Module 3: Answer Key - Open School

Module 3: Answer Key

Section 1: Geometry

Lesson 1 Vocabulary of Circles 265

Lesson 2 Geometry of Circles and Chords 267

Lesson 3 Angles and Arcs of a Circle 273

Lesson 4 Properties of Tangents 281

Lesson 5 Exploring Polygons 287

Review 291

Section 2: Trigonometry

Lesson 1 The Trigonometric Ratios of Angles 0° ≤ θ ≤ 360° 295

Lesson 2 Related Angles and Solving Trigonometric Equations of Linear Form 301

Lesson 3 Solving Triangles 307

Lesson 4 The Ambiguous Case 313

Review 223

Section 3: Analytic Geometry I

Lesson 1 Circle 329

Lesson 2 Distance Between Points and Lines 337

Lesson 3 Systems of Linear Equations 345

Lesson 4 Solving Systems of Equations Algebraically 351

Lesson 5 Systems of Equations Containing Three Variables 363

Section 4: Analytic Geometry II

Lesson 1 Applications of Systems 369

Lesson 2 Non-Linear Systems 375

Principles of Mathematics 11 Answer Key, Contents 263

Module 3

264 Answer Key, Contents Principles of Mathematics 11

Module 3

Lesson 3 Graphing Linear Inequalities 389

Lesson 4 Quadratic, Rational, and Absolute Value Inequalities in Two Variables 399

Lesson 5 Verifying and Proving Assertions in Coordinate Geometry 415

Review 423

Module 3

Lesson 1

Answer Key

1.

2. a)D

F

O

E

G

l

D

F

E

G

l

D

F

E

G

lO

D

F

E

G

lO

¼ ¼ ¼

» » ¼ » » » » » »

¼ ¼ ¼ ¼ ¼ ¼ ¼ ¼

∠ ∠ ∠ ∠ ∠

∠ ∠ ∠ ∠ ∠ ∠ ∠

suur

a) ACb) AOE, EOD, DOC, AOD, COE

c) ABC, AEC, or ADC

d) AB

e) AB, AD, BE, AC, BD

f) AB, AE, AD, DE, CD, BD, CE, BE, BCg) EBD, ABD, BAD, ABE, DAC, CAB, ADB

h) BCA, ABE, ABD, EBD, DAC, BAD, CBE, BCE, B¼EC

Principles of Mathematics 11 Section 1, Answer Key, Lesson 1 265

Module 3

b)

c) ∠ DOE, ∠ EOG, ∠ FOG, ∠ DOF

d) ∠ DEF, ∠ DGF, ∠ EDG, ∠ EFG

e) B, C

f) i) the interior region bounded by chord and minor arc DE

ii) the interior region bounded by chord and minor arc FG

g) AB, AC

FG

DE

D

F

E

G

l

D

F

E

G

l

D

F

E

G

lO

D

F

E

G

l

266 Section 1, Answer Key, Lesson 1 Principles of Mathematics 11

Module 3

Lesson 2

Answer Key

1. a) A line segment from the centre of the circle to themidpoint of the chord is perpendicular to the chord.

b) Chords equidistant from the centre of the circle to thechord are congruent or are equal in length.

c) Equal chords are equidistant from the centre.d) A line segment from the centre perpendicular to the chord

bisects the chord.

e) The right bisector of a chord passes through the centre ofthe circle.

2. a) CD = 12 units; chords equidistant from the centre of thecircle are equal

b) Because , , and are all radii and OE = 4.2,then OA = OB = 4.2. The diameter will be twice theradius. ∴ AB = 8.4 units.

c) Because chord equals chord , then must be 5as equal chords are equidistant from the centre.

3 a) BC = 3. A line from the centre perpendicular to the chordbisects the chord (i.e., OC bisects AB).

b) AB = 6. AC + BC = 3 + 3 = 6

c) Join OB.

The radius is 5.

d) If the radius of the circle is 5, the diameter is 10, becausethe diameter is twice the length of the radius.

O

B

C

4

3

BC OC OB (Pythagorean Theorem)

OB

OBOB

2 2 2

2 2 2

2

3 4

9 165

+ =

+ =

+ ==

OECDAB

ABOEOBOA


Module 3

4. a)

b)

c)

OE = 8 because it is a radius and ∆ EXO is a right triangleso the problem is solved using the Pythagorean theorem.

5. a) is the perpendicular bisector of and OF = DF,making ∆ OFD an isosceles right triangle with OD = 10 units.

∴

b)

The length of the radius is half the length of thediameter.

EO DE units= ( ) =12

10

OD OF FD

units

OF units

2 2 2

2 2 2

2

2

10

100 2

50

50 25 2 5 2

5 2

= +

= +

=

=

= = =

=

x x

x

x

x

CDOF

E

O

X

8

6EO EX OX

OX

OX

OX

( ) = ( ) +( )

= +( )

= + ( )

= ( )

2 2 2

2 2 2

2

2

8 6

64 36

28

228

4 7

2 7

=

=

=

OX

OX

units OX

OX is the perpendicularbisector of EFEX

12

EF

12

12 6 units

=

= =bg

OF is a radius, which is12

the length of CD

OF =12

(CD) =12

16 8 unitsbg=


12

16 8( ) = units

12

12 6( ) = units

c)

d)

6. CS = SD = 18 cmis the perpendicular bisector of

OC = 24 cmAlso, ∆ CSO is a right triangle

7.

∴ = +

= +

= =

OA

OA

OA = 4 13 cm (radius)

diameter cm

2 2 2

2

8 12

64 144

2 8 13OAb g

Line segment OD from thecentre of a circle is theperpendicular bisector of AB∴ AD = 12 cm∆ OAD is also a right ∆

l

A

O

D B12

8

OS CS OC

OS 18 24

OS 24 18

OS

SB OS

2 2 2

2 2 2

2 2 2

252 6 7

+ =

+ =

= +

= == ++

= +

OB

cm6 7 24

CDAB

OF is the perpendicularbisector of CD

CF DF

CD DF

=

= ( )

= ( )2

2 5 2

OF DF 5 2

DF 5 2 units

= =

=


Module 3

diameter OA cm= =2 8 13( )

8. a)

b) Diameter = 24Chord = 16Chord = 20

The shortest distance between the two chords will be theperpendicular distance between the two chords. If

is joined to form ∆ CXO, CX = 8, OC = 12, and byusing the Pythagorean theorem

∴ = +

= +

the total distance OX OY

units4 5 2 11

OF 12 (radii)

FY12

20 10

=

= =bgOY FY OF

OY 12

OY 1

OY

( ) +( ) = ( )

( ) + =

+ =

( )

2 2 2

2 2 2

2 2

2

10

100 44

==

=

44

2 11OY

CX OX OC

OX

OX

OX

Similarly, by joining OF to form right OYF

b g b g b gb g

2 2 2

2 2 2

2 2 2

8 12

12 8

4 5

+ =

+ =

= −

=

∆

OCXY

EFCD

AB

l

E

O

Y F

A B

XC D


Module 3

(20) = 10

CX OX OC

8 OX 12

OX 12 8

OX

( ) +( ) = ( )

+( ) =

= −

=

2 2 2

2 2 2

2 2 2

4 5

9. Since chords and are equal, they are equidistantfrom the centre meaning OE ≅ OF. Since and

, ð OEP ≅ ð OFP = 90°. OP is the identity side to∆ OEP and ∆ OFP. The two triangles are congruent byhypotenuse — leg (H.L.) and because ð EPO and ð FPO arecorresponding parts, OP bisects ð APD.

10. Perimeter of AHIBP AP AH HI IB PB

312

8 612

6 4

3 4 6 3 4

20 units

= + + + +

= + + + +

= + + + +=

bg bg

OF PD⊥OE PA⊥

DCAB


Module 3

= + ( )+ + ( )+312

8 612

6 4

∠ ∠

∠ ∠∠


Module 3

Notes

Module 3

Lesson 3

Answer Key

1. a) any angles that measure less than 90° are acute angles∠ AOB, ∠ BOC, ∠ DOC, ∠ DOE

b) any angles that measure between 90° and 180° are obtuse∠ AOC, ∠ BOD, ∠ COE, ∠ AOE

c)

d) ∠ BOC = 180° – 75° – 40° = 65°

e) ∠ AOB = 75° Because ∠ AOB and ∠DOE are verticallyopposite angles, ∠ DOE = 75°.

f) Any arc with a measure less than 180° is a minor arc

g) The largest central angle will be 360° – (smallest centralangle). Since the smallest central angle is 40°, ∠ CAD(major) = 320°.

2. a) Yes, is a semicircle. is a diameter so ∠ C

is 90°.AB

l

E

O

B

A

C

D

45 °

45°

35 °

25°ACB

l

E

B

A

C

D

40°

65°

75°105°

75°

AB, AC, BC, BD, CD, CE, DE, EA

EAB, EDB or ECB, DEA, DCA or DBA


b) ∠ ABE and ∠ ADE

c) The inscribed angle ABE subtended by has a measureof 35°. The measure of central angle AOE subtended by

is twice the measure of ∠ ABE. ∴ ∠ AOE = 70°.

d) It is indicated that the two sides are congruent in ∆ CBA.Because the angle at C measures 90°, then base anglesCBA and CAB are 45°.

e) ∠ CBD and ∠ CAD are supplementary because CBDA is acyclic quadrilateral.

∠ CBD = ∠ CBA + ∠ DBA∠ DBA = 60° (given)∠ CBD = 45° + 60° = 105°∠ CAD = 75°∠ BAD = ∠ CAD – ∠ CAB

∴ ∠ BAD = 75 – 45° or 30°

f) subtends the inscribed angle BAD and the centralangle BOD. Because ∠ BAD = 30°, then the central ∠ BODis 60°, as the measure of the central angle is alwaysdouble the inscribed angle.

3. a) The radius of the circle is 6.5 because OC is a radius.Because is a diameter, its length is 2(OC) or 2(6.5) =13 units.

b) is a diameter subtending ∠ C so its measure is 90°.Because the triangle is right angled

c)

d) Area of circle

units

=

=

=

π

π

r2

2

2

6 5

132 7

.

.

b g

Area of ABC AC BC

units

∆ =

=

=

1212

5 12

30 2

b gb gbgbg

BC AC AB

BC 13

BC

BC

BC

( ) +( ) =

( ) + =

= −

==

2 2 2

2 2 2

2 2 2

2

5

13 5

144

112 units

AB

AB

BD»

AE»

AE»


Module 3

=

=

1212

( )( )

( )( )

AC BC

5 12

(6.5)2

e)

Do not approximate π if the exact value is required.

4. Let the diameter = c units.Because is a diameter, the inscribed angle C = 90°.∴ c2 = a2 + b2 by the Pythagorean theorem

5. a)

∠ + ∠ = °

° + ∠ = °

∠ = °

∠ = ∠

∠

DAB BCD

BCD

BCD

DOB BCD

D

180

90 180

90

2

OOB or= ⋅ °

= °− ° = °

2 90 180

180 50 130x

∴ + =

=

69 180

111

y

y

o

o

∴ z = 69°

Opposite angles ofa cyclicquadrilateral aresupplementary

∠ = ° + = °

∠ = ∠

= ( ) °

∠ + ∠

AOC

ABC AOC

or

ABC A

50 88 138

1212

138 69

DDC = °180

Area of the circle

but

so

=

= FHGIKJ=

= +

=+

π

π π

π

r

Ac c

c a b

Aa b

2

2 2

2 2 2

2 2

2 4

4

d i

Radius =c2

AB

C

units

=

=

=

2

2 6 5

13

π

π

π

r

.b g


Module 3

(6.5)

A= πc2

2

= πc2

4

A= πa2 + b2( )

4

b) i) Draw up an equation to solve for x.

∠ A + ∠ C = 180° because the opposite angles of a cyclicquadrilateral are supplementary

∴ ∠ C = 30°

ii) ∠ A = (5x)° or 5 x 30° = 150°

iii) ∠ D = 180° – ∠ B or 180° – 100° = 80°

c) = 180° because is a diameter

i)

ii)

iii)

iv) ∠ ( ) = °−∠

∠ = ∠ = °

BOC major BOC

BOC AOD Vertically Opp

360

43 oosite Angles

BOC major or

( )∴ ∠ ( ) = °− ° °360 43 317

∠ = −( )°

= ⋅ °− °

= °

AOD 2 1

2 22 1

43

x

∠ = +( )°

= ⋅ ° + °

= °

DOB 6 5

6 22 5

137

x

2 1 6 5 180

8 4 180

8 180

22

x x

x

x

x

−( )° + +( )° = °

+( )° = °

° = °

° = °

AB∠AOB

5 180

6 180

30

x x

x

x

( )° + ° = °

( )° = °

° = °


Module 3

d) i) The sum of the measures of the central angles in acircle is 360°.

ii)

iii)

e) i)

ii) Similarly,

iii) ð = − ð

= −

=

Q So

o o

o

180

180 120

60

ð + ð =

ð + ð =

ð =

ð =

ð =

S Q

S S

S

S

S

o

o

o

o

o

18012

180

32

180

3 360

120

Because opposite angles ofa cyclic quadrilateral aresupplementary.

ð + ð =

∴ + ð =

ð =

P R

R

R

o

o o

o

180

114 180

66

∠ ( ) = ∠ + ∠

= + + += ⋅ + + ⋅ +

=

AOC major AOD DOC

5 8 4 4

5 20 8 4 20 4

1

x x

000 8 80 4

192

+ + +

= °

ð = + +

= + +

= + +

=

DOB

o

4 4 3

4 20 4 3 20

80 4 60

144

x x

bg bg

ð + ð + ð + ð =+ + + + + + =

+ =

=

=

AOB BOC COD DOA o

o

o

360

5 8 3 4 4 5 8 360

17 20 360

17 340

20

x x x x

x

x

x


Module 3

= (20) + 4 + 3(20)

∠ ∠ ∠ ∠

∠

∠ ∠

∠

∠

∠

∠

∠

∠

∠

∠

∠

∠∠

f) i)

ii)

iii) Similarly,

6. a) The circular face of the hydro meter is divided into 10equal arc lengths.

∴ The central angle subtended by each arc length is

.

b) When the dial moves from 3 to 8, it goes through 5 arclengths or 5 × 36° = 180°.

7. a) In 60', the minute hand moves 360°

∴ In 1', it moves or 6°

∴ In 55', it moves 55 × 6° or 330°

or • 360 = 330°5560

36010

36or °

ð + ð =

+ ð =

+ ð =

ð = −

=

D B

B

B

B

o

o

o

o

o

180

13 180

13 5 180

180 65

115

x

ð = +

= +

= +

=

C

o

2 10

2 5 10

50 10

60

2

2

x

bg

Opposite angles of a cyclicquadrilateral aresupplementary.

x cannot be negativebecause it is a measure

ð + ð =

∴ + + + =

+ =

=

== ±

=

A C o

o

180

4 20 2 10 180

6 30 180

6 150

25

5

5

2 2

2

2

2

x x

x

x

x

x

x


Module 3

∠ ∠

∠

∠ ∠

∠

∠

∠

= (5)2 + 10

13 5⋅

36010

b)

8. a) Number of People Percent

b) Item Percent Central Angle Measurei) Beef 35% 0.35 x 360° = 126°

ii) Pork 15% 0.15 x 360° = 54°

iii) Lamb 5% 0.05 x 360° = 18°

iv) Chicken 25% 0.25 x 360° = 90°

v) Fish 20% 0.20 x 360° = 72°Total = 360°

l

Beef 35%

Fish 20%Chicken 25%

Lamb 5%

Pork 15%

i) Beef

ii) Pork

iii) Lamb

iv) Chicken

v) Fish

Total

5601600

100 35%

2401600

100 15%

801600

100 5%

4001600

100 25%

3201600

100 20%

100%

× =

× =

× =

× =

× =

In 12 hours the hour hand moves 360 .

In 1 hour, it moves 36012

or 30 .

In 212

hours, it moves 30 2.5 or 75 .

o

o

o×


Module 3


Module 3

Notes

Module 3

Lesson 4

Answer Key

1.

2.

Similarly, ∠ DEF = ∠ FDA and ∠ FDA = 60°

Because the sum of the measures of the angles in ∆ DAF is180°, ∠ BAC = 60°.

3. a) AP = BP because tangents drawn from the same externalpoint to the circle are equal. Because AP = 5, BP = 5.Similarly, BR = RC = 5 and AQ = CQ = 9.

∴ PQ = QA + AP = 9 + 5 = 14 units

PR = PB + BR = 5 + 5 = 10 units

QR = CQ + RC = 9 + 5 = 14 units

Total: 38 units∴ The perimeter is 38 units.

The angle between tangent andchord equals the inscribed angle onthe opposite side of the chord.

∠ = ∠

∴ ∠

DFA DEF

DFA = 60o

The angle between tangent andchord equals the inscribed angle onthe opposite side of the chord.

∠ = ∠

∴ ∠

ACD CBA

CBA = 50o

∠ == ⋅

=

ACD

o

22 25

50

x

Tangent is a straightline

∠ + ∠ + ∠ =

− + + =

+ =

=

=

BCE BCA ACD o

o

o

o

o

180

3 5 60 2 180

5 55 180

5 125

25

x x

x

x

x


Module 3

b) , , and because the radius isperpendicular to the tangent at the point of contact.

This means that QAO + OCQ = 180° making AOCQ a cyclic quadrilateral. If AOC = 140°, then

AQC = 180° – 140° = 40°.

Similarly, PAO + PBO = 180° making APBO acyclic quadrilateral.

If AOB = 110°, then APB = 180° – 110° = 70°.Similarly, RBO + RCO = 180° making COBR acyclic quadrilateral.

If BOC = 110°, then BRC = 180° – 110° = 70°.

The measures of the angles of PQR are 70°, 40°, and70° respectively.

4.

Angles on a line add up to 180º.

∴ ∠ =∠ + ∠ + ∠ =

∴ + ∠ + =∠ =

3 42

1 2 3 180

56 2 42 180

2 82

o

o

o o

o

o

Because the angle between thetangent and chord is equal to theinscribed angle on the oppositeside of the chord.

∠ = ∠3 BFD

∠ = =112

112 56o o

Because the angle betweenthe tangent and the chordis half the measure of thecentral angle.

( )∠ + ∠ =

∠ + =∴∠ =

∠ = ∠

o

o o

o

FOB FOB major 360

FOB 248 360

FOB 112

11 FOB

2

∠ = − ∠ − ∠

= − −

=

AOB AOC BOC

360 140 110

110

o

o o o

o

360

OB PR⊥OC QR⊥AO QP⊥


∠ ∠

)(

∠∠

∠ ∠∠∴

∴

∠ ∠

∠∠∠

∠

bevc

Typewritten Text

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bevc

Typewritten Text

bevc

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Typewritten Text

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Typewritten Text

5. a) ∠ PAC = ∠ ABP Because the angle between the tangent and chord is equal to the inscribed angle on the opposite side of the chord.

