round corners”.

39

Activity 9.31. (a) – (d) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y = –x

+ 1

y =

–2x

+ 1

y =

–3x

+ 1

y =

–4x

+ 1

2. The graphs are getting steeper.

3. (a) – (d) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y = –x + 1

y = –2x + 1

y = –3x + 1

y = –4x + 1

4. The graphs are getting steeper.

5. They are travelling in a different direction.

Exercise 9.41. (a) Gradient = 2; Graph slopes upwards (b) Gradient = –1; Graph slopes downwards (c) Gradient = 4; Graph slopes upwards (d) Gradient = 0; Graph is horizontal (e) Gradient = – 3 __

2 ; Graph slopes downwards

(f) Gradient = 1 __ 2 ; Graph slopes upwards

2. (a) y–int: 6, x–int: 3, gradient: –2 (b) y–int: –5, x–int: 5, gradient: 1 (c) y–int: 1 __

2 , x–int: –1 __

8 , gradient: 4

(d) y–int: 0, x–int: 0, gradient: 1 (e) y–int: 0, x–int: 0, gradient: 5 (f) y–int: 8, x–int: 0, gradient: 0

3. (a) y = 3x + 5 (b) y = –4x + 8 (c) y = 1 __

2 x – 6 (d) y = 2

(e) y = x (f) y = –5x – 10

4. Graph y–intercept Point 1 Point 2 Equation

A 5 (4,13) (1,7) y = 2x + 5

B 2 (3,2) (6,–1) y = –x + 2

C 0 (1,4) (–3,–12) y = 4xD –8 (6,–2) (3,–1) y = – 1 __ 3 x – 8

5. P: y = –x + 5 Q: y = 2x – 1

R: y = –2 S: y = –2x –4

6. (c) P(–1, 0), Q(0,–3)

7. Answers will vary but any graphs whose value of m is 3, will be parallel.

8. (a) H: y = 1 __ 2 x – 1 and I: y = –2x + 2

(b) The product of their gradients is –1. (c) Obviously there are many correct answers. One

example: The graphs y = x + 1 and y = –x + 3 are perpendicular. The product of the gradients of two perpendicular graphs is always –1.

y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y = x

+ 1

y = –x + 3

Activity 9.41. y = x²

x –3 –2 –1 0 1 2 3

y 9 4 1 0 1 4 9

(x,y) (–3,9) (–2,4) (–1,1) (0,0) (1,1) (2,4) (3,9)

y

x0 1 2 3 4 5 6 7–1–2–3–4–5–6–7

1

2

3

4

5

6

7

–1

–2

–3

–4

–5

–6

–7

y = x2

(0,0)

40

y = 2x² + 1

x –3 –2 –1 0 1 2 3

y 19 9 3 1 3 9 19

(x,y) (–3,19) (–2,9) (–1,3) (0,1) (1,3) (2,9) (3,19)

y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y =

2x2 +

1

3. The graph y = –x² is upside down. The others are right side up.

The larger the coefficient of x², such as fory = 2x² + 1, the narrower the graph is.

The graph y = 2x + 1 sits higher up, in fact its turning point is one unit up from the origin. This would appear to relate to the “+ 1” after the “x²” part of the equation.

Exercise 9.51. (a) y = x² + 1

x –3 –2 –1 0 1 2 3

y 10 5 2 1 2 5 10

(x,y) (–3,10) (–2,5) (–1,2) (0,1) (1,2) (2,5) (3,10)

y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

(–3,10)

(–2,5)

(0,1)

(0,1)

y =

x2 +

1

2. y = –x²

x –3 –2 –1 0 1 2 3

y –9 –4 –1 0 –1 –4 –9

(x,y) (–3,–9) (–2,–4) (–1,–1) (0,0) (1,–1) (2,–4) (3,–9)

y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y = –x2

y = 2x²

x –3 –2 –1 0 1 2 3

y 18 8 2 0 2 8 18

(x,y) (–3,18) (–2,8) (–1,2) (0,0) (1,2) (2,8) (3,18)

y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

9101112131415161718

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y =

2x2

41

(d) y = –2x² + 1

x –3 –2 –1 0 1 2 3

y –17 –7 –1 1 –1 –7 –17

(x,y) (–3,–17) (–2,–7) (–1,–1) (0,1) (1,–1) (2,–7) (3,–17)

y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

–11

–12

–13

–14

–15

–16

–17

y = –2x2 + 1

(e) y = x² – x

x –3 –2 –1 0 1 2 3

y 12 6 2 0 0 2 6

(x,y) (–3,12) (–2,6) (–1,2) (0,0) (1,0) (2,2) (3,6)

y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

9

10

11

12

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y =

x2 –

x

(b) y = –x² – 2

x –3 –2 –1 0 1 2 3

y –11 –6 –3 –2 –3 –6 –11

(x,y) (–3,–11) (–2,–6) (–1,–3) (0,2) (1,–3) (2,–6) (3,–11)

y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

(0,2)

y =

– x2

– 2

(c) y = 2x² – 1

x –3 –2 –1 0 1 2 3

y 17 7 1 –1 1 7 17

(x,y) (–3,17) (–2,7) (–1,1) (0,1) (1,1) (2,7) (3,17)

y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

11

12

13

14

15

1617

y =

2x2 –

1

42

3. (a) y = –x² + 4

x –3 –2 –1 0 1 2 3

y –5 0 3 4 3 0 –5

(x,y) (–3,–5) (–2,0) (–1,3) (0,4) (1,3) (2,9) (3,–5)

(b) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y = –x2 + 4

4. (a) y = –x² –2x + 8

x –3 –2 –1 0 1 2 3

y 5 8 9 8 5 0 –7

(x,y) (–3,5) (–2,8) (–1,9) (0,8) (1,5) (2,0) (3,–7)

(b) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y = –x2 – 2x + 8

Extension activityHandshake puzzle:1. (a) 1 (b) 2 (c) 4 people: 6 handshakes; 5 people:

10 handshakes; 6 people: 15 handshakes.

2. Answers will vary.

(f) y = x² + x – 3

x –3 –2 –1 0 1 2 3

y 3 –1 –3 –3 –1 3 9

(x,y)

y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

9

10

–1

–2

–3

–4

–5

–6

–7

y =

x2 –

x –

3

2. (a) y = –x² + 4x – 4

x 0 1 2 3 4

y –4 –1 0 –1 –4

(x,y) (0,–4) (1,–1) (2,0) (3,–1) (4,–4)

(b) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y = –4

y = –x2 + 4x – 4

(c) Yes, there are four square blocks between the parabola at the line y = –4 which makes 4 m in real life. The bridge is wide enough.

43

Revision exercise1. (a) – (b) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y = x

+ 3

y = – x + 5

(c) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

–3 2

y = – 2x – 3

(d)

y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y =

3x –

7

73

3. (a) Ordered pairs: (2,1); (3,2); (4, 6); (5, 10); (6,15) y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

9101112131415

–1

–2

–3

–4

–5

(b) The graph is beginning to form a parabola.

4. y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y =

x2

–

x1 2

1 2

5. 1 __ 2 (20)² – 1 __

2 (20)

= 200 – 10 = 190 handshakes

44

2. y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

13

23

(0,2)

(1,5)

(0,–1)

(–1,–4)

y =

3x +

1y

= 3x

– 1

(b) They have the same gradient. They are parallel lines

3. A: y = 1 __ 2 x – 2

B: y = –x + 1 C: y = 3x + 3

4.

5. y

x0 1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10

1

2

3

4

5

6

7

8

910

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

(0,–2)

y =

x2 –

2

Line of symmetry: x = 0; TP: (0,–2)