∴ Because ∠ ABP = 35°, ∠ PAC = 35°.

b) ∠ PAB + ∠ ABP + ∠ BPA = 180°Because the measures of the three angles add up to 180°

∴ ∠ PAB + 35° + 27° = 180°

In ∆ BCA, ∠ ABC + ∠ ACB + ∠ CAB = 180°

Alternate Solution

∠ PAC + ∠ APC + ∠ PCA = 180°Because the sum of the angles in a triangle is 180º

c) ∠ BAQ = ∠ ACB The angle between the tangent and chord AB is equal to theinscribed angle ACB.

∴ ∠ BAQ = 62°

d)

If ∠ BAC = 83°, then ∠ COB = 2 • 83° or 166°.

The measure of inscribed angleis half the measure of thecentral angle.

∠ = ∠BAC COB12

∠∠

∠ ∠∠

∠

35º + 27º + PCA = 180° PCA = 118°

ACB + PCA = 180° (Angles on a line) ACB + 118° = 180°

ACB = 62°

35 83 180

62

o o o

o

ACB

ACB

+ ∠ + =

∠ =

∠

∠ ∠ =

+ ∠ =

∠ =

PAB = 118

PAC+ CAB 118

CAB 118

CAB 83

o

o

o o

o

35


Module 3

6. Draw a line segment (radius) from the centre of the circle toa point on the circle. Then construct a line perpendicular tothis radius at the point on the circle.

7. ∠ RPD = ∠ PCD

∠ PCD = ∠ BAD

∴ ∠ BAD = ∠ RPD Both equal ∠ PCD.

But these are alternate interior angles and so AB || RQ.

8.

Because tangents drawn to the same circle from the sameexternal point are congruent, PQ = PR. The radii areperpendicular to the tangents at the point of contact, so∠ Q = ∠ R and is the identity side. ∆ POQ ≅ ∆ POR byhypotenuse-leg. This makes ∠ QPO = ∠ RPO because ofcorresponding angles. These two angles have the samemeasure so you can conclude that bisects ∠ QPR.OP

OP

Join OQ and OR.

l

B

O

A

R

Q

P

Exterior ∠ PCD has the samemeasure as the interioropposite ∠ BAD.

∠ RPD, the angle between thetangent and chord is equal tothe inscribed ∠ PCD on theopposite side of chord DP.


Module 3

Alternate Solution:

Join OQ and OR

Because tangents to the same circle from the same externalpoint are congruent, PQ = PR. Radii QO and RO are equal.PO is the identity side. ∆ POQ ≅ ∆ POR by SSS. ∠ QPO = ∠ RPO because of corresponding angles. Since thetwo angles are equal, OP bisects ∠ QPR.


Module 3


Module 3

Notes

Module 3

Lesson 5

Answer Key

1. (a) and (b) are polygons with (a) being convex. (c) is not apolygon because one side is not a line segment.

2. a) b) c)

3. a) 2b) none

c) 9

d) 14

4. The sum of the measures of the interior angles of a hexagonis (6 – 2)(180°) = 720°.The three angles indicated by the marking are supplementsof the given angles.

The Supplement of y is: 720 – 138 – 120 – 95 – 150 – 120 =97°.

y = 180 – 97 = 83°

15 0°

85 °

12 0°

42°

60 °

y

95 °

138 °

138 °

1 20°

l

l

l

l

l

l

ll

ll

ll

ll


Module 3

5. a) Sum = (n – 2)180°= (15 – 2)180° = 2340°

b) 360° regardless of the number of sides

6. (n – 2)180° = 2700° where n is the number of sides

7. a) Measure of each interior angle of a regular polygon =

b)

8. Exterior angle measure =

a)

b)72

360

72 360

5

=

==

o

o

sides

nn

n

45360

45 360

8

=

==

o

o

sides

nn

n

360o

n

1602 180

160 180 360

20 360

18

=−

= −− = −

=

n

nn n

n

n

o

sides

1202 180

120 180 360

360 60

6

=−

= −==

n

nn n

n

n

o

sides

nn

− 2 180o

n n n− = − = =22700180

2 15 17 sides


(

(

(

)

)

)

bevc

Typewritten Text

9. a)

b)

10.

The angles are 70°, 90°, 110°, 130°, 150°, and 170°.

11. a) There are five diagonals.

b) BE

CE

AD

AC

BD

= − + − =

= − + − =

= − + − =

= − + − =

= − + − =

2 3 6 0 37

4 3 5 0 26

0 5 2 2 5

0 4 2 5 5

2 5 6 2 5

2 2

2 2

2 2

2 2

2 2

x x x x x x n

x

x

x

x

+ + + + + + + + + + = −

+ = −

+ ==

=

20 40 60 80 100 2 180

6 300 6 2 180

6 300 720

6 420

70

o

o

o

15 2 180

15156

36024

−=

=

oo

oo

(interior)

15(exterior)

sum of the exterior anglesexterior angle measure

exterioro

o

n=

=3609

40

sum of the interior anglesof a regular polygon

interior angle measure

interioro

o

n=

−=

9 2 1809

140


Module 3

( )( )

( )

( )

( )

( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( )

bevc

Typewritten Text


Module 3

Notes

Review

Answer Key

1. Refer to Lesson 1 of Module 3.

2. a) AP = OP and APO is a right angle triangle, so use a

2+ b

2= c

2

b)

c) OP is a perpendicular bisector of AB

d)

3. a) AB = 2 • 6.5 = 13

b) AB AC BC

BC

BC

BCunits BC

2 2 2

2 2 2

2

2

13 5

169 25

14412

= +

= +

− =

==

O

B

A C5

6.5

ll

llll

AP OP

P

=

∴ =A 4 2

AB 2 AP= ⋅

=

=

2 4 2

8 2

radius is 12

diameterOC12

AC=

= ⋅

=

12

16

8

A

O

P

B

C

8

8

ll

ll

AO AP OP

OP

2 2 2= +

= +

=

=

=

=

8

64 2

32

4 2

4 2

2 2 2

2

2

x x

x

x

x

Principles of Mathematics 11 Section 1, Answer Key, Review 291

Module 3

( )

∆

bevc

Typewritten Text

c) r

4. a) BOD = 2x

b) COD = 2y

c) BAC = x + y

d) BOC = 2x + 2y

5.

6. a) ODE = 90° OD CE — radius tangentline at point of contact

b) ADB = 82° inscribed angle is one-half the central angle subtended by thesame chord

c) BOD = 2 • 48 = 96° inscribed angle is one-halfcentral angle subtended by thesame chord

d) OBD = base angle of isosceles triangle

e) ∠ = ∠ ∠∠ =

oAOD 360 – AOB – BODAOD 100º

180 962

42o o

o−=

∠ = =

∠ =−

=

∠ =−

=−

=

118

360 45

2180 45

267

12

32 180 8 2 180

8135

of o o

o o o

o oon

n

A

units

∆ =

= ×

=

1212

5 12

30 2

bh

292 Section 1, Answer Key, Review Principles of Mathematics 11

Module 3

( )

( ) ( )

∠

∠

∠

∠ ⊥ ⊥

∠

∠∠

∠

d) area of circle =2

= (6.5)2

= 42.25 units2

ππ

π

f) ∠ =ADE

=12

12

AOD ∠

( )100º = 50°

Angle between tangent andchord is equal to half thecentral angle subtended by thechord.

bevc

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bevc

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Typewritten Text

g)

h)

i) ∠ BOD (major) = 360° – 96° = 264°

7. AD = BD

∠ 1 = 70°

∴ ∠ ADB = 180° – 70° = 110°

∠ 4 = ∠ 5 (AD = BD)

∴ ∠ 4 + ∠ 5 = 70°

2∠ 5 = 70°

∠ 4 = ∠ 5 = 35°

∠ 2 = ∠ 4 = 35° (chord tangent)

∠ 3 = 180° – (70° + 35°) = 75°

8. a) ∠ ODC = 40° (base angle of isosceles triangle)

b) ∠ OCB = 180° – 80° – 40° = 60° (opposite angles of a cyclic quadrilateral are supplementary)

c) ∠ ADC = 180° – 130° = 50° (opposite angles of a cyclicquadrilateral are supplementary)

d ) ∠ OCD = 40° (base angles)

9. ∠ 3 = ∠ 9 (tangent-chord theorem)

∠ 2 = ∠ 1 (angle is bisected)

∠ 2 + ∠ 3 = ∠ 1 + ∠ 9

but ∠ 7 = ∠ 1 + ∠ 9 (exterior angle equals sum of twointerior opposite angles)

∠ = °− ° = °ODA 90 50 40

Angle between tangent andchord is equal to the inscribedangle on the other side of thechord.

or : ∠ = ∠ = °BDC BAD 48

Angle between tangent andchord is equal to half thecentral angle subtended by thechord.

∠ =

= °( ) = °

BDC BOD1212

96 48


Module 3

bevc

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bevc

Typewritten Text

bevc

Typewritten Text

∴ ∠ 7 = ∠ 2 + ∠ 3

∴ ∆ BCE is isosceles

BC = CE (definition of isosceles)

10.

11.

12.

13.

selgna lacitrev)a.41

b)

c)

=

AOC is supplementary to COB

AOC o o

o

180 63

117

=

COBo

2 19 25

63

− = +==

AOD COB4 13 2 25

2 38

19

x x

x

x

36030

30 360

12

oo

o

sides

nn

n

=

==

n − = −

=

2 180 19 2 180

3060

oo

o

n

nn n

n

n

−=

− =

==

2 180168

180 360 168

12 360

30

o

o

o

sides

n

n

n

n

− =

− =

− ==

2 180 3960

23960180

2 22

24

o o

o

o

sides

11 scitamehtaM fo selpicnirPweiveR ,yeK rewsnA ,1 noitceS492

Module 3

∠ ∠

∠

∠ ∠

∠

( )

( )

( )

•

bevc

Typewritten Text

bevc

Typewritten Text

Lesson 1

Answer Key

1. a)

b) r x y

yr

xr

yx

= +

= − + −

=

=

= = −

= = −

= = −−

=

2 2

2 25 12

169

13

12135

13125

125

sin

cos

tan

θ

θ

θ

x

y

P( 5 , 12)

r12

5

r x y

yr

xr

yx

= +

= − +

=

=

= =

= = −

= =−

2 2

2 25 12

169

13

1213

51312

5

sin

cos

tan

θ

θ

θ

x

yP( 5,12)

r

12

5


Module 3

θ

θ

( )

( )( )

– –

–

bevc

Typewritten Text

bevc

Typewritten Text

c)

d) r x y

yr

xr

yx

= +

= − + −

=

=

= = −

= = −

= = −−

=

2 2

2 24 3

25

5

354

534

34

sin

cos

tan

θ

θ

θ

x

y

P( 4, 3)

r3

4

r x y

yr

xr

yx

= +

= + −

=

=

= = −

= =

= = −

2 2

2 25 12

169

13

12135

13125

sin

cos

tan

θ

θ

θ

x

y

P(5, 12)

r 12

5


Module 3

θ

( )( )

( )( )

−

θ

bevc

Typewritten Text

bevc

Typewritten Text

bevc

Typewritten Text

e)

f) r x y

yr

xr

yx

= +

= − +

=

=

= =

= = −

= =−

= −

2 2

2 28 15

289

17

1517

81715

8158

sin

cos

tan

θ

θ

θx

yP( 8 ,15)

r

15

8

sin

cos

tan

θ

θ

θ

= = −

= =

= = −

yr

xr

yx

725

2425

724

r x y= +

= + −

=

=

2 2

2 224 7

625

25

x

y

P(24, 7)

r 7

24


Module 3

( )( )

( )( )

θ

θ

bevc

Typewritten Text

bevc

Typewritten Text

g)

h) r x y

yr

xr

yx

= +

= +

=

= = =

= = =

= =

2 2

2 25 3

34

334

3 3434

534

5 3434

35

sin

cos

tan

θ

θ

θ

x

y

P(5, 3)

r

3

5

5

r x y

yr

xr

yx

= +

= + −

=

=

= = −

= =

= = −

2 2

2 28 15

289

17

15178

17158

sin

cos

tan

θ

θ

θ

x

y

P(8, 15)

r

15

8


Module 3

θ

θ( )

bevc

Typewritten Text

i)

j) r x y

yr

xr

yx

= +

= − + −

=

= = − = −

= =−

=−

= =−−

=

2 2

2 23 7

58

758

7 5858

358

3 5858

73

73

sin

cos

tan

θ

θ

θ

x

y

P( 3, 7)

r

3

7

sin

cos

tan

θ

θ

θ

= = − = −

= = =

= = − = −

yr

xr

yx

2

2 17

1717

8

2 17

4 1717

28

14

r x y= +

= + −

=

=

2 2

2 28 2

68

2 17

x

y

P(8 , 2 )

r

8

2


Module 3

θ

θ

−−

−

−

( )( )

( )

bevc

Typewritten Text

bevc

Typewritten Text

2. a) cos 181° = –0.99985x-coordinate is negative in Quadrant III

b) sin 255° = –0.96593y-coordinate is negative in Quadrant III

c) tan 340.4° = –0.35608y-coordinate is negative in Quadrant IV

d) sin 261° = –0.98769y-coordinate is negative in Quadrant III

e) cos 224° = –0.71934x-coordinate is negative in Quadrant III

f) tan 152.2° = –0.52724x-coordinate is negative in Quadrant II

g) cos 121.5° = –0.52250x-coordinate is negative in Quadrant II

h) cos 332° = 0.88295x-coordinate is positive in Quadrant IV

i) tan 271.6° = –35.80055y-coordinate is negative in Quadrant IV

3. a) Quadrants I and II

b) Quadrants I and IV

c) Quadrants I and IIId) Quadrants III and IV

e) Quadrants II and III

f) Quadrant IV


Module 3

Lesson 2

Answer Key

1. a)

b)

c) θ r = −

=

360 352

8

o o

o

x

y

θr = 8 °

352°

θ θ

θr

r

= −

= −

=

180

180 120

60

o o

o

x

y

θ r = 60° 120°

θ θ

θr

r

= −

= −

=

180

180 98

82

o o

o

y

θ r = 82° 98°


Module 3

d)

2. a)

b)

c)

d)

Because sin θ > 0 inQuadrant II

θ r = −

=

==

180 150

30

150 3005

o o

o

o osin sin.

Because tan θ < 0 inQuadrant II

θ = −

=

= −= −

o o

o

o o

180 162.8

17.2

tan162.8 tan17.20.30955

r

Because cos θ < 0 inQuadrant II

θ r = −

=

= −= −

180 123 8

56 2

123 8 56 2055630

o o

o

o o

.

.

cos . cos ..

θ

θ

= −

=

= >=

o o

o

o o

180 136.4

43.6

sin 136.4 sin43.6 Because sin 0 in0.68962 Quadrant II

r

θ θ

θr

r

= −

= −

=

180

263 180

83

o

o o

o

y

θ r = 83°

2 6 3°


Module 3

e)

f)

3. a) θr = sin–1

0.78615= 51.8°

sin θ > 0 in Quadrants I and IIIn Quadrant I, θ = 51.8°In Quadrant II, θ = 180° – 51.8º = 128.2°

b) θr = cos–1 0.43214= 64.4°

cos θ > 0 in Quadrants I and IVIn Quadrant I, θ = 64.4°We only evaluate the angle on I since angles in Quadrant IV are > 180º

c) θr = tan–1 1.28728= 52.2°

tan θ > 0 in Quadrants I and IIIIn Quadrant I, θ = 52.2°

d) θr = cos–1 0.81673= 35.2°

cos θ < 0 in Quadrants II and IIIIn Quadrant II, θ = 180° – 35.2° = 144.8°

e) θr = tan–1

2.81763= 70.5°

tan θ < 0 in Quadrants II and IVIn Quadrant II, θ = 180° – 70.5° = 109.5°

Because tan θ < 0 inQuadrant II

θ r = −

=

= −= −

180 1002

798

1002 798555777

o o

o

o o

.

.

tan . tan ..

Because cos θ < 0 inQuadrant II

θ = −

=

= −= −

o o

o

o o

180 98.3

81.7

cos98.3 cos81.70.14436

r


Module 3

4. a) r = cos–1

= 48.2°

cos < 0 in Quadrants II and IIIIn Quadrant II, = 180° – 48.2° = 131.8°

b) tan – 2 = 5tan = 7

r = tan–1 7= 81.9°

tan > 0 in Quadrants I and IIIIn Quadrant I, = 81.9°

c)

tan = 10r = tan–1 (10)

= 84.3°tan > 0 in Quadrants I and IIIIn Quadrant I, = 84.3°

d) 5 cos – 2 = 0

cos > 0 in Quadrants I and IVIn Quadrant I, = 66.4°

cos

cos

.

θ

θ

=

=

=

−

25

25

664

1r

o

tan θ2

5=


Module 3

23

θ

θθ

θθθ

θθ

θθ

θ

θ

θθ

θ

bevc

Typewritten Text

bevc

Typewritten Text

e) 5 tan + 4 = 0

tan < 0 in Quadrants II and IVIn Quadrant II, = 180° – 38.7° = 141.3°

f)

tan > 0 in Quadrants I and IIIIn Quadrant I, = 80.5°

g)

cos < 0 in Quadrants II and IIIIn Quadrant II, = 180° – 75.5° = 104.5°

h) 4 tan – 7 = 5 tan – 6–tan = 1tan = –1

r = tan–1 (1) = 45°tan < 0 in Quadrants II and IVIn Quadrant II, = 180° – 45° = 135°

2 112

21214

14

755

1

cos

cos

cos

cos

.

θ

θ

θ

θ

+ =

= −

= −

=

=

−r

o

tan

tan

tan

tan

.

θ

θ

θ

θ

61 0

61

6

6

805

1

− =

=

=

=

=

−r

o

tan

tan

.

θ

θ

= −

=

=

−

45

45

387

1r

o


Module 3

θ

θθ

θθ

θ

θθ

θθ

θθ

θ

bevc

Typewritten Text


Module 3

Notes

Lesson 3

Answer Key

1. a)

b)

c)

d)

( ) ( )

2 2 2

2 o2

2 cos A (SAS)

4.5 10 2 4.5 10 cos110

151.0318

151.0318 12.29

a b c bc

a

= + −

= + −

=

= =

c a b ab2 2 2

2 2 2

2 2 2

2 2 2

1

2

10 5 7 2 5 7

2 5 7 5 7 10

5 7 102 5 7

03714285714

03714285714

1118

= + −

= + −

= + −

= + −

= −

= −

=

−

cos

cos

cos

cos

cos .

cos .

.

C (SSS)

C

C

C

C

C

C o

∠ = − +

=

=

=

=

=

B (ASA)

A B

o o o

o

o o

o

o

180 150 20

10

200150 10

200 10150

6946

a b

b

b

sin sin

sin sin

sinsin

.

a b

a

a

sin sin

sin

sinsin

.

A B(AAS)

asin 67o o

o

o

=

=

=

=

4273

42 6773

4043


Module 3

)(

)( ( )

)( ( )

)(

)( ( )

bevc

Typewritten Text

bevc

Typewritten Text

2. Distance that sailboat A is from lighthouse is b.

Distance that sailboat, B, is from lighthouse is a.

3.

B

C

A60 km ( s t a r t i n g p o i n t )

( e n d i n g p o i n t )

80 km

1 5 °

N

S

EW

a c

a

a

a

sin sin

sin sin

sinsin

.

A C

m

o o

o

o

=

=

=

=

100310

50

310 10050

39853

∠ = − +

=

=

=

=

=

C

B C

m

o o

o

o o

o

o

180 100 30

50

30310

50

310 3050

20234

o

b c

b

b

b

sin sin

sin sin

sinsin

.


Module 3

)(

bevc

Typewritten Text

bevc

Typewritten Text

bevc

Typewritten Text

From the diagram:

4.

b a c ac

b

2 2 2

2 2

2

198 15 2 2 19 8 152 45

197 4582863

14 05

= + −

= + −

=

=

cos

. . . . cos

.

.

B

m

o

b a c ac

b

2 2 2

2 2

2

80 60 2 80 60 165

19272 88793

1927288793

138 8

= + −

= + −

=

=

=

cos

cos

.

.

.

B

km

o

∠ = −

=

B o o

o

180 15

165


Module 3

A

C

B

45°

15.2 m

19.8 m

Pitcher's Mound

Home

First Base

)( )(

)()( )( )(

bevc

Typewritten Text

bevc

Typewritten Text

bevc

Typewritten Text

5. a)

In BCK, using the sine law

the kite is approximately 1.2 km above the ground

Alternate Solution:

∆ =

=

=

CKIn BCK, sin75º

BKCK

0.965931.287

1.243 CK

CK

CK

CK

o o

o

o

sin.

sin

. sinsin

. .

.

751287

90

1287 7590

1287 0 965931

1243

=

= =

=

BK

BK

BK

o o

o

o

sin sin

sinsin

..

.

25380

3 2580

3 042262098481

1287

=

= =

=

∠ = − + =BKA

In BKA, by the sine law

o o o180 75 25 80

∆

CB A

K ( k i t e )

3 k m

75° 25°


Module 3

)(

)(

)(

∆

∴

bevc

Typewritten Text

b)

In ABK, using the sine law

In CKB, using the sine law

In this case, the kite is 1.6 km above the ground.

CK

CK

o o

o

o

sin.

sin

. sinsin

. ..

751655

90

1655 7590

1655 0 965931

16

=

= = =

BK

BK

o o

o

o

sin sin

sinsin

..

.

25350

3 2550

3 0 42262076604

1655

=

= = =

∠ = − =

∠ = − + =

KBA

BKA

o o o

o o o o

180 75 105

180 105 25 50

C B A

K ( k i t e )

3 k m

75° 25°


Module 3

∆

∆

( )

( )

( )

bevc

Typewritten Text

bevc

Typewritten Text

bevc

Typewritten Text

6. a)

The ship is approximately 154.5 km from port.b)

Thus, the port is S 75° E of the ship.

Alternative Solution:

Since SAP is isosceles, it follows that

BSP = 60° + 15° = 75°Thus, the port is S 75° E of the ship.

∠ = ∠ =− ∠

=−

=

ASP APS

o

1802

180 1502

15

SAP

Since SB NA BSA

ASP

ASP

ASP

BSP

o

o o

o

o

o o

/ / ,

sin sin

.

sinsin

...

.

∠ =

∠=

+

∠ = = =

∠ =

∴ ∠ = + =

60

80

60 90

1545

80 1501545

80 05154 5

02589

15

60 15 75o

SP

SP

o o2 2 280 80 2 80 80 90 60

6400 6400 12800 086603

12800 11085

23885

154 5

= + − +

= + − −

= +==

cos

.

.

ABP

N

S

2 x 4 0

2 x 40 60°


Module 3

∠

∆

∴

)()( )(

)(

)(

)(

bevc

Typewritten Text

Lesson 4

Answer Key

1. a)

undefined as sin B > 1orthe height of the triangle = 10 sin 35

= 5.7because the height > the length of a, the triangle does notexist and so there is no solution.

b) Because b > a, and ∠ B is given to be obtuse, there is onlyone possible triangle.

c)

d)

e) B = 60°, b = 14, c = 15Two solutions when b < c and ∠ B is given.

f) The triangle does not exist because the shorter side isopposite the obtuse angle.

Two solutionsresult whenheight < a < c.

50

18°

38

38

A

B C

There is only one ∆possible when the longerside is opposite the givenangle.43

60°

3 2

A

BC

Because B , the angle is undefisinsin

.= =10 35

5114715


Module 3

2. a)

Because b < a, there are two possibilities:

∠ BA2C has the same sine but it is obtuse

∴ ∠ BA2C = 180° – 48.2° = 131.8°

In ∆ A1CB

When A

BCA

o

1o o o

o

o o

o

o

∠ =

∠ = − +

=

=

=

=

1 482

180 48 2 34

978

978334

3 97834

53

.

.

.

sin . sin

sin .sin

.

d i

c

c

c

−

=

=

=

=

=

=

o

o

o1

o

o1

sin A sin B

sin A sin344 3

4 sin 34sin A

34sin34

A sin3

A 48.2

A 48.2

a b

34°A1A 2

C

B

a = 4b = 3


Module 3

In A2CB

b)

Since x < y, there are two possibilities:Find Y1:

If Y1 = 16.3°, then Y = Y1 = 16.3° and

XY2Z = 180° – 16.3° = 163.7°.

In XY1Z:

In XY2Z:

If Y

then Z

o

o o o

o

∠ =

∠ = − +

=

2 163 7

180 13 163 7

33

.

.

.

If Y

then Z

o

o o o

o

∠ =

∠ = − +

=

1 163

180 163 13

1507

.

.

.

=

=

=

=

1

o1

o

1

o1

sin X sin Y

12 15sin13 sin Y

15 sin 13sin Y

12Y 16.3

x y

13°Y 1Y2

Z

X

y = 15x = 12

When A

BCA

o

o o o

o

o o

o

o

∠ =

∠ = − +

=

=

=

=

2 1318

180 1318 34

14 2

334 14 2

3 14 234

13

.

.

.

sin sin .

sin .sin

.

c

c

c


Module 3

∠

∠ ∠ ∠∠

∆

∆

∆

( )

( )

( )

bevc

Typewritten Text

In XY1Z: In XY2Z:

For z: For z:

c)

Since b > a, only one triangle exists.

Note: You can verify that there is only one value for A.

If there were two triangles, then the second A = 180° – 26.5° = 153.5°.

C = 180° – (34 + 153.5)° = –7.5° (impossible)

side

B C

o o

o

o

c

b c

c

c

c

sin sin

sin sin .

sin .sin

.

=

=

=

=

534 1195

5 119534

7 8

a bsin sin

sin sin

sinsin

.

.

.

A B

A

A

A

C

o

o

o

o o

o

=

=

=

∠ =

∠ = − +

=

4 534

4 345

265

180 34 265

1195

34°A

C

B

a = 4b = 5

o o

o

o

sin X sin Z12

sin13 sin 3.312 sin 3.3

sin133.1

x z

z

z

z

=

=

=

=

o o

o

o

sin X sin Z12

sin13 sin150.712 sin150.7

sin1326.1

x z

z

z

z

=

=

=

=


Module 3

∆ ∆

∠

∠∠

( )

d)

Because r > t, there is only one solution.To find the measure of T:

To find the measure of s:

s r

s

s

s

sin sin

sin . sin

sin .sin

.

S R

o o

o

o

=

=

=

=

12220130

20 12 2130

55

r tsin sin

sin sin

sinsin

.

.

.

R T

T

T

T

S

o

o

o

o o o

o

=

=

=

∠ =

∠ = − +

=

20130

16

16 13020

378

180 37 8 130

122

130°

S

R

Tr = 20

t = 16

( n o t t o s c a l e )


Module 3

∠

( )

e)

Because m > t, there is only one solution:

To find the measure of T:

To find the measure of b:

b m

b

b

b

sin sin

sin . sin

sin .sin.

B M

o o

o

o

=

=

=

=

4 219170

19 4 2170

80

t msin sin

sin sin

sinsin

.

.

.

T M

T

T

T

B

o

o

o

o o o

o

=

=

=

∠ =

∠ = − +

=

11 19170

11 17019

58

180 170 5 8

4 2

170°

T

M

B

t = 11

m = 19

( n o t t o s c a l e )


Module 3

( )

∠

f)

To find the measure of ∠ A:

∠ A does not exist, because sin A > 1

3. a)

Because a < c, there are two possible C values.

The other possible measure for ∠ C = 180° – 23° = 157°.

a csin sin

sin sin

sinsin

.

A C

C

C

C

o

o

o

=

=

=

∠ =

2519

30

30 1925

230

19°

A C

B

a = 2530

2 C 1

a bsin sin

sin sin

sinsin

sin .

A B

A

A

A

o

o

=

=

=

=

4 234

4 342

1118

34°A

C

B

a = 4b = 2


Module 3

b)

Because h > d, there is only one value for D.

c)

Because x > z, there is only one solution.

x zsin sin

.sin

.sin

sin. sin

..

X Z

Z

Z

Z

o

o

o

=

=

=

∠ =

9358

75

75 5893

431

58°

Y

X

x = 9.3z = 7.5

Z

d hsin sin

sin sin

sinsin

.

D H

D

D

D

o

o

o

=

=

=

∠ =

20 5028

20 2850

10 8

28°D

Jd = 20h = 50

H


Module 3

d)

Because b < g, there are two solutions.

For G:g b

sin sin

sin sin

sinsin

.

.

.

.

. .

.

.

G B

G

G

G

G

I

G

I

o

o

o

o

o o o

o

o o o

o o o

o

=

=

=

∠ =∠ =

∠ = − +

=∠ = − =

∠ = − +

=

1000 90039

1000 39900

44 4

44 4

180 39 44 4

966

180 44 4 1356

180 135 6 39

5 4

1

2

39°

I

B

b = 900g = 1000

G1GG 2


Module 3

∠

( )

( )


Module 3

Notes

Review

Answer Key

1. a)

b)

sin

cos

tan

θ

θ

θ

= =

= = −

= =−

yrxryx

45

3543

r = + −

= +

==

4 3

16 9

25

5

2 2

x

y

P( 3, 4)

4

3

sin

cos

tan

θ

θ

θ

= = −

= =

= = −

yrxryx

5131213

512

r = + −

= +

==

12 5

144 25

169

13

2 2

5

x12

y

P(12, 5)


Module 3

( )

( )

−

−

−

−

c)

2. a) tan < 0 in Quadrants II and IV

b) sin > 0 and tan < 0 is in Quadrant IIc) cos > 0 and sin < 0 is in Quadrant IV

3. a) r = sin–1(0.78164)

= 51.4°

sin > 0 in Quadrants I and II

In Quadrant I, = 51.4°In Quadrant II, = 180° – 51.4° = 128.6°

b) r = cos–1(0.42316)

= 65°cos > 0 in Quadrants I and IV

In Quadrant I, = 65°

c) r = tan–1(1.46271)

= 55.6°tan < 0 in Quadrants II and IV

In Quadrant II, = 180° – 55.6° = 124.4°

sin

cos

tan

θ

θ

θ

= =−

=−

=−

= =−

=−

=−

= = −−

=

yr

xryx

22 10

110

1010

62 10

310

3 1010

26

13

r = − + −

= +

=

=

6 2

36 4

40

2 10

2 2

x

y

P ( 6, 2)

6

2


Module 3

( )( )

−

−

−−

θθθ

θ

θθ

θθθ

θ

θθ

θ

θθ

4. a) a = 7, b = 4, and c = 9.

Largest angle is∠C opposite longest side

To find ∠ A:

b) b = 8 cm, c = 11 cm, ∠ A = 57°To find side a

( ) ( )

2 2 2

2

2

2 cos A

64 121 2 8 11 cos 57º

185 95.8565 89.1435

9.4 cm

a b c ab

a

a

a

= + −

= + −

= − =

=

( )( )( )

o

o

B 180 A C

180 106.6º 48.2º

180º 154.8º

25.2º

∠ = − ∠ + ∠

= − +

= −

=

( )

[ ]

o

o

1

o

sin A sin A

sin A sin106.67 9

7 sin106.6sin A

9.7453559

A sin .7453559

A 48.2

a c

−

=

=

=

=

=

∠ =

( ) ( )

2 2 2

1 o

2 cos C

81 49 16 2 7 4 cos C

81 65 56 cos C

56 cos C 65 81

56 cos C 16

–16 –2cos C

56 7

2C cos 106.6

7

c a b ab

−

= + −

= + −

= −

= −

= −

= =

= − =


Module 3

To find ∠ C:

c)

6. a)

Since sin B ≥ 1, there is no solution to the triangle.

b a

a bsin sin

sin sin

sinsin

sin .

B AB A

B

B

o

=

=

=

=

47 5223

1610282744

( ) ( ) o

o

2

1sin A

21

8 11 sin 57244 sin 57

36.9 cm

A bc=

=

=

=

( )( )

o

o

B 180 A C

180 57º 78.9º

180º–135.9ºB 44.1º

∠ = − ∠ +∠

= − +

=∠ =

Find the larger of the twomissing angles first.

o

o

o1

o

sin C sin A

sin C sin 5711 9.4

11sin 57sin C

9.4

11sin 57C sin

9.4

C 78.9

c a

−

=

=

=

=

∠ =


Module 3

b)

Case 1 Case 2

c)

a bsin sin

sin sin

sinsin

sinsin

.

A B

A

A

A

A

o

o

o

o

=

=

=

∠ =

∠ =

−

5 634

5 346

5 346

278

1

Since b > a, there isone triangle.

C

AB

5 6

34 °

∠∠

∴ ∠ −

∴ ∠ = − −∠ =

AC B is the supplement

of AC C

AC B = 180 56.8

= 123.2

BAC

BAC

1

1

1o o

o

1o o o

1o

180 1232 42

14 8

.

.

b c

cb

sin sin

sinsin

sinsin

sinsin

.

.

.

B C

CB

C

C

C

A

o

o

o

o o o

o

=

=

=

∠ =

∠ =∠ = − −

=

−

20 4216

20 4216

568

180 42 568

812

1

Since b < c, there aretwo cases.

CB

A

161 6

20

42 °

C 1


Module 3

sin o20 4216

sin o5 346

∴ ∠ C = 180° – 34° – 27.8°

∠ C = 118.2°

b c

c

c

c

sin sin

sin sin .

sin .sin

.

B C

o o

o

o

=

=

=

=

634 118 2

6 118234

95


Module 3

Module 3

Lesson 1Answer Key

1. The equation of the circle in standard form is (x – h)2 + (y – k)2 = r2, where C(h, k) is at the centre and r isthe radius.

2. C(h, k) represents the coordinates of the centre and r is theradius for (x – h)

2+ (y – k)

2= r

2.


x y

x y

− + − =

+ =

0 0 3

3

2 2 2

2 2

a) C r0 0 3, =

x y

x y

− − + − =

+ + − =

2 3 5

2 3 25

2 2 2

2 2

c) C r− =2 3 5,

x y

x y

− − + − =

+ + − =

2 12

2 114

2 22

2 2

b) C r− =2 112

,

a) x y

C r

+ + − =

− = = =

2 1 12

2 1 12 4 3 2 3

2 2

,

b) x y

C r r

+ + + =

− = =

0 119

0 119

13

2 2

,

c) x y

C r

+ + + =

− − =

1 1 64

1 1 8

2 2

,

( )

( ) ( ) ( )

( )

( )( ) ( )

)( )(

1

)(

( )( ) ( )

)()(

)(

)(

)(

)(

)(

)(

)(

)(

)(

Module 3

3. a) C( 0, 0) r = 16

b) C(–1, –2) r = 3

4. Each one of these questions involves completing the squareand finding C(h, k) + radius (r) from (x – h)2 + (y – k)2 = r2.

b) x2

+ y2

= –4This equation does not define a circle because r2 –4 is impossible


x y

x y

− + − =

+ =

0 0 16

256

2 2 2

2 2

x y

x y

− − + − − =

+ + + =

1 2 3

1 2 9

2 2 2

2 2

a) x x y y

x x y y

x y

2 2

2 2 2 2 2 2

2 2

4 2 4

4 2 2 1 4 2 1

2 1 9

+ + + − + =

+ + + − + = + +

+ + − =

C r− = =2 1 9 3, ,

c) x y y

x y

2 2 2 2

2 2

0 6 3 12 3

0 3 21

+ + + + = +

− + + =

C r0 3 21, ,− =

d) x x y y

x y

2 2 2 2 2 2

2 2

10 5 4 2 0 5 2

5 2 29

− + + − + = + +

− + − =

C r5 2 29, , =

=

( ) ( )

( ( )) ( )( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( )

( () )

( ) ( )

( )

5. a) C(–2, 4) r = 3


Module 3

x x y y

x y

2 2 2 2 2 2

2 2

6 3 2 1 6 3 1

3 1 4

− + + + + = − + +

− + + =

C r3 1 2, − =

b)

x

y

( 2, 4)

r = 3

x

y

(3, 1)

r = 2

( ) ( )

( )

6. a) C(0, 2), passes through (0, 0).

r = 2, vertical line segment

b) C(5, 0), diameter = 10, radius = 5


Module 3

x

y

r = 2

(0, 0)

(0, 2) }x h y k r

x y

x y

− + − =

− + − =

+ − =

2 2 2

2 2 2

2 2

0 2 2

2 4

x h y k r

x y

x y

− + − =

− + − =

− + =

2 2 2

2 2 2

2 2

5 0 5

5 25

( ) ( )

( ) ( )

( )

( ) ( )

( ) ( )

( )

∴


c) C(4, 3), passing through (1, 2)

The distance from the centre (4, 3) to the point on thecircumference of the circle (1, 2) is the radius.

x1 = 4, y1 = 3, x2 = 1, y2 = 2

( ) ( )

( ) ( )

( ) ( )

( )

= − + −

= − + −

= − + −

= +

=

2 22 1 2 1

2 2

2 2

Length of radiu

1 4 2 3

3 1

9 1

4, 3 10

s x x y y

C r

x h y k r

x y

x y

− + − =

− + − =

− + − =

2 2 2

2 2 2

2 2

4 3 10

4 3 10

Module 3

( ) ( )

( ) ( )

( ) ( )

( )

Module 3


d) Because the radius is perpendicular to the tangent at thepoint of contact, the centre of the circle lies at (3, 2).

e) Because the radius is perpendicular to the tangent at thepoint of contact, the centre of the circle lies at (3, 1) andr = 3.

f) C(0, 0), A = 12

Because the area of a circle = r2, r2 = 12.

x

y

C (3, 1)

( 0 , 1 )

x h y k r

x y

x y

− + − =

− + − =

− + − =

2 2 2

2 2 2

2 2

3 1 3

3 1 9

x h y k r

x y

or

x y

− + − =

− + − =

+ =

2 2 2

2 2

2 2

0 0 12

12

x

y

y = 2C (3, 2)

2 4

(3, 0)

4

r

x h y k r

x y

x y

=

− + − =

− + − =

− + − =

2

3 2 2

3 2 4

2 2 2

2 2 2

2 2

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

π

π

Module 3


g) C(2, –1), C = 20

Because the circumference of a circle = 20 r

2 20

10

π πr

r

==

x h y k r

x y

x y

− + − =

− + − − =

− + + =

2 2 2

2 2 2

2 2

2 1 10

2 1 100

x

y

( 0 , 4 00)

x y

x y

2 2 2

2 2

400

160 000

+ =+ =

7.

The corresponding x-valueif the y-coordinate is 400,would be 0.

π

π

( ) ( )

( ) ( )

( ) ( )

( )

Module 3


8. Given circle:

Three units to the left of 1 is 1 – 3 or –2 and two units downfrom –4 is –4 – 2 or – 6.

Therefore, new centre (–2, –6). The radius remains the same.

New equation:

These are examples of congruent circles when they have thesame radius.

9. C(3, 3), radius = 3.

10. For the large circle the coordinates of the centre are A(6, 0) and the radius is 4 units. For the small circle, thecentre is at C(1, 0) and radius is 1.

Equation:

x y

x y

− − + − − =

+ + + =

2 6 16

2 6 16

2 2

2 2

x y

x y

− + + =

− + − − =

1 4 16

1 4 16

2 2

2 2Rewritten

x h y k r

x y

− + − =

− + − =

2 2 2

2 23 3 9

x

y

C B

A

x h y k r

x y

− + − =

− + =

2 2 2

2 21 1

( ) ( )( )

( )

( )( )

( ) ( )

( ) ( )

( )( )

( )

( ) ( )

( ) ( )

( )

Module 3

Lesson 2Answer Key

1. a) A(4, 6) and B(6, 5)

Let x1 = 4, y1 = 6, x2 = 6, y2 = 5

b) C(–4, –2) and D(2, 2)

Let x1 = 2, y1 = 2, x2 = –4, y2 = –2

2. Make certain that all the lines are written in the form Ax + By + C = 0 because the formula to determine thedistance from a point to a line requires it.

a) P(2, 3) Line: 4x + 3y – 10 = 0

x1 = 2, y1 = 3, A = 4, B = 3, C = –10


d x x y y= − + −2 12

2 12

d = − + − = + − =6 4 5 6 2 1 52 2 2 2

d x x y y= − + −2 12

2 12

d = − − + − − = − + − = + =

=

4 2 2 2 6 4 36 16

52 2 13

2 2 2 2

dx y

=+ +

+

A B C

A B1 1

2 2

d =+ + −

+

=+ −

+

=

=

4 2 3 3 10

4 38 9 10

16 9725

75

2 2

( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( )( ) ( )( ) ( )

Module 3

b) P(–2, –1) Line: x – y – 2 = 0

x1 = –2, y1 = –1, A = 1, B = –1, C = –2

c) P(6, 2) Line: x + 1 = 0

x1 = 6, y1 = 2, A = 1, B = 0, C = 1

d) P(0, 8) Line: 6x + y = 0

x1 = 0, y1 = 8, A = 6, B = 1, C = 0


d =− + − − + −

+ −

=− + −

=−

=

1 2 1 1 2

1 1

2 1 2

23

2

32

3 22

2 2

or Rationalized

d =+ +

+

=+

=

1 6 0 2 1

1 06 1

17

2 2

d =+ +

+

=

=

6 0 1 8 0

6 18

37

837

8 3737

2 2

or

( ) ( )

( ) ( ) ( ) ( )( )

( )( )

( ) ( )( )( )

( )

3. Midpoint of A(3, –4) and B(–15, 2):

x1 = 3, y1 = –4, x2 = –15, y2 = 2

Midpoint:

4. Parallel lines y = 3x + 1 and y = 3x – 9

Vertical distance is the distance between the y-intercepts:= |1– (–9)| = 10

Horizontal distance is the distance between the x-intercepts:

The shortest distance is the perpendicular distance betweenthe two lines. P(0, 1) lies on y = 3x + 1. Find the distance tothe other line, 3x – y – 9 = 0.

x1 = 0, y1 = 1, A = 3, B = –1, C = –9


Module 3

x x y y1 2 1 2

2 2+ +,

=+ − − +

= − −

3 152

4 22

6 1

,

,

= − − =13

3103

dx y

=+ +

+

A B C

A B1 1

2 2

d =+ − + −

+ −

=− −

+

=

3 0 1 1 9

3 1

0 1 9

9 11010

10

2 2

or

( ) ( )( )

( )

( )( )

( )( )

)(

x x y y1x1x 2 1 2y2y2 2+ +,

+ − − +3 152

4 22

,(−(− )

5. a) (3, 0) represents a point on 2x – 3y = 6. 2x – 3y – 9 = 0 is the equation of the line.

x1 = 3, y1 = 0, A = 2, B = –3, C = –9

b) (2, 0) represents a point on 3x – 4y – 6 = 0. You will findthe distance from that point to the line 3x – 4y – 12 = 0.

x1 = 2, y1 = 0, A = 3, B = –4, C = –12


Module 3

dx y

=+ +

+

A B C

A B1 1

2 2

dx y

=+ +

+

A B C

A B1 1

2 2

d =+ − + −

+ −

= −

=

2 3 3 0 9

2 3

6 913

313

2 2

or3 13

13

d =+ − + −

+ −

=−+

=

3 2 4 0 12

3 4

6 129 16

65

2 2

( ) ( )( ) ( )( )

( )

( ) ( )( ) ( )( )

( )

6. a) Determine the equation of BC and then find the distancefrom A to BC. The slope of line BC is

Equation of BC:

x1 = –1, y1 = 2, A = 6, B = 1, C = –17

b) Area of a triangle =

The altitude to

Length of BC =

Area = 12

3721

37

212

⋅ = square units

2 3 5 1 1 6 372 2 2 2− + − − = − + =

BC =21

37

12

bh


Module 3

MBC =− −

−=

−= −

5 12 3

61

6

y y m x x

y x

y xx y

− = −

− = − −− = − +

+ − =

1 1

5 6 2

5 6 126 17 0

dx y

d

=+ +

+

=− + + −

+

=− + −

=

A B C

A B

or21 37

37

1 1

2 2

2 2

6 1 1 2 17

6 16 2 17

37

2137

(

)(

)

( )

( ) ( )( ) ( )( )

) ( )(( ( )

7. Choose a point on 3x – 4y – 6 = 0, for instance (2, 0). Fiveunits away from this line are two parallel lines of the form3x – 4y + C = 0. (Remember the slope is the same as thegiven line. Find the value for C.

x1 = 2, y1 = 0, A = 3, B = –4, d = 5

the equations are 3x – 4y + 19 = 0 or 3x – 4y – 31 = 0

8. The given lines will be of the form x – 4y + C = 0. The lineshave to be three units away from (4, 1) which does not lie onthe given line.

x1 = 4, y1 = 1, A = 1, B = –4, d = 3. Find C.


Module 3

dx y

C

=+ +

+

=+ − +

+ −

=+

= ++ = + = −

= = −

A B C

A BC

C

C or CC or C

1 1

2 2

2 25

3 2 4 0

3 4

56

2525 6

6 25 6 2519 31

dx y

=+ +

+

A B C

A B1 1

2 2

31 4 4 1

1 4

34 4

17

3 17

3 17 3 17

2 2=

+ − +

+ −

=− +

=

= −

C

C

C

C or

∴ − + = − − =The equations are orx y x y4 3 17 0 4 3 17 0

( ) ( )( )( )

( )

( )

( )

( )( )( )

∴

9. a)

b)

The line segment connecting the midpoints is half thelength of AB.

c) The median from C goes to the midpoint of AB.

Length of the median is measured from (–3, 4) to (6, 1).


Midpoint of AC =+ − +

=

=

5 32

4 42

22

82

1 4

,

,

,

Midpoint of BC =+ − − +

=

=

7 32

2 42

42

22

2 1

,

,

,

Distance = − + −

= − +

=

1 2 4 1

1 3

10

2 2

2 2

AB = − + − −

= − +

= +

=

=

5 7 4 2

2 6

4 36

40

2 10

2 2

2 2

Midpoint of AB = + + −

=

5 72

4 22

6 1

,

,

d = − − + −

= − +

=

=

3 6 4 1

9 3

90

3 10

2 2

2 2

Module 3

+ − +5 32

4 42

,

22

82

,

+ − − +7 32

2 42

,

42

22

,

+ + −5 72

4 22

,

( )

( )

( )

( )

( )

( ) ( )( )

)(

)(

)( )(

)(

)(

)(

)(


Module 3

Module 3

Lesson 3

Answer Key

1. a) — iii) one solution — independent systemb) — i) no solution — inconsistent systemc) — ii) infinitely many solutions — dependent system

2. a) infinitely many — dependent systemb) no solution — inconsistent systemc) one solution — independent system

3. To check the given point as a solution algebraically,substitute the coordinates into each equation.

a) Point (2, –2)4x – y = 10 4(2) – (–2) = 10 8 + 2 = 10

2x + 3y = –2 2(2) + 3(–2) = –2 4 – 6 = –2

The coordinates of the point is the solution.

b) Point (4, 0)–x + 6y = 4 –4 + 6(0) ≠ 4

3x + y = 12 3(4) + 0 = 12

Because the coordinates do not satisfy the first equation,(4, 0) is not the solution.

c) Point (5, 2)2k – 3m = 4 2(5) – 3(2) = 4 10 – 6 = 4

3k + m = 17 3(5) + 2 = 17 15 + 2 = 17

The coordinates satisfy both equations, so (5, 2) is asolution.


Module 3

4. a)

Point (0, 3) is not a solution of the linear equation.

b)

Point (4, 4) is the solution of the linear equation.


l

l

x

y

2x + y = 3

l

x + 2y = 8

ll x

y

x + 2 y = 4l

−3x + 4

y = 4

c)

Point (2, –1) is the correct solution.

The graphing calculator may be used to check the points.Enter each equation and then use (CALC)5: Intersect.

5. a) Independent system. Pointof intersection is at (2, –1).

The solution is the orderedpair (2, –1).

Check:

b) Dependent system. Becausethe two lines coincide, thesolution set is all realnumbers that satisfy theequation y = 2 – 3x.

y x− = −− − = −

− = −

3

1 2 3

3 3

2 3

2 2 1 3

4 1 3

x y+ =+ − =

− =

2nd


Module 3

x

y

x + 2 y =4

y = 1

)( )( )( )(

c) Inconsistent system. Becausethe two lines have the sameslope, the two lines areparallel. They do not intersectso there is no solution.

d) Independent system. Pointof intersection is at (2, 3).The solution is the orderedpair (2, 3).

Check:

6. Set up the two equations.x = number of months y = total costClub A: y1 = 5x + 400Club B: y2 = 30xEnter the equations in the function register to graph theequations.Window settingXmin = 0Xmax = 20Xscl = 5Ymin = 0Ymax = 600Yscl = 100Xres = 1

Use (CALC) 5: Intersect.Point of intersection: (16, 480)Interpretation — After 16 months the charges to belong toeither club would be $480.

2nd

y x− =− =

=

1

3 2 11 1

x y2 322

33

1 1

=

=

=


Module 3

7. Set up the two equations.Let x = part of $4500 invested at 4%Let y = part of $4500 invested at 5%x + y = 45000.04x + 0.05y = 210Enter the equations and graph:

Window SettingXmin = 500Xmax = 2000Xscl = 800Ymin = 500Ymax = 4000Yscl = 800Xres = 1

Use (CALC) 5: Intersect.

Point of intersection: (1500, 3000)Interpretation — $1500 is invested at 4% and $3000 isinvested at 5%.

2nd

y x

yx

1

2

4500

210 0 04005

= −

=− ..


Module 3

8. For a new business, the break-even point occurs whenrevenue (the income from sales) is equal to cost (the amountthe business has spent).

Let R = revenue brought in during each week

R = 8000t represents the revenue for t weeksC = cost for the first t weeks

C = 7400t + 85 000

Use your graphing calculator to find the break even point.

Enter the equations and graph: y1 = 8000x

y2 = 7400x + 85 000Window SettingXmin = 0Xmax = 200Xscl = 10Ymin = 0Ymax = 2 000 000Yscl = 600 000Xres = 1

Use (CALC) 5: Intersect.

Point of intersection: (141.67, $1 133 333.30)Interpretation — The break even point occurs at 141.67weeks at which time you will have spent and taken in thesame amount of $1 133 333.30

9. a) y = x + 4y = 2Substitute the value y = 2 into the first equation2 = x + 4– 2 = x(–2, 2) represents the solutionCheck your answer using the graphing calculator.

b) y = 2x – 4x = 2Replace x with 2 in the first equation.y = 2(2) – 4y = 0∴ the point (2, 0) represents the solution.

2nd


Module 3

Module 3

Lesson 4Answer Key

1. a) 2x + y = 3 Equation 12y – 3x = 6 Equation 2Solve Equation 1 for y in terms of x. Replace y in thesecond equation with the expression involving x.

Substitute 0 for x in Equation 1.

The ordered pair is (0, 3), which is the point ofintersection. Check in each equation:Equation 1: 2x + y = 3 2 (0) + 3 = 3 3 = 3

Equation 2: 2y – 3x = 6 2(3) – 3(0) = 6 6 = 6

(0, 3) is the solution.

b) 2y – 5x = 1 Equation 1x + y = –3 Equation 2

In Equation 2, solve for x in terms of y.

x = –3 – y

Replace x in Equation 1 with –3 – y


y x

y x

x x

x x

x

x

= −

− =− − =− − =

− ==

3 2

2 3 6

2 3 2 3 6

6 4 3 6

7 0

0

2 0 3

0 3

3

+ =+ =

=

y

y

y

2 5 3 1

2 15 5 17 15 1

7 14

2

y y

y yy

y

y

− − − =+ + =

+ == −= −

)(

)(

)(

∴

Module 3

Substitute y = –2 into Equation 2.

Check the ordered pair (–1, –2) in both equations.

2y – 5x = 1 2(–2) – 5(–1) = 1 –4 + 5 = 1 1 = 1

x + y = –3 –1 + (–2) = –3 –3 = –3The solution is (–1, –2), which represents the point of

intersection.


x

x

+ − = −= −

2 3

1

x yy

+ = −22

14

c)

Equation 1Equation 2

Solve Equation 2 for y

y = x + 7

Substitute x + 7 for y in Equation 1

Ordered pair is (–5, 2).

Check:

Equation 1: 2x + 5y = 0 2(–5) + 5(2) = 0 0 = 0

Equation 2: y – x = 7 2 – (–5) = 7 7 = 7

The solution is at (–5, 2) which represents thepoint of intersection.

Clear denominator by multiplying eachside by 4

2 2 1

2 4 12 5 0

7

x y y

x y yx y

y x

+ = −

+ = −+ =

− =

2 5 7 0

2 5 35 0

7 355

7

5 7

2

x x

x x

xx

y x

y

y

+ + =+ + =

= −= −= += − +=

∴

∴

)(

)(


Module 3

53

23

1

x yy

x y

+ =

+ =

d)


Solve Equation 2 for y in terms of x

y = –2x + 3

Replace in Equation 1

Substitute into

Ordered pair is:

Check:

Equation 1

Remove denominators by multiplyingeach side by yRemove denominator by multiplyingeach side by 3

5 3

2 35 2 0

2 3

x y y

x yx y

x y

+ =+ =

− =+ =

5 2 2 3 0

5 4 6 09 6

23

x x

x xx

x

− − + =

+ − ==

=

y x

y

y

y

= − +

= − +

= − +

=

2 3

223

3

43

93

53

23

53

,

5 2 0x y− =

Equation 2

2 3x y+ =

2 5 10 105 2 0 0 0 0

3 3 3 3 − = − = =

2 5 4 52 3 3 3 3

3 3 3 3 + = + = =

23

53

,

( )


Module 3

e)


Solve Equation 2 for y in terms of x and replace y by thatvalue in Equation 2.

This is interpreted as a true statement, which meansEquations 1 and 2 coincide with each other.

The solution is written {(x, y)| 2x – y = 3}, which meansinfinitely many ordered pairs make up the equation.

Note: In the beginning if you had divided the firstequation by 2 you would have arrived at the answer.

The equation 4x – 2y = 6 becomes 2x – y = 3, which is thesame as Equation 2.

4 2 6

2 3

2 3

4 2 2 3 6

4 4 6 66 6

x y

x y

y x

x x

x x

− =− =

= −− − =

− + ==

2. Use the addition-subtraction method (linear combinationmethod) to solve the following:

a)

Substitute the y-value into either equation:

(1, 0) is the ordered pair to be checked.

(1, 0) is the solution. It represents the point ofintersection of the two equations.

− + = −+ =

==

x y

x yy

y

1

12 0

0

Equation 1 Use addition to solve for y

Equation 2

− + = −− + = −

=

x y

xx

1

0 11

− + = −− + = −

− = −

x y 1

1 0 1

1 1

x y+ =+ =

=

1

1 0 1

1 1

( )

( ) ( ) ( )

∴

∴


Module 3

b)

Solve for y by substituting 2 in for x in Equation 2.

(2, –1) is the ordered pair to be checked.

The solution is (2, –1).

3 2 4

3

3 2 4

2 2 6

5 102

x y

x y

x y

x y

xx

+ =− =

+ =− =

==

Equation 1 Equation 2: Multiply Equation 2 by 2and add

x y

yy

y

− =− =− =

= −

3

2 31

1

3 2 4

3 2 2 1 4

4 4

x y+ =+ − =

=

x y− =− − =

=

3

2 1 3

3 3

c)

These lines are parallel as they have the same slope.There is no solution, which is represented by �.

If you did not realize the above, multiply Equation 1by 2 and subtract.

This is a false statement, which means there is nosolution because the lines are parallel and do notintersect.

5 3 110 6 0

x y

x y

+ =+ =

5 3 110 6 0

x y

x y

+ =+ =


Equation 1 x 2Equation 2Subtract Equation 2 from(Equation 1)x 2

10 6 6

10 6 00 6

x y

x y

+ =+ =

=

∴

∴

∴

( ) ( ) ( )


Module 3

d) 2 3 17

9 5 16

18 27 153

18 10 32

37 185

52 3 17

2 3 5 17

2 15 17

2 2

1

x y

x y

x y

x y

y

yx y

x

x

x

x

+ = −− + = −

+ = −− + = −

= −= −

+ = −

+ − = −− = −

= −= −

Equation 1 x 9Equation 2 x 2

Add Equations 1 and 2

Ordered pair is (–1, –5).

Check

The solution is (–1, –5).

2 3 17

2 1 3 5 17

17 17

x y+ = −− + − = −

− = −

− + = −− − + − = −

− = −

9 5 16

9 1 5 5 16

16 16

x y

Replace by –5 in Equation 1

e) 8 3 3

3 2 5

16 6 6

9 6 15

7 21

3

3 2 5

3 3 2 5

9 2 5

2 14

7

x y

x y

x y

x y

x

x

x y

y

y

y

y

− =− = −− =− = −

==

− = −− = −− = −− = −

=


Subtract

Substitute x = 3 into Equation 2

∴

( )

( ) ( ) ( ) ( )

( )


Module 3

The ordered pair is (3, 7).

Check

The solution is (3, 7).

8 3 3

8 3 3 7 3

3 3

x y− =− =

=

3 2 5

3 3 2 7 5

5 5

x y− = −− = −

− = −

f)

3 2 3

23 0

3 0

x y y

x y xx y

x y

+ =− = −− =− =

Remove denominators in Equations 1 and 2

The system contains the same line.

The solution is {(x, y)|3x – y = 0}, meaning there areinfinitely many solutions

g)


Remove denominator by multiplying eachside by (2x + 1) (y – 3)

32 1

43x y+

=−

8 3 10x y= +

3 3 4 2 1

3 9 8 48 3 13

8 3 10

y x

y xx y

x y

− = +

− = +− + =

− =

Add Equations 1 and 20 0 230 23

+ ==

Because this is a false conclusion, there is no solutionbecause the lines are parallel.

∴

∴

∴

( ) ( ) ( ) ( )

( ) ( )


Module 3

3. a) Substitution is most appropriate.

2 33 2 6

3 2

3 2 3 2 6

3 6 4 600

2 3

2 0 3

3

x y

x y

y x

x x

x x

x

x

x y

y

y

+ =+ =

= −+ − =

+ − =− =

=+ =+ =

=

Equation 1Equation 2Solving for y in Equation 1Replacing y in Equation 2 andsolving for x

Check the ordered pair (0, 3).

The solution is (0, 3) for this independent system.

2 3

2 0 3 3

3 3

x y+ =+ =

=

3 2 6

3 0 2 3 6

6 6

x y+ =+ =

=

b) Simplify the system first by multiplying Equation 1 by 2

x y

x y

+ = −− = −

3

2 3 1

Substitution may be preferable by solving for y inEquation 1.

y x

x x

x x

x

x

x y

y

y

= − −− − − = −

+ + = −= −= −

+ = −− + = −

= −

3

2 3 3 1

2 9 3 15 10

23

2 31


Replace y in Equation 2

Substitute x = –2 into Equation 1

∴

( )

( )

( ) ( )

( )


Module 3

c) Addition is probably preferable.

4 3 13 6 3

12 9 312 24 12

15 151

4 3 1

4 3 1 1

4 41

x y

x y

x y

x y

y

y

x y

x

x

x

+ =− − =

+ =− − =

− == −

+ =+ − =

==


Add the two equations

Substitute y = –1 into Equation 1

Ordered pair (1, –1).Check:

The solution is (1, –1) for this independent system.

Ordered pair to be checked is (–2, –1).

The solution is (–2, –1) for this independent system.

x y+ = −− + − = −

− = −

3

2 1 3

3 3

2 3 1

2 2 3 1 1

1 1

x y− = −− − − = −

− = −

d) Addition-subtraction is preferable

2 3 5

4 6 3

4 6 104 6 3

0 7

x y

x y

x yx y

− =− =− =− =

=

Equation 1 x 2Equation 2

SubtractionFalse statement

Lines are parallel and there is no solution for thisinconsistent system.

4 3 1

3 6 3

x y

x y

+ = →− − = →

4 1 3 1 1

3 1 6 1 3

+ − =

− − − =

4 3 13 6 3

− =− + =

1 13 3

==

∴

( )( )( ) ( )

( )

( )( )

( )( )

∴


Module 3

4.

Solve Equation 2 for x in terms of y. Substitute the resultingexpression involving y into Equation 1. Proceed to solve theresulting equation involving the variable y. Once the value ofy is known, substitute into either equation and solve for thex-value. Check the resulting ordered pair in both equations. Ifthe check works, then the ordered pair is the solution for theindependent system.

12 4 13 9

x yx y

+ =− =


5.

To use addition to solve for y, multiply Equation 1 by 2 andEquation 2 by 3. Add the equations and solve the resultingequation for y. Once you know the y-value, substitute intoEquation 1 to obtain the x-value. Check the ordered pair inthe two equations. If it checks out, then that ordered pair isthe solution of the independent system. (You could alsomultiply Equation 1 by 2 and Equation 2 by 5 and solve for x.)

6 5 304 2 7

x yx y

− =− + =


6. If you obtained the solution –6 = –6, the system was made upof two equations that were identical to each other. The solutionconsists of infinitely many ordered pairs for the dependentsystem.

7. If you obtained the solution 10 = 0, the system is composed oftwo parallel lines for which no solution exists.

e) Subtraction method

3 6 33 6 3

0 0

x yx y

+ =+ =

=

Equation 1 x 3

Because this is a true statement, the system isdependent and an infinite number of ordered pairs is thesolution, written as {(x, y)|x + 2y =1}.


Module 3

9. Let x = number of 2-point basketsy = number of 1-point free throws

+ =+ =

== − =

2 1 3005

1919

1086 (2 point baskets)

1919 1086 833 (1 point freethrows)

x y

x y

x

y

10.

2 2 3 3 135 4 2 3 6

5 137 8 5

7 35 917 8 5

43 862

5 13

5 2 13

10 133

3

x y x y

x y x y

x y

x y

x y

x y

y

yx y

x

x

x

x

− − − = −− + = −

− − = −− + = −

− − = −− + = −

− = −=

− − = −− − = −

− − = −− = −

=

Simplify and rearrange theterms of the equations


Substitute y = 2 into Equation 1

The solution is (3, 2). Check your answer.

2 3 13

5 2 2 3 2

x y x y

x y x y

− − + = −

− − = −

8. a) False. If you solve a system by addition-subtraction orsubstitution, you will find the exact solution to thesystem.

b) True.c) False. If, while solving a system of linear equations, you are

left with an equation that can never be false, youknow that the system has infinitely many orderedpairs in the solution of the dependent system.

d) False. A dependent system of linear equations hasinfinitely many solutions or an inconsistent systemof linear equations has no solution.

e) True.f) True.


( )

( )

( )

( ) ( )


Module 3

12. Ax + By = 10 P(2, –1) and Q(6, 2)

When you are given an ordered pair, you know the x- andy-values. Substitute the values for the points into Ax + By = 10 and form a system of equations

A 2 B 1 10

A 6 B 2 10

6A 3B 306A 2B 10

5B 20B 4

6A 3B 30

6A 3 4 30

6A 12 306A 18

A 3

+ − =

+ =− =+ =− =

= −− =

− − =+ =

==


Subtract

Substitute B = –4 into Equation 1

The equation is 3x – 4y = 10.

11. x y

x y

x

xx y

y

y

+ =− =

==

+ =+ =

=

8

4

2 12

68

6 8

2

Coordinates of the island are (6, 2).

Add Equations 1 and 2


∴

( )( )

( )( )

( )

Lesson 5Answer Key

1. Substitute the point into the equation to test whether it is asolution.


a)

(1, –2, –3) is asolution


Point(1, –2, –3)

(0, –2, –4)

x y z+ − =

− − − =

− + ==

2

1 2 3 2

1 2 3 22 2

0 2 4 2

2 4 2

2 2

− − − =− + =

=

b)

(1, –1, –1) is not asolution


Point(1, –1, –1)

(2, –3, –1)

x y z− + =− − + − =

+ − =≠

2 3

1 1 2 1 3

1 1 2 3

0 3

2 3 2 1 3

2 3 2 3

3 3

− − + − =+ − =

=

c)

(0, 0, 4) is not a

solution

(0, –2, 8) is not a

solution

Point(0, 0, 4)

(0, –2, 8)

x y z+ + =+ + =

≠

4 2 0

0 4 0 2 4 0

8 0

0 4 2 2 8 0

8 16 08 0

+ − + =− + =

≠

d)

(–1, –1, 6) is not asolution

(2, –6, 4) is asolution

Point(–1, –1, 6)

(2, –6, 4)

5 0

5 1 1 6 0

5 1 6 0

12 0

x y z+ − =− + − − =

− − − =− ≠

5 2 6 4 0

10 6 4 0

0 0

+ − − =− − =

=

∴

∴

∴

∴

∴

∴

∴

∴

( )

( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( )( )

Module 3


2. Evaluate.

2 3 3 2 2 5

2 2 3 2 3 5 4 6 15 13

x y z x y z+ + = = − =+ − + = − + =

if , ,

3. Substitute the coordinates of the ordered triple into each of thegiven equations.

4. a)

Solution

(1)(2)(3)

(4)(5)(6)

Point

(–1, 2, –3)

(–1, 2, –3)

Equation

x y z a

a

aa

+ − =

− + − − =

− + + ==

2 3

1 2 2 3 3

1 4 912

− − + =− − − + − =

− − =− =

x y z b

b

b

b

1 2 3

1 2 3

4

(–1, 2, –3) 2 3 2

2 1 3 2 2 3

2 6 610

x y z c

c

cc

+ − =

− + − − =

− + + ==

3 2 2 11

2 12 2 4

x y z

x y zx y z

+ − =+ + = −− + = −

4 3 10

2 3

4 8 1211 22

2

x y

x y

x yy

y

+ =− + =

− + ===

Eqn (1) + Eqn (2):Eqn (2) – Eqn (3):Eqn (5) x 4:Eqn (4) + Eqn (6):

( )( )

( )

( )

( )

( )( )


Module 3

b)

Solution

(1)(2)(3)

(4)(5)(6)(7)(8)(9)(10)

3 4 5 2

4 5 3 55 3 2 11

x y z

x y zx y z

− + =+ − = −− + = −

8 10 6 1015 9 6 33

23 439 12 15 6

20 25 15 2529 13 19

299 13 559270 540

2

x y z

x y z

x y

x y z

x y zx y

x y

x

x

+ − = −− + = −

+ = −− + =+ − = −

+ = −+ = −

− == −

Eqn (2) x 2:Eqn (3) x 3:Eqn (4) + Eqn (5):Eqn (1) x 3:Eqn (2) x 5:Eqn (7) + Eqn (8):Eqn (6) x 13:Eqn (9) – Eqn (10):

Substitute y = 2 into Eqn (5):

The check is left for you.The solution is (1, 2, –2).

− + =− = −

=

x

x

x

2 2 3

1

1

Substitute x = 1, y = 2 into Eqn (1): 3 1 2 2 2 11

3 4 2 112 4

2

+ − =+ − =

− == −

z

z

z

z

Substitute x = –2 into Eqn (6):

The check is left for you.The solution is (–2, 3, 4).

23 2 43

46 433

− + = −

− + = −=

y

yy

Substitute x = –2, y = 3 into Eqn (1): 3 2 4 3 5 2

6 12 5 25 20

4

− − + =− − + =

==

z

z

z

z

∴

∴

( )

( )

( )

( ) ( )

( )


Module 3

c)

5. Circle passes through (1, 1) (–2, 3), and (3, 4).

x2 + y2 + Dx + Ey + F = 0

For (1, 1): 12 + 12 + D(1) + E(1) + F = 0

D + E + F = –2 ----------------------------- (1)

For (–2, 3): (–2)2 + 32 + D(–2) + E(3) + F = 0

–2D + 3E + F = –13 ---------------------- (2)

For (3, 4): 32 + 42 + D(3) + E(4) + F = 0

3D + 4E + F = –25 ----------------------- (3)

Solution

(1)(2)(3)

(4)

(4)(5)(6)(7)

x y z

x y zx y z

+ + =+ + =− + =

3 12

2 3 134 11

3 7 24

11

x z

xx

+ =− = −

=

Eqn (2) + Eqn (3):Eqn (1) – Eqn (2):

Eqn (1) – Eqn (2):Eqn (1) – Eqn (3):Eqn (4) x 2:Eqn (5) x 3:Eqn (6) + Eqn (7):

Substitute x = 1 into Eqn (4):

The check is left for you.The solution is (1, 2, 3).

3 1 7 24

7 21

3

+ ===

z

z

z

Substitute x = 1, z = 3 into Eqn (1): 1 3 3 12

10 122

+ + =

+ ==

y

yy

3D 2E 112D 3E 236D 4E 226D 9E 69

13E 91E 7

− =− − =

− =− − =

− == −

∴

( )

( )


Module 3

Substitute E = –7 into Eqn (6):

The equation of the circle is x2 + y2 – x – 7y + 6 = 0.

6D 4 7 22

6D 28 226D 6

D 1

− − =+ =

= −= −

Substitute D = –1, E = –7 into Eqn (1): − + − + = −= +

1 7 F 2

F 6

6. Set up the three equations.

Triple(4, 1, 2)(3, 2, 1)

(–6, –1, –2)

(1)(2)(3)

(4)

(5)

(6)(7)

Eqn (1) + Eqn (3):

Eqn (6) x 2:Eqn (5) + Eqn (7):

Substitute A = –1 into Eqn (2):

Substitute A = –1, C = 2 into Eqn (1):

Substitute A = –1 into Eqn (3):

The equation is –1x + 1y + 2z = 1

− + + =+ =

3 2B 1C 12B 1C 4

6 1B 2C 1

1B 2C 5

2B 4C 103C 6

C 2

− − =− − = −− − = −

− = −=

A B C 14A 1B 2C 13A 2B 1C 16A 1B 2C 1

x y z+ + =+ + =+ + =

− − − =

− == −

2A 2A 1

4 1 B 2 2 1

B 1

− + + ==

∴

∴

( )

( )( )


Module 3

7. R = ap2

+ bp + c

If (p, R), then (30, 6000), (40, 6000), (50, 5000) are the threepoints.Set up the three equations.

Eqn (1) – Eqn (3):Eqn (1) – Eqn (2):Eqn (5) x 2:Eqn (4) – Eqn (6):

(4)(5)(6)

Substitute a = –5 into Eqn (5):

The equation is R = –5p2 + 350p.

6000 30 30

6000 900 30

6000 40 40

6000 1600 40

5000 50 50

5000 2500 50

2

2

2

= + += + +

= + +

= + +

= + += + +

a b c

a b c

a b c

a b c

a b c

a b c

(1)

(2)

(3)

1000 1600 20

0 700 10

0 1400 201000 200

5

= − −= − −= − −= −

− =

a b

a b

a ba

a

0 700 5 10

3500 10

350

= − − −− = −

=

b

b

b

Substitute a = –5, b = 350 into Eqn (1):

6000 900 5 30 350

6000 4500 10500

0

= − + += − + +=

c

c

c

∴

( )

( )

( )

( ) ( )

( ) ( )

( ) ( )

Review

Answer Key

1.

C(–6, 2), r = 8

2. a)

x h y k r

x y

− + − =

∴ + + − =

2 2 2

22

132

1494

centre midpoint of diameter=

= − + + −

= −

6 42

5 22

132

,

,

radius12

diameter=

= − − + − −

= +

=

12

6 4 5 2

12

100 49

12

149

2 2

x x y y

x x y y

x y

2 2

2 2

2 2

12 4 24

12 36 4 4 24 36 4

6 2 64

+ + − =

+ + + − + = + +

+ + − =


Module 3

( ) ( )

( ) ( )( )

( )

( ) ( )

( ) y −2

32

−132

,

b)

C(h, k)

h coordinate is radius = 5

(x – 7)2

+ (y – k)2

= 25

use (4, 0) to find k in the above equation

We have two possible answers:

If C(7, 4), (x – 7)2 + (y – 4)2 = 25

If C(7, –4), (x – 7)2

+ (y + 4)2

= 25

c) C C= −∴ =

=

∴ − + + =

2 2 1

2 105

2 1 252 2

ππ π

r

r

r

x y

,

4 7 0 25

9 25

16

4

2 2

2

2

− + − =

+ =

== ±

k

k

k

k

4 102

7+ =

(4, 0) ( 1 0 , 0 )


Module 3

∴

( )

( ) ( )

( ) ( )

3. Centre of circle1, (4, 0) Centre of circle2, (10, 0)

Radius of both circles = 3

Radius of larger circle is the diameter of the smaller circle.r = 2 • 3 or 6

Centre of new circle = midpoint of line segment (4, 0) and (10, 0)

Equation: (x – 7)2

+ y2

= 36

4. A = 2, B = –2, C = 7x1 = 5, y1 = –4

which is approximately 8.8

This is within the ten kilometre radius, therefore, the shipwill see the light.

dx y

d

=+ +

+

=+ − − +

+ −

=

=

A B C

A B1 1

2 2

2 2

2 5 2 4 7

2 2

25

8

25 24

(5, 4 )

2x 2y + 7 = 0

x

y

=+ +

=

4 102

0 02

7 0

,

,


Module 3

( )

∴∴

( ) ( )( )

( )

–

–

•

•

5. a)

b) Median from C to AB.

Midpoint of AB

Length of median = distance from C(–3, 4) to midpoint (6, 1)

= − − + −

= +

=

=

3 6 4 1

81 9

90

3 10

2 2

= + + −

=

5 72

4 22

6 1

,

,

d x x y y= − + −

= − + − = + =

2 1

2

2 1

2

2 21 2 4 1 1 9 10

Midpoint of AC

Midpoint of BC

: , ,

: , ,

5 3

24 4

21 4

7 3

22 42

2 1

+ − + =

+ − − + =

(7, 2 )

x

y

(5 , 4 )( 3, 4 )

CA

B


Module 3

–

–

,7 3

22 42

+ − − +

,5 3

24 4

2

+ − +

+ + −5 72

4 22

,

( )

( )

( ) ( )

( ) ( )

( )

( ) ( )

( )

( )

( )

c) The altitude would be the distance from B to line AC.

The equation of AC is the horizontal line y – 4 = 0.

A = 0, B = 1, C = –4, x1 = 7, y1 = –2

The altitude from B to AC is 6 units in length.

d) If AC is the base, its length

6. a) Choose a point on one of the lines:

3x – 4y = 12

If x = 0, 3 • 0 – 4y = 12y = –3

x1 = 0, y1 = –3, A = 3, B = –4, C = –24

Note: Always choose the other line that was not used for x1 and y1 values.

dx y

=+ +

+

=+ − − + −

+ −

=−

=

A B C

A B

units

1 1

2 2

2 2

3 0 4 3 24

3 4

12

5125

Area of triangle

Area units

=

∴ = ⋅ =

12

12

8 6 24 2

bh

= − − + −

= −

=

3 5 4 4

8

8

2 2

2

d =+ − −

= − =

0 7 1 2 4

16 6


Module 3

∴

( ) ( )( )

∴

( ) ( )

( )

( )( ) ( )( ) ( )( )

b) Horizontal distance = distance between x-intercepts

x-intercept, let y = 0 x-intercept, let y = 0

Horizontal distance = |4 – 8| = 4 units.

c) Vertical distance = distance between y-intercept

y-intercept, let x = 0 y-intercept, let x = 0

Vertical distance = |–3 – (–6)| = 3 units.

7.

Solve equation 2 for y: y = 2x + 4Substitute that value into Equation 1 for y.

Find y:

Point of intersection is (–2, 0).

This is an independent system.

y

y

= − += − +=

2 2 4

4 4

0

2 3 2 4 4

2 6 12 48 16

2

x x

x x

x

x

+ + = −+ + = −

= −= −


2 3 4

2 4

x y

y x

+ = −− =

3 4 24

4 246

x y

yy

− =− =

= −

3 4 12

4 123

x y

yy

− =− =

= −

3 4 24

3 248

x y

xx

− ===

3 4 12

3 124

x y

xx

− ===


Module 3

( )

∴

( )

8.

There is no solution. Therefore, it is an inconsistent system.

9. The two lines must coincide when graphed, e.g.,

There are infinite possibilities.

10. y = Ax2 + Bx and Q(2, 2)

Substitute P(–1, 5) into the equation.

Equation 1: 5 = A(–1)2

+ B(–1)

A – B = 5Substitute Q(2, 2) into the equation

Equation 2: 2 = A(2)2 + B(2)

4A + 2B = 2

2 x Equation 1

Equation is y = 2x2 – 3x

A 2A B 52 B 5

B 3B 3

=− =− =− =

= −

2A 2B 104A 2B 26A 12

− =+ =

=

x yx y

− =− =

62 2 12

Using a table ofvalues or a graphingcalculator, you getthe accompanyinggraph.

l

l x

y

l

2x − 3y = 1 2

l

2x − 3y = − 6


Module 3

11.

Substitute (Equation 5)

Substitute

Solution: − −119

23

19

, ,

− − − = −

− − = −

− = − +

− = −

=

119

23

2

179

2

2179

19

19

z

z

z

z

z

x y= − = −119

23

and into Equation 3

623

8

18 2 24

18 22

119

x

x

x

x

− = −

− = −= −

= −

y x y= − + = −23

6 8in

(4)(3)(2)(5)(6)

3 2 53 3 3 63 2 3 2

6 86 4 10

3 223

x y

x y z

x y z

x y

x y

y

y

+ = −+ − = −− + = −

+ = −+ = −− =

= −

Equation 1 – Equation 3Equation 3 x 3Equation 2Equation 2 + Equation 3Equation 4 x 2Equation 5 – Equation 6

(1)(2)(3)

4 3 7

3 2 3 22

x y z

x y zx y z

+ − = −− + = −

+ − = −


Module 3

− −119

23

19

, ,

Module 3

Lesson 1

Answer Key


1. Let x = length of the base of the isosceles triangle

y = length of each of the congruent sides of the isosceles triangle

(1)

(2)(1)

(2)

y x

x yx yx y

=

+ =− =+ =

32

2 603 2 0

2 60

4 6015452

225

xx

y

==

= =

cm is length of base

cm is the length of each congruent side.

2. Let x = hundreds digit

y = tens digit

z = units digit

Note: Remember a number like 576 means 100 • 5 + 7• 10+ 6 when you interpret its place value.

Accordingly, the number in this question is 100x + 10y + z.

Rearrange equationswhere necessary.

Eqn (2) ÷ 90:Eqn (3) ÷ 99:

Eqn (1) + Eqn (4):Eqn (5) + Eqn (6):

(4)(5)(6)

(1)(2)

(3)

x y zx y z x y zx y z x y z

+ + =+ + = + + ++ + = + + +

13100 10 10 100 90100 10 10 100 99

x y zx yx z

+ + =− =− =

1390 90 9099 99 99

x yx zx z

xx

− =− =+ =

==

11

2 143 15

5


Substitute x = 5 into Eqn (5):

Substitute x = 5, z = 4 into Eqn (1):

Rearrange

(1)(2)(3)

(4)(5)(6)(7)

Eqn (1) – Eqn (2):Eqn (2) – Eqn (3):Eqn (4) x 3:Eqn (5) x 2:Eqn (6) + Eqn (7):

Ordered triple: (5, 4, 4)Interpretation: The number is 544.

5 1

44

− =− = −

=

z

zz

5 4 13

4

+ + ==

y

y

Substitute z = 5 into Eqn (4):

Substitute y = 7, z = 5 into Eqn (1):

Ordered triple: (11, 7, 5)Interpretation: Melissa has 11 loonies, seven $5 bills, and five$10 bills

− − = −

− = −=

4 9 5 73

4 287

y

yy $5 bills

x

x

+ + ==

7 5 23

11 loonies

3. Let x = number of loonies

y = number of $5 bills

z = number of $10 bills$1x = value of the loonies

$5y = value of the $5 bills

$10z = value of the $10 bills

x y z

x y zx y z

+ + =+ + =

+ = +

23

1 5 10 961

x y z

x y zx y z

+ + =+ + =

− − = −

23

5 10 961

− − = −+ =

− − = −+ =

− = −=

4 9 73

6 11 97

12 27 219

12 22 194

5 25

5

y z

y z

y z

y z

z

z $10 bills

( )


Module 3

(1)(2)(3)

4. Let x = cost of a cauliflower

y = cost of a bunch of celery

z = cost of one bag of radishes

x y zx y zx y z

+ + =+ + =+ + =

$3.$8.$7.

672 4 53

2 3 1 04

Substitute z = $1.29 into Eqn (6):

Substitute y = $0.99, z = $1.29 into Eqn (1):

Eqn (2) – Eqn (1):Eqn (1) x 2:Eqn (3) – Eqn (5):Eqn (4) – Eqn (6):

y zx y z

y zzz

+ =+ + =

− = −==

3 4 862 2 2 734

034 16

29

.

..

$5.$1. for one bag of radishes

(4)(5)(6)

− = −=

1.29 0.3$0.99 for one bunch

of celery

yy

xx

+ + ==

0 99 129 3 67. . .$1.39 for one cauliflower


Module 3


Substitute y = 10, z = 15 into Eqn (1):

Ordered triple: (50, 10, 15)

The box is 50 cm long, 10 cm wide, and 15 cm high.

(1)(2)(3)

Rearrange

Eqn (1) – Eqn (2):Eqn (4) 3:Eqn (3) + Eqn (5):

5. Let x = length of the rectangular box

y = width of the rectangular box

z = height of the rectangular box

x y z

x y z

y z

x y z

x y z

y z

+ + == +

− =

+ + =− − =

− =

75

2

2 5

752 2 0

2 5

3 3 7525

3 30

10

y zy z

y

y

+ =+ =

==

(4)(5)

10 2515

+ ==

z

z

x

x

+ + ==

10 15 7550

( )

÷


Module 3


Substitute x = 10, y = 10 into Eqn (1):


There are 10 nickels, 10 dimes, and 5 quarters.

(1)(2)(3)

Divide by 0.05Divide by 0.05

Eqn (3) – Eqn (1):Eqn (1) 5:Eqn (2) – Eqn (5):Eqn (4) 4:Eqn (6) + Eqn (7):

6. Let x = number of nickels

y = number of dimes

z = number of quarters$0.05x = value of the nickels

$0.10y = value of the dimes

$0.25z = value of the quarters

x y z

x y zx y z

+ + =

+ + =+ + =

25

05 10 25 7510 25 05 75

$0. $0. $0. $2.$0. $0. $0. $3.

x yx y z

x y

x y

y

y

+ =+ + =− − = −

+ ===

4 505 5 5 125

4 3 70

4 16 200

13 130

10

(4)(5)(6)(7)

x

x

+ ==

4 10 50

10

10 10 255

+ + ==

z

z

x y z

x y zx y z

+ + =+ + =

+ + =

25

2 5 552 5 75

( )


Module 3

Substitute z = 25 into Eqn (3):

Substitute x = 11, z = 25 into Eqn (1):


The sides of the triangle are 11 cm, 16 cm, and 25 cm.

(1)(2)(3)

RearrangeRearrange

Eqn (1) + Eqn (2):

7. Let x = length of shortest side of the triangle

y = length of second side of the triangle

z = length of the longest side of the triangle

x y zz x y

z x

+ + =+ = +

− =

522

2 3

2 5025

zz

==

− + =− = −

=

2 25 32 22

11

xxx

11 25 5216

+ + ==

yy

x y zx y z

x z

+ + =− − + = −

− + =

522

2 3

Module 3

Lesson 2Answer Key


1. a) Graph Algebraic solution:

Because both equations areequal to y:

Substitute the x-values intothe linear equation.

The two ordered pairs are (0, 0) and (2, 16) meaningthere are two points ofintersection.

or

(2, 16 )

(1, 4)( 1, 4)

( 0 , 0 )

y = 8x

y = 4x 2

x

y y x

y x

==

48

2

4 8

4 8 0

4 2 0

2

2

x x

x x

x x

=

− =− =

4 00

x

x

==

x

x

− ==

2 02

y x

yy

== ⋅=

8

8 00

y x

yy

== ⋅=

8

8 216

( )

Module 3


b) Graph Algebraic solution:

Solve Eqn (2) for y in termsof x and substitute thatexpression into Eqn (1).

This quadratic equation doesnot factor. Use b2 – 4ac todetermine if it has a solution.

Because b2 – 4ac < 0, there is nosolution. This means the twographs do not intersect.

(1)

(2)

2(x 2) + (y 3) = 4

2

x + y + 3 = 0

( 3 , 0 )

x

y

C (2, 3)

r = 2

(0, 3)

x y

x

− + − =

+ + =

2 3 4

3 0

2 2

y x

x x

x x x x

x x

x x

= − −

− + − − − =

− + + + + =+ + =+ + =

3

2 3 3 4

4 4 12 36 4

2 8 36 0

4 18 0

2 2

2 2

2

2

a b c

b ac

= = =

− = −= −

1 4 18

4 4 4 1 18

56

2 2

, ,

c) Graph Algebraic solution:

Solve for y in Eqn (2) andsubstitute that expressioninto Eqn (1).

(1)

(2)

x + y = 2

x

y

(1, 1)

xy = 1

xy

x y

=+ =

1

2

y x

xy

x x

x x

x x

x

x

= −=

− =

− =− + =

− ==

21

2 1

2 1

2 1 0

1 0

1

2

2

2

y

( )

( )

( )( )

( )( )

( )( )


Module 3

The ordered pair (1, 1)represents the solution to thesystem.

If x = 1, y x= −= −=

22 11

d) Graph Algebraic solution:

Replace y by x in Eqn (1).

The two ordered pairs (3, 3)and (–3, –3) represent thesolution or the points ofintersection of the graph.

(1)

(2)

l

x + y = 182 2

x

y

l

(3, 3)

y = x

( 3 , 3 )

x yy x

2 2 18+ ==

x x

x

xx

2 2

2

2

18

2 18

93

+ =

=

== ±

If y xy

== ±3


Module 3

e) Graph

Algebraic solution:

Solve Eqn (2) for x and replace x in Eqn (1).

This does not factor easily so check the value of b2 – 4ac .

Because b2 – 4ac < 0, there is no solution and, therefore,no point of intersection.

(1)

(2)

x + y = 162 2

x

y

2y = x 1 0

x y

y x

2 2 162 10

+ == −

x y

x y

y y

y y y

y y

= +

+ =

+ + =

+ + + =

+ + =

2 10

16

2 10 16

4 40 100 16

5 40 84 0

2 2

2 2

2 2

2

a b c= = =− = −5 40 84

40 4 5 84 802

, ,( )

( )

( )


Module 3

2. a)

If x = 0, y = 0

If x = 3, y = 3

The coordinates of Q are (3, 3).

y x

y x x

x x x

x x

x x

== −= −

− =− =

4

4

3 0

3 0

2

2

2

x x= =0 3or

Replace y by x

b) You can use to find the x-coordinate of P.

The coordinates of vertex P are (2, 4).

xba

= −2

y x x

a b

= − += − =

2 41 4,

x

y x x

= −−

= =

= − +

= − += − +=

42 1

42

2

4

2 4 2

4 84

2

2

c) Distance from O(0, 0) to Q(3, 3):

= − + − = = =3 0 3 0 18 9 2 3 22 2

.

( )

( ) ( )

( ) ( )

( )


Module 3

d) Distance from P to OQ:

Equation of OQ: y = x or x – y = 0

Coordinates of P: (2, 4)x1 = 2, y1 = 4, A = 1, B = –1, C = 0

dAx By C

A B=

+ +

+1 1

2 2

d =+ − +

+ −

=−

=

1 2 1 4 0

1 1

2

222

2

2 2

or

e) Area of triangle:

A

A OQ

units

=

= ⋅

=

=

121212

3 2 2

3 2

bh

d

( )( )

( )( ) ( )( )


Module 3

3. a) (1)(2)

Substitute 3y for x2 in Eqn (1).

Find the corresponding x-values:

The ordered pairs are (3, 3) and (–3, 3).

x y

y x

2 2

2

18

3

+ ==

3 18

3 18 0

6 3 0

2

2

y y

y y

y y

+ =+ − =

+ − =y y= − =6 3or

3

3 6

18

2

2

2

y x

x

x

=− =

− =

3

3 3

93

2

2

2

y x

x

x

x

==

== ±

b) (1)

(2)

Solve Eqn (2) for x in terms of y:

Substitute for x in Eqn (1):

This has no solution.

x y

xy

2 22 22 0

− =+ =

xy

=−2

− − =

− =

− == + −= + −

= + −

22 2

42 2

4 2 2

0 2 2 4

0 2

0 2 1

22

22

4 2

4 2

4 2

2 2

yy

yy

y y

y y

y y

y y

y

y

2

2

2 0

2

+ == −

y

y

2 1 01

− == ±

This is notpossible becausex

2cannot be

negative.

( )( )

−2y

( )( )

( )( )


If y = 1 and

(–2, 1)

The solution is the ordered pairs (–2, 1) and (2, –1).

(2, –1)

xy

x

x

= −

= −

= −

2

212

If y = –1 and xy

x

x

= −

= −−

=

2

21

2

c) (1)

(2)

In Eqn (2) xy = 0

x = 0 or y = 0

Replace x = 0 in Eqn (1):

Points (0, 3) and(0, –3) are solutions.

The four solutions are (0, 3), (0, –3), (3, 0), and (–3, 0).

x xy y

xy

2 2 90

+ + ==

x xy y

y y

y

y

2 2

2 2

2

9

0 0 9

93

+ + =+ ⋅ + =

== ±

Replace y = 0 in Eqn (1):

Points (3, 0) and(–3, 0) are solutions.

x xy y

x x

x

x

2 2

2 2

2

9

0 0 9

93

+ + =+ ⋅ + =

== ±

d)

and

(1)

(2)

Eqn (1) + Eqn (2):

2 4

2 4

2 2

2 2

x y

x y

+ =− =

4 8

2

2

2

2

x

x

x

==

= ±

Substitute x = 2

2 2 4

4 4

00

22

2

2

+ =

+ ===

y

y

y

y

x = − 2 into Eqn (1)

∴ 2 0, is the ordered pair.

The solution is and2 0 2 0, , .−

∴ − 2 0, is the ordered pair.

2 2 4

4 4

00

22

2

2

− + =

+ ===

y

y

y

y

( )

∴ ∴

( )

( )( )

( )( )


e) (1)

(2)

Eqn (1) – Eqn (2): 0 = 5

This gives a false statement because 0 � 5. Therefore,there is no solution for this system.

x y

x y

2 2

2 2

9

4

+ =+ =

f) (1)

(2)

Eqn (2) 2 x2

+ y2

= 16 (3)Because the two equations are the same, the solution setis {(x, y)|x2 + y2 = 16}.

x y

x y

2 2

2 2

16

2 2 32

+ =+ =

4. Let x = width of the rectangley = length of the rectangle

x y

x y

2 2 2172 2 46

+ =+ =

x17

x

y

(1) (Pythagorean theorem)

(2) (Perimeter)

Eqn (2) 2 x + y = 23 (3)

Solve for y in terms of x in Eqn (3) and substitute into Eqn (1)

x = 15 or x = 8

If x = 15, y = 23 – 15 or 8.If x = 8, y = 23 – 8 or 15.

The width is 8 cm and the length is 15 cm.

y x

x y

x x

x x x

x x

x x

x x

= −+ =

+ − =

+ − + =− + =− + =− − =

23

289

23 289

529 46 289

2 46 240 0

23 120 0

15 8 0

2 2

2 2

2 2

2

2

÷

÷

( )( )

( )

Module 3


5. Let x = width of the rectangley = length of the rectangle

Perimeter: 2x + 2y = 26

Area: xy = 12Eqn (1) 2 x + y = 13 (1)

xy = 12 (2)

Solve Eqn (1) for y in terms of x and substitute that expressioninto Eqn (2).

y x

xy

x x

x x

x x

x x

= −=

− =

− =− + =

− − =

1312

13 12

13 12

13 12 0

12 1 0

2

2

If x = 12, then y = 1.

If x = 1, then y = 12.

The width of the rectangle is 1 cm and the length is 12 cm.

6. Let x = one numbery = second number

Sum: x + y = 27

Product: xy = 126

x + y = 27 (1)xy = 126 (2)

Solve for x in terms of y in Eqn (1) and then substitute thatexpression for x in Eqn (2):

( )

( )( )


Module 3

If y = 21, then x = 6.If x = 6, then y = 21.

The two numbers are 6 and 21.

x y

xy

y y

y y

y y

y y

y

= −=

− =

− =− + =− − =

=

27126

27 126

27 126

27 126 0

21 6 0

21 6

2

2

or


Sum: x + y = 3

Sum of Squares: x2+ y

2= 17

x + y = 3 (1)

x2 + y2 = 17 (2)

Solve for y in terms of x in Eqn (1) and substitute thatexpression for y in Eqn (2):

The two numbers that satisfy the equation are –1 and 4.

Check

y x

x y

x x

x x x

x x

x x

x x

= −+ =

+ − =

+ − + =− − =− − =

− + =

3

17

3 17

9 6 17

2 6 8 0

3 4 0

4 1 0

2 2

2 2

2 2

2

2

x x= = −4 1or

x y

x y

+ = →+ = →

3

172 2

− + = →

− + =

1 4 3

1 4 172 2

3 317 17

==

( )

( )( )

( )

( )( )

( )


Module 3


Since we are looking for positive numbers, x = 7. Substitutex = 7 into Eqn (2):

Eqn (1) + Eqn (2):

x y

x y

2 2

2 2

65

33

+ =− =

2 98

497

2

2

x

x

x

=== ±

(1)

(2)

x y

y

y

y

y

y

2 2

2 2

2

2

33

7 33

33 49

164

4

− =

− =

− = −− = −

= ±= The numbers are 7 and 4.

9. Let x = width of rectangley = length of rectangle

Solve Eqn (1) for y in terms of x and substitute thatexpression into Eqn (2) for y.

xy

x y

=+ =

60

1692 2

yx

x y

xx

xx

x x

x x

x x

=

+ =

+ =

+ =

+ =− + =

− − =

60

169

60169

3600169

3600 169

169 3600 0

144 25 0

2 2

22

22

4 2

4 2

2 2

(1)

(2)

x60

( )( )

( )


Module 3

Because the length and width cannot be negative, discard–12 and –5.

If x = 12, then y = 5.If x = 5, then y = 12.The dimensions of the rectangle would be 5 cm by 12 cm.

x x= ± = ±12 5or


Module 3

Notes

Module 3

Lesson 3Answer Key

1. The graph of y x + 2 is the half plane below and including

the line y = x + 2.

The graph of y 0 is the half plane above and including thex-axis. The graph of x 0 is the half plane to the left andincluding the y-axis.

2. a) — iii)

b) — iv)

c) — i)d) — ii)

3. Upper boundary is AB. AB is a horizontal line.

m = 0

Point online: (0, 3)Equation:

Inequality: The graph represents the part of the half planebelow and including the line y = 3, that is, y 3.


121

2

y x

yy

− = −

− ==

3 0 0

3 03

≤

≤≥

≤

( )

••

•

•••••••••

x

y

Module 3

Lower boundary is line segment CD (horizontal line).

m = 0Point on line: (–4, –1)

Equation:

Inequality: The graph represents the part of the half planeabove and including the line y = –1, that is, y –1.

The right boundary is the line segment BC.

Point on line: (9, 3)

Equation:

Inequality: The graph represents the part of the half planeabove and including the line y = x – 6, that is, y x – 6.

The left boundary is the line segment AD.

Point on line: (0, 3)

Equation:

Inequality: The graph represents the part of the half planebelow and including the line y = x + 3, that is, y x + 3.

The system of inequalities is:


y x

y

− − = − −

+ =

1 0 4

1 0

m =− −

−= =

3 19 5

44

1

y x

y xy x

− = −

− = −= −

3 1 9

3 96

m =− −− −

= =3 10 4

44

1

y x

y xy x

− = −

− == +

3 1 0

33

yy

y x

y x

≤≥ −≥ −≤ +

31

6

3

( )( )

( )

( )

≥

( )( )

( )

≤

≥

4. a) 2x – y > 2 (1)

y – 2x < 1 (2)

Rewrite in the form y = mx + b.

The graph of y < 2x – 2 is the half plane below the line y = 2x – 2.

The graph of y < 2x + 1 is the half plane below the line y = 2x + 1.

Algebraically this would be written as

b) 2x – y 2

5x – y 2Rewrite in the form y = mx + b.


Module 3

− > − +< −< +

2 2

2 22 1 (2)

y x

y xy x

Multiply by –1 involves changing thedirection of the inequality.

x y y x x y y x x y y x, | , | , | .< − ∩ < + = < −2 2 2 1 2 2

− ≥ − +≤ −

− ≥ − +≤ −

y xy x

y x

y x

2 22 2

5 2

5 2

. . . . . . . . . . Inequality 1

. . . . . . . . . . Inequality 2

The solution is theregion below y = 2x – 2.

≥

≥

( ){ } ( ) ( ){ } { }

The graph of y 2x – 2 is the line y = 2x – 2 and the halfplane below that line.

The graph of y 5x – 2 is the line y = 5x – 2 and the halfplane below that line.

The solution is the area denoted by .

Algebraically, this would be written as

c) x + y > 3–4x + y < 4


The graph of y > –x + 3 is the half plane above the line y = –x + 3.

The graph of y < 4x + 4 is the half plan below the line y = 4x + 4.



Module 3

x y y x x y y x, | , | .≤ − ∩ ≤ −2 2 5 2

y x

y x

> − +< +

3

4 4(1)(2)

x y y x x y y x, | , | .> − + < +3 4 4

{ } { }( ) ( )

≤

≤

{ }( ) ( ){ } ∩

The graph of y < –2x + 5 is the half plane below the line y = –2x + 5.

The graph of is the line and the

half plane above the line.

Algebraically this would be written as


Module 3

+ <

≥ −

2 5 (1)5

3 (2)4

x y

y x

d) Rewrite in the form y = mx + b.

< − +

≥ −

2 5 (1)5

3 (2)4

y x

y x

y x≥ −54

3 y x= −54

3

x y y x x y y x, | , | .< − + ∩ ≥ −2 554

3{ }( ) { }( )


Module 3

x y x y

y x

yx

+ > + <> − +

< − +

6 2 3 56

23

53

and5.

The graph of y > –x + 6 is the half plane above the line y = –x + 6.

The graph of is the half plane below the line


The word “and” means the intersection of the two sets ofpoints.

y x< − +23

53

y x= − +23

53

.

x y y x x y y x, | , | .< − + ∩ > − +23

53

6( ){ } ( ){ }


Module 3

x y x y− < + >2 5 2 3or

− < − +

> −

> − +

2 512

52

2 3

y x

y x

y x

6.

The graph of is the half plane above the line

The graph of y > –2x + 3 is the half plane above the liney = –2x + 3.


The word “or” means the union of the two sets of points.Therefore, the solution is all the shaded parts.

y x> −12

52

y x= −12

52

.

x y y x x y y x, | , | .> − ∪ > − +12

52

2 3


(1)

(2)

( ) { }( ){ }


Module 3

5 6 6

3 43

x y

x yy

− ≥+ ≤

≥ −

− ≥ − +

≤ −

≤ − +≥ −

6 5 656

1

3 43

y x

y x

y x

y

7.

The graph of is the line and the half

plane below the line.

The graph of y –3x + 4 is the line and the half plane below the line.

The graph of y –3 is the line y = –3 and the half planeabove the line.

y x≤ −56

1 y x= −56

1

y x= − +3 4


(1)

(2)

(3)

≤

≥


Module 3

x

y

y = 3x + 4

y = 156

x

2

4

2

22

y = 3

xyx y

y x

≥≥+ ≤

− > −

00

3 42 1

xyy xy x

≥≥≤ − +> −

00

3 42 1

8. a)

b)

The graph of x ≥ 0 is the line x = 0 and the half plane orthe right side of the line.

The graph of y ≥ 0 is the line y = 0 and the half planeabove that line.

The graph of y ≤ –3x + 4 is the graph of y = –3x + 4 andthe half plane below the line.

The graph of y > 2x – 1 is the half plane above the line y = 2x – 1.

The coordinates of the vertices are (0, 4) (0, 0), (1, 1), and(0.5, 0).



Module 3

x

y

y = 3 x + 4

2

4

22

y = 2x 1

2

9.

The graph of y x is the line y = x and the half plane belowthat line.

The graph of x 6 is the line x = 6 and the half plane on theleft side of the line.

The graph of y 0 is the graph of the line y = 0 and the halfplane above the line.

Area of base height

units

∆ = ×

=

=

1212

6 6

18 2

≥

≤

≤

–

–

–

–

( )( )

Module 3

Lesson 4Answer Key


1. a) y > –x2

+ x + 4 (1, 6)Substitute x = 1, y = 6

y x x> − + +

> − + +> − + +>

2

2

4

6 1 1 4

6 1 1 46 4

This results in a true statement, so (1, 6) is a solution ofthe inequality.

b) y 2x2

– x + 1 (–1, 5)Substitute x = –1, y = 5

y x x≤ − +

≤ − − − +≤ + +≤

2 1

5 2 1 1 1

5 2 1 15 4

2

2

This results in a false statement, so (–1, 5) is not asolution of the inequality.

2. a) — iv)b) — ii)c) — iii)d) — i)

( ) ( )

≤

( ) ( )

Module 3


3. a)

x

y

1

2

3

4

123

For y x x= +2 4

x-coordinate of vertex

y-coordinate

Zeros: Let y = 0

= −

=−

= −

ba24

2 1

2

= +

= − + −= −= −

x x2

2

4

2 4 2

4 84

x x

x x

2 4 0

4 0

+ =+ =

x x= = −0 4

The parabola is represented by a solid line and the regionbelow the parabola is shaded.

b)

x

y

1

2

3

4

123

1

2

For y x x= + +12

3 22

x-coordinate of vertex

y-coordinatey-intercept: 2

= −

= −⋅

= −

ba23

2 12

3

= + +

= − + − +

= + − +

= −

12

3 2

12

3 3 3 2

92

9 2

52

2

2

x xThe region above

is shadedy x x= + +12

3 22

and the outline of the curveis broken.

( )

( )

( )

( )

( ) ( )


Module 3

c)

x

y

1

2

3

4

123

1

2

3

( 2 , 0 )

(0, 0)

( 1, 3)For y x x< − −3 62

Graph y = – 3x2

– 6x anduse a broken line curve.

y-coordinate

x-coordinate

Zeros:

= −

= −−−

= −

ba2

62 3

1

= − −

= − − − −= − +=

3 6

3 1 6 1

3 63

2

2

x x− − =

− + =

+ =

3 6 0

3 2 0

2 0

2

2

x x

x x

x x

x x

x x

= + == = −

0 2 0

0 2

and

and

The shading is below the curve.

4. Find the solution set for the following:

a)

The critical numbers are –4 and 1.

• Draw the number line showing the intervals.

• Find the signs of the products in the three intervals.Interval: x –4 or (– , –4] Test: x = –5

x x

x x

2 3 4 0

4 1 0

+ − ≤+ − ≤ 14

+ +

f x x x

f

t

= + −

= − + − −

− − +

4 1

5 5 4 5 1

signs of factors sign of produc

( )( )

( )( ) ( )

( )( ) ( )

( )

( )( ) ( )

( )

( ) ( ) {=

≤


Module 3

Interval: –4 x 1 or [–4, 1] Test: x = 0

The quadratic is negative or zero in the interval–4 x 1 or [–4, 1]

f x x x

f

= + −

= + −

4 1

0 0 4 0 1

Interval: x 1 or [1, ] Test: x = 2

f x x x

f

= + −

= + −

4 1

2 2 6 2 1

b) 2 3 5 0

2 5 1 0

2x x

x x

+ − >+ − > 15

2

+ +

f x x x

f

= + −

− = − + −

2 5 1

3 6 5 3 1

Zeros or critical numbers:−52

1,

Interval: or Test pointx x< − −∞ − = −52

52

3, :

f x x x

f

+ −

= ⋅ + −

2 5 1

0 2 0 5 0 1

Interval:5

2or Test point

− < < − =x x15

21 0, . :

t

− +signs of factors sign of produc

( )( ) ( )

( )( ) ( )

( ) {=

≤

t

+ − –signs of factors sign of produc

( ) ( ) {=

( )( ) ( )

( )( ) ( )

≥

t

+ − +signs of factors sign of produc

( ) ( + ) {=

≤

( )( ) ( )

( )( )

−

∞

( )( ) ( )

−( )

−∞ −52

,

−52

1,

( )

( )( )

( )

( )

( )

=

t

− –signs of factors sign of produc

( ) {=+( )


Module 3

f x x x

f

= + −

= ⋅ + −

2 5 1

2 2 2 5 2 1

Interval: or Test pointx x> ∞ =1 1 2, . :

Solution: The quadratic is positive in the interval

x x x x< − > ∞ − ∞52

15

21or or , , .

c)

Zeros or critical numbers: –4, 5

Interval: x < –4 or (– , –4) Test: x = –5

x x

x x

2 20 0

5 4 0

− − <− + < 5

+ +

4

f x x x

f

= − +

− = − − − +

5 4

5 5 5 5 4

Interval: –4 < x < 5 or (–4, 5) Test: x = 0

f x x x

f

= − +

= − +

5 4

0 0 5 0 4

Interval: x > 5 or (5, ) Test: x = 6

f x x x

f

= − +

= − +

5 4

6 6 5 6 4

The quadratic will be negative in the interval{x|–4 < x < 5} or (–4, 5).

( )( ) ( )

t


( ) ( + ) {=

( )( ) ( )

( ){ } { } ∞ −52

,

( )( )

( )( )( )

( ) ( )( )

∞

t

– +signs of factors sign of produc

( ) {=−( )

( )( )( )

( )( )( )

t

+ –signs of factors sign of produc

( ) {=−( )

∞

( )( )( )

( )( )( )

t


( ) ( + ) {=

( )


Module 3

d)

Zeros or critical numbers: –6, 3

Interval: x –6 or (– , –6] Test: x = –7

x x

x x

2 3 18 0

6 3 0

+ − ≥+ − ≥ 36

+ +

f x x x

f

= + −

− = − − − −

6 3

7 7 6 7 3

Interval: –6 x 3 or [–6, 3] Test: x = 0

f x x x

f

= + −

= + −

6 3

0 0 6 0 3

Interval: x 3 or [3, ) Test: x = 6

f x x x

f

= + −

= + −

6 3

6 6 6 6 3

The quadratic is positive or zero where x –6or x 3 or on (– , –6] [3, ).

( )( ) ( )

( )( ) ( )

∞≤

( )( )

( )( )( )

( )( )( )

t

– –signs of factors sign of produc

( ) {=−( + )

≤

∞≥

( )( )( )

( )( )( )

t

+ +signs of factors sign of produc

( ) {=−( + )

∞≥ ∞≤

≤

t

– +signs of factors sign of produc

( ) {=( – )


Module 3

5. a)

b)

Zeros or critical numbers: –3, –1

Interval: x < –3 or (– , –3) Test: x = –4

xx

++

<31

0

f x x x

f

= + +

− = − + − +

3 1

4 4 3 4 1

Interval: –3 < x < –1 or (–3, –1) Test: x = –2

f x x x

f

= + +

− = − + − +

3 1

2 2 3 2 1

Interval: x > –1 or [–1, ) Test: x = 0

f x x x

f

= + +

= + +

3 1

0 0 3 0 1

The expression is negative in the interval –3 < x < –1or (–3, –1).

xx x

++ −

≥1

1 20

In this instance, cancel x + 1 from the numerator anddenominator, which now means x –1. Our inequalitybecomes

12

0x −

≥

2

+

13

+ +

( )( )

( )( )( )

( )

( )( )

( )( )( )

( )

( )( )

( )( )( )

( )

( )( )

∞

∞

− − +signs of factors sign of

( ) ( ) {=expression

+ − –signs of factors sign of


+ + +signs of factors sign of


≠

( )

( )( )


Module 3

Critical number: 2

Interval: x < 2 or (– , 2) Test: x = 1

f x x

f

= −

= −

2

1 1 2

Interval: x > 2 or (2, ) Test: x = 3

f x x

f

= −

= −

2

3 3 2

The expression is positive in the interval (2, ).

c)

Critical points: (x + 2)(x – 2) x = –2, x = 2

Interval: (– , –2) Test: x = –3

x

x

x x

x x

+−

≤

+ +− +

≤

24

0

2 2

2 20

2

22

+

2

+

f x x x

f

= + −

− = − + − −

2 2

3 3 2 3 2

(x + 2) can be cancelled meaning now x –2

Because the denominator cannot equal zero, x 2.

xx

+−

≤22

0

∞

∞

∞

≠

∞

≠

( )

( )( )

( )

( )

( )( )

( )

( )( )

( )( )

( )( )

( )( )( )

( )

( )

−signs of factor sign of

( ) {=expression

+ +signs of factor sign of

{=expression

– – +signs of factors sign of

{=expression

–{

{


Module 3

Interval: (–2, 2) Test: x = 0

f x x x

f

= + −

= + −

2 2

0 0 2 0 2

Interval: (2, ) Test: x = 3

f x x x

f

= + −

= + −

2 2

3 3 2 3 2

The solution is where

The solution is the interval

xx

+−

≤

∴ −

22

0

2 2

.

, .

d)

Critical numbers: –3, –1, 1, 2

Use solid dots for –3 and 2 because the numerator canequal 0. Use open dots for –1 and 1 since the denominatorcan never be zero.

2 51

11

0

2 5 11 1

1 11 1

0

2 3 5 2 11 1

0

61 1

0

3 2

1 10

2 2

2

xx

xx

x x

x x

x x

x x

x x x xx x

x xx x

x x

x x

++

−+−

≤

+ −+ −

−+ +− +

≤

+ − − − −+ −

≤

+ −+ −

≤

+ −+ −

≤

Simplify the rationalexpression

21

+ +

13

+

( )( )

( )( )

( )( )

( )( )

( )( ) ( )( )

( )( ) ( )( )

( )( )

( )( )

( )( )

( )( )( )

( )

( )( )

( )( )( )

( )

( )( )+ – –signs of factors sign of

{=expression

( )( )+ + +signs of factors sign of

{=expression

∞


Module 3

The solution is in the intervals where the expression isnegative; [–3, –1) (1, 2].

f x x x x x

f

= + − + −

− = − + − − − − − −

3 2 1 1

4 4 3 4 2 4 1 4 2

/

/

Interval: [–3, –1) Test point: x = –2

f x x x x x

f

= + − + −

− = − + − − − + − −

3 2 1 1

2 2 3 2 2 2 1 2 1

/

/

f x x x x x

f

= + − + −

= + − + −

3 2 1 1

0 0 3 0 2 0 1 0 1

/

/

Interval: (–1, 1) Test point: x = 0

f x x x x x

f

= + − + −

= + − + −

3 2 1 1

3 3 3 3 2 3 1 3 1

/

/

Interval: [2, ) Test point: x = 3

f x x x x x

f

= + − + −

= + − + −

3 2 1 1

32

32

332

232

132

1

/

/

Interval: (1, 2] Test point: x =

Interval: (– , –3] Test point: x = –4

32

∞

∞

( )( )

( )( )

( )( )

( )( )

( )( )( )( )

( )( )

( )( )

( )( )

( )( )

( )

( )

( )

( )

( )

( )

( )

( )( )( )( )

( )( )( )( )

( )

( )

( )( )( )( )

( )( )( )( )

32

+32

3 −32

2 +32

1 −32

1


Module 3

f x x x x x

f

= + − − +

− = − + − − − − − +

3 4 5 5

6 6 3 6 4 6 5 6 5

/

/

Interval: (–5, –3] Test point: x = –4

f x x x x x

f

= + − − +

− = − + − − − − − +

3 4 5 5

4 4 3 4 4 4 5 4 5

/

/

f x x x x x

f

= + − − +

= + − − +

3 4 5 5

0 0 3 0 4 0 5 0 5

/

/

Interval: [–3, 4] Test point: x = 0

f x x x x x

f

= + − − +

= + − − +

3 4 5 5

92

92

392

492

592

5

/

/

Interval: [4, 5) Test point: x =

Interval: (– , –5) Test point: x = –6

92

e)

Critical numbers: –5, –3, 4, 5

Because x –5 or 5 but f(x) can be zero, –5 and 5 haveopen dots and –3 and 4 have solid dots.

x x

x x

+ −− +

≥3 45 5

053

+ +

45

+( )( )( )( )

≠

∞

( )( )( )( )( )

92

+92

3 −92

4 −92

5 +92

5

( )( )

( )( )

( )( )

( )( )

( )

( )

( )

( )

( )( )( )( )

( )( )( )( )

( )

( )

( )( )( )( )

( )( )( )( )


Module 3

Therefore, x belongs to (– , –5) [–3, 4] (5, )

f x x x x x

f

= + − − +

= + − − +

3 4 5 5

6 6 3 6 4 6 5 6 5

/

/

Interval: (5, ) Test point: x = 6

f x x x x x

f

= + − − +

− = − + − − − − − +

2 1 3 3 1 2

3 6 1 3 3 9 1 3 2

/

/

f x x x x x

f

= + − − +

= + − − +

2 1 3 3 1 2

0 0 1 0 3 0 1 0 2

/

/

Interval: (– , –2) Test point: x = –3

f) 2 1 33 1 2

0x x

x x

+ −− +

≤31

2

+ +

13

2

+

Critical numbers:

Solid dots at and , open dots at and

− −

− −

212

13

3

12

3 213

, , ,

.

Interval: Test point :− =12

13

0, x

( ) ( ) ( ) ( ) ( )( )( ) ( ) ( )( ) ( )( )

1Interval: 2, Test point: 1

22 1 3 / 3 1 2

2 1 1 1 3 / 3 1 1 1 2

x

f x x x x x

− − = − = + − − +

= − + − − − − − +

∞

∞

∞

( )( )

( )( )

( )( )

( )( )

( )

( )

( )( )

( )( )

∞

( )

( )

( )( )( )

( )( )( )( )

( )

( )

( )

( )( )( )

( )( )( )( )

( )


Module 3

f x x x x x

f

= + − − +2 1 3 3 1 2

1 2 1 1 3 3 1 1 2

/

/

f x x x x x

f

= + − − +

= + − − +

2 1 3 3 1 2

4 8 1 4 3 12 1 4 2

/

/

Interval: [3, ) Test point: x = 4

Interval: Test point:13

3 1, =x

The solution is in the intervals where the expression is

negative − −212

13

3, , .

6. a)5

85

2

85

25

x

x

− ≤

− ≤

This means that the distance from to85

is less than orx

10 5

85

65

25 5

Algebraically:

equal to 25

.

− ≤ − ≤− + ≤ − + ≤ +

≤ ≤

≤ ≤

2 5 8 22 8 5 8 8 2 8

6 5 1065

2

65

2

x

x

x

x

,

Add 8 to each part

Divide by 5

65

2,

Factor 5 out

Divide by 5

65

2,

65

2,

+ − − +

2

∞

( )( )

( )( )

( )( )

( )( )

( )

( )

( )( )

( )( )

( )( )

( )( )( )

( )


Module 3

b) 2 4x ≤

2

10

22

Algebraically:

2 1 5

2 4

4 2 42 2

2 2

x

x

xx

+ ≤

≤

− ≤ ≤− ≤ ≤

− ,

Divide by 2

Factor out –4 and take itsabsolute value

Divide by 4

Subtract 1 from each side

Geometrically, this means that the distance from x to 0is 2 units or less.

−2 2,

c) 3 4 9

434

9

434

9

34

94

− >

− − >

− >

− >

x

x

x

x

This means that the distance from to 34

is greaterx

12 4

34

64

94

94

Algebraically, either

than94

units.

3 4 9

4 123

− < −− < −

>

x

xx

3 4 94 6

32

− >− >

< −

x

x

x

or

or

−∞ − ∞, ,32

3

−∞ − ∞, ,32

3

2

2

−∞ −,32

∞,3

−∞ −,32

∞,3

–

4 4


Module 3

d) 4 8 4

4 2 4

4 2 4

4 2 4

2 1

2 1

x

x

x

x

x

x

+ >

+ >

+ >

+ >

+ >

− − >

2 13

11

Algebraically,

Either:

4 8 4

4 123

x

xx

+ < −< −< −

4 8 4

4 41

x

xx

+ >> −> −

or

or

x x< − > −−∞ − − ∞

3 1

3 1

or or

, ,

−∞ − − ∞, ,3 1

Geometrically, this means that the distance from x to –2is greater than 1 unit.

4 8 4x + >

e) − + >

− >

< −

3 4 10

3 6

2

x

x

x

This is impossible when dealing with absolute valueinequalities, so there is no solution.

Take |–4|

Factor out 4

Divide by 4

Subtract 4 from each sideDivide by –3

( )

( )

( )( )

− − −23

1

2 1

1

2

( )( )

Notes


Module 3

Module 3

Lesson 5 Answer Key

1. a)

You need the coordinates of D.Since the diagonals of a rectangle bisect each other, themidpoint of AC and BD have the same coordinates.

Coordinates of D(–7, –7) constitute the fourth vertex.

b) Perimeter of the rectangle = 2l + 2w.

Midpoint of AC:

A C

Midpoint of AC:

Midpoint of BD:

B D

x x y y

x y

x y

x y

1 2 1 2

2 2

2 2

2 2

2 29 0 7 3

9 72

0 32

132

132

5 4

15

232

42

5 2 4 37 7

+ +

− −

∴− + + −

∴ − −

− −

∴ − = + − = +

+ = − + = −= − = −

,

( , ) ( , )

,

,

,

( , ) (?, ?)

x

y

B(5, 4 )

C(7, 3 )

D(?, ? )

A( 9, 0 )


∴

132

− −,

132

− −,

9 72

0 32

− + + −,

x x y y1 2 1 2

2 2+ +

,

Module 3

Let AB = lengthBC = width

c)

2.

The right angle (if there is one) will be opposite the longestside.

If mAC • mAB = –1, AC AB and A is a right angle.

x

y

C(4, 6 )

B(2, 8 )

A( 4, 2)

Area

units

=

= ⋅

=

lw

2 53 53

106 2

Perimeter:

units

2 2 53 2 53

4 53 2 53

6 53

⋅ + ⋅

= +

=

BC = − + −

= − + − −

= +

=

x x y y2 1

2

2 1

2

2 25 7 4 3

4 49

53

AB = − + −

= − − + −

= +

=

=

=

x x y y2 12

2 12

2 29 5 0 4

196 16

212

4 53

2 53


(Equation of AC)

⋅2 53

( )( ) ( )( )

( )( ) ( ( ))( )

3. A parallelogram is a quadrilateral that has opposite pairs ofsides parallel.

Because the slopes ofopposite sides are equal,opposite sides are parallel.

The quadrilateral is a parallelogram.

m

m

m m

m

m

m m

AB

CD

AB CD

BC

AD

BC AD

=− − −

− −= −

−=

= −− −

=

∴ = =

=− −

−=

=− −

− − −=

∴ = =

2 15 1

16

16

4 34 2

16

16

4 14 1

53

3 22 5

53

53

x

y

C(4, 4 )

B(1, 1 )

A( 5, 2)

D( 2 , 3 )

m

m

m m

AC

AB

AC AB

=− −− −

= =

=− − −

− −=

−= −

∴ ⋅ = − = −∴ ⊥ ∠

6 24 4

88

1

2 84 2

66

1

1 1 1

AC AB and A is a right angleTherefore, ABC is a right triangle.∆


Module 3

∴

( )

( )

( )

( )( )

( )

( )

( )

( )( )

4.

5.

22

32

3 2 4

3 6 43 10

103

rr

r

r

r

+= −

− + =− − =

− =

= −

mr

ml l3 4

3 12

4 21 5

64

32

= −+

=− −

−=

−= −

If then the slopes of the two lines are equal.l l3 4|| ,

22

32

1

62 2

11

2 2 6

2 4 62 2

1

x

x

x

x

x

x

+⋅ − = −

−+

= −

− + = −− − = −

− = −=

If , then l l m m

mx

m

l

l l

1 2 11 2

1 2

1

3 12

4 2

1 564

32

⊥ ⋅ = −

= −− −

=− −

−=

−= −

.


Module 3

( )

( )

( )

( )

( )

( )

6. First determine the slopes of the lines.

The opposite sides are || and adjacent sides areperpendicular.

Then prove that the sides are equal in length. To do this,find the four points of intersection and then find the distancebetween these points to ensure that all four sides have thesame length.

l l x y l l x y

x y x y1 2 2 33 4 6 1 4 3 33 1

4 3 33 2 3 4 19 2

: :− = + =

+ = − = −

Since m m l l

m m l l

m m l l

m m l l

1 2 1 2

2 3 2 3

3 4 3 4

4 1 4 1

1

11

1

⋅ = − ⊥⋅ = − ⊥⋅ = − ⊥⋅ = − ⊥

Since , then Since , then

m m l l

m m l l1 3 1 3

2 4 2 4

==

|| .|| .

ly x

y x m

l x y

y x y x m

ly x

y x m

l x y

y x y x m

1 1

2

2

3 3

4

2

44

3 64

34

32

34

4 3 33

3 4 3343

114

344

3 194

34

194

34

4 3 8 0

3 4 84

383

43

⇒ =−

= − =

⇒ + =

= − + =−

+ =−

⇒ = + = + =

⇒ + − =

= − + = − + = −

or

or

or

or


Module 3

( )

( )

( )

( )

∴

Now find the lengths of each side:

d

d

d

d

AB

BC

CD

AD

= − + − = + = =

= + + − = + = =

= − − + − = + = =

= − + − = + = =

6 3 3 7 9 16 25 5

3 1 7 4 16 9 25 5

1 2 4 0 9 16 25 5

6 2 3 0 16 9 25 5

2 2

2 2

2 2

2 2

∴ D 2 0,∴ −C 1 4,

3 x Eqn (1)4 x Eqn (2)Subtract

12 9 2412 16 24

25 00

4 3 0 8

4 82

x y

x y

y

y

x

x

x

+ =− =

==

+ ===

3 x Eqn (1)4 x Eqn (2)Add

9 12 5716 12 32

25 251

3 1 4 19

4 164

x y

x y

x

x

y

y

y

− = −+ =

= −= −

− − = −− = −

=

l l x y l l x y

x y x y3 4 4 13 4 19 1 4 3 8 1

4 3 8 2 3 4 6 2

: :− = − + =

+ = − =

∴ B 3 7,∴ A 6 3,

3 x Eqn (1)4 x Eqn (2)Subtract

12 9 9912 16 76

25 1757

4 3 7 33

4 123

x y

x y

y

y

x

x

x

+ =− = −

==

+ ===

3 x Eqn (1)4 x Eqn (2)Add

9 12 1816 12 132

25 1506

3 6 4 6

18 4 64 12

3

x y

x y

x

x

y

y

y

y

− =+ =

==

− =− =− = −

=


Module 3

( )

( )

( )

( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

It is a square because all four sides are equal, oppositesides are parallel, and adjacent sides are perpendicular.

Comment: Other solutions are possible. Showing adjacentsides are perpendicular shows that the quadrilateral is arectangle. To find the width, select any point on l1 and findits distance from l2. To find the length, select any point on l3and find its distance from l4. In this case, it turns out thatthe length and width are both 5, showing that the rectangleis a square.

7. Midpoint of RQ = M (0, 3).

To show that the median from P to RQ is an altitude, youmust show that PM RQ by showing that mPM • mRQ = –1.

PM is bothperpendicular to RQand a bisector of RQ.

m

m

RQ

PM

=

= −

∴ ⋅ − = −

5

15

515

1

x

y

Q(1, 8 )

P( 5 , 4)

R( 1, 2)

x

y

B(3, 7 )

C( 1, 4 ) A(6, 3 )

D(2, 0 )


Module 3

∴

− 15

∴

5

8.

Because the opposite sides have equal slopes, ABCD is aparallelogram as opposite sides are parallel.

m m

m m

AB CD

BC AD

= − −−

− − =− − −

− −=

−

=− −− −

= = =− − −

−= =

3 14 1

43

1 53 0

43

1 11 3

24

12

3 54 0

24

12

A: midpoint of PQ

B: midpoint of RQ

C: midpoint of RS

D: midpoint of PS

: , ,

: , ,

: , ,

: , ,

5 32

0 62

4 3

1 32

2 02

1 1

1 52

2 42

3 1

5 52

6 42

0 5

+ + −= −

− + + =

− + − + −= − −

+ − − + −= −

x

y

Q(3, 0 )

R( 1, 2)

S( 5, 4)

P(5, 6)

A

B

C

D


Module 3

( )

( )

( )

( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

,5

26

2+ − +( ) 5( ) 5−( ) − ( ) 4( ) 4−( ) −

,1

22

2− + + ( ) 4( ) 4−( ) −( ) 5( ) 5−( ) −

,5 3

20

2+ + ( ) 6( ) 6−( ) −

,1 32

2 02

− + +

ReviewAnswer Key

1. a)

Solve Equation 2 for y: y = –x + 2

Substitute into Equation 1 for y

When x = 3, y = –3 + 2 = –1

When x = –1, y = –(–1) + 2 = 3

Therefore, (3, –1) and (–1, 3) are the solutions.

b)

Add Equation 1 and Equation 2

When x = 4 When x = –4

Points of intersection

(4, 3) (4, –3) (–4, 3) (–4, –3)

x y

y

y

y

2 2

2 2

2

25

4 25

9

3

+ =

− + =

== ±

x y

y

y

y

2 2

2 2

2

25

4 25

9

3

+ =

+ =

== ±

x y

x y

x

x

x

2 2

2 2

2

2

25

7

2 32

164

+ =

− ==== ±

x x

x x x

x x

x x

x x

x

2 2

2 2

2

2

2 10

4 4 10

2 4 6 0

2 3 0

3 1 0

3 1

+ − + =

+ − + =− − =− − =

− + == −or

(1)(2)

x y

x y

2 2 102

+ =+ =


Module 3

( )

( )

( )

( )


Module 3

c)

Solve Equation 1 for y.

Substitute –3x – 3 for y in Equation 2.

Ordered pair: (0, –3)

Check solution in Equations 1 and 2:

The solution is (0, –3), which represents the point ofintersection.


Clear denominators and rearrange if necessary

Clear denominators and rearrange if necessary

33

1

31

3 33

3 33

x y

xy

x y

x y

x y

x y

+ = −

+ = −

+ = −+ = −+ = −+ = −

y x= − −3 3

x x

xx

x

y x

+ − − = −− − = −

− === − − = −

3 3 3

2 3 32 0

0

3 3 3

3 33

x y

x y

+ = −+ = −

3 0 3 3

0 3 3

+ − = −

+ − = −− = −− = −

3 3

3 3( )

( )

∴

( )


Module 3

d) Solve for y in Equation 1 and substitute that value intoEquation 2.


Remove denominator bymultiplying by 3

Substitute the x-value into Equation 1.

Clear denominator Multiply by 5

Ordered pair is

intersection.

Check answer in original to confirm the solution, which is

, which represents the point of

2 3 63 2 25

3 2 62 6

3

3 22 6

325

9 2 2 6 75

9 4 12 755 87

875

x y

x y

y x

yx

xx

x x

x x

x

x

+ = −+ =

= − −

= − −

+− −

=

+ − − =− − =

=

=

872 3 65

174 15 3015 204

20415685

y

yy

y

y

+ = − + = −

= −−=

−=

875

685

,−

875

685

, .−

2 63x− −

875

685

,−

875

685

, −

2. a) Choose test points to determine shading.

Algebraic Solution

b) Choose test points to determine shading.

Algebraic Solution

x y x y x y x y, | , |2 6 2 4+ ≤ − ≥

{ } ( ){ }− > + <, 2 4 , 2 6x y x y x y x y


Module 3

x( ) ( ) ( ) x( ) x6 4

3.

4. a) f(x) = (x + 4)(x – 1)

Interval: –x –4 or (– , –4) Test: x = –5

f(–5) = (–5 + 4)(–5 – 1)

(–) (–) = + productInterval: –4 x 1 or [–4, 1] Test: x = 0

f(0) = (0 + 4)(0 – 4)

(+) (–) = – product

Interval: x 1 or [1, ) Test: x = 2

f(2) = (2 + 4)(2 – 1)(+) (+) = + product

[–4, 1] or –4 x 1 is the solution.

b)

Solution: (2, ) x 2

21

+xx x

++ −

≥1

1 20

104

= − −

= − += = −

= −

< − −< −

2

2

20To find -intercepts0 ( 5)( 4)

5 4-intercept 20

Test point 0,0

0 0 0 200 20is not true so shade outside

y x x

x

x x

x x

y x

y

� 4 5


Module 3

∴

>

– ,

≤ ≤

≥

( ) ( )

≤ ≤

≤

–

–

––

c)

d)

5.

Since PM bisects QR, M(0, 3) is the midpoint of QR.

If PM QR, the slopes of these lines will be negativereciprocals.

Since the slopes of RQ and PM are negative reciprocals,PM QR, and therefore the median from P to QR is alsoperpendicular to QR.

m

m

of RQ

of PM

=− −− −

= =

=−

− −= −

8 21 1

102

5

3 40 5

15

Midpoint of QR =− + + −1 1

28 2

20 3

,( )

( , )

x

y

Q(1, 8)

P ( 5 , 4)

R ( 1, 2 )

M

− ≤ + ≤− ≤ ≤

− ≤ ≤

8 3 1 89 3 7

373

x

x

x

x

x xx x

+ >

+ < − + >< − >

2 7

2 7 2 79 5

oror


Module 3

−

− −

( )

( )

( )

Module 3: Answer Key - Open School

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Transcript of Module 3: Answer Key - Open School