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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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 2 267
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=
268 Section 1, Answer Key, Lesson 2 Principles of Mathematics 11
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
= =
=
Principles of Mathematics 11 Section 1, Answer Key, Lesson 2 269
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
270 Section 1, Answer Key, Lesson 2 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 2 271
Module 3
= + ( )+ + ( )+312
8 612
6 4
∠ ∠
∠ ∠∠
272 Section 1, Answer Key, Lesson 2 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 3 273
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»
274 Section 1, Answer Key, Lesson 3 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 3 275
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
( )° + ° = °
( )° = °
° = °
276 Section 1, Answer Key, Lesson 3 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 3 277
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
278 Section 1, Answer Key, Lesson 3 Principles of Mathematics 11
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×
Principles of Mathematics 11 Section 1, Answer Key, Lesson 3 279
Module 3
280 Section 1, Answer Key, Lesson 3 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 4 281
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⊥
282 Section 1, Answer Key, Lesson 4 Principles of Mathematics 11
∠ ∠
)(
∠∠
∠ ∠∠∴
∴
∠ ∠
∠∠∠
∠
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 4 283
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.
284 Section 1, Answer Key, Lesson 4 Principles of Mathematics 11
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.
Principles of Mathematics 11 Section 1, Answer Key, Lesson 4 285
Module 3
286 Section 1, Answer Key, Lesson 4 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 5 287
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
288 Section 1, Answer Key, Lesson 5 Principles of Mathematics 11
(
(
(
)
)
)
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 5 289
Module 3
( )( )
( )
( )
( )
( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( )
290 Section 1, Answer Key, Lesson 5 Principles of Mathematics 11
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
( )
∆
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.
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
Principles of Mathematics 11 Section 1, Answer Key, Review 293
Module 3
∴ ∠ 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
∠ ∠
∠
∠ ∠
∠
( )
( )
( )
•
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 1 295
Module 3
θ
θ
( )
( )( )
– –
–
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
296 Section 2, Answer Key, Lesson 1 Principles of Mathematics 11
Module 3
θ
( )( )
( )( )
−
θ
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 1 297
Module 3
( )( )
( )( )
θ
θ
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
298 Section 2, Answer Key, Lesson 1 Principles of Mathematics 11
Module 3
θ
θ( )
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 1 299
Module 3
θ
θ
−−
−
−
( )( )
( )
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
300 Section 2, Answer Key, Lesson 1 Principles of Mathematics 11
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°
Principles of Mathematics 11 Section 2, Answer Key, Lesson 2 301
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°
302 Section 2, Answer Key, Lesson 2 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 2 303
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=
304 Section 2, Answer Key, Lesson 2 Principles of Mathematics 11
Module 3
23
θ
θθ
θθθ
θθ
θθ
θ
θ
θθ
θ
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 2 305
Module 3
θ
θθ
θθ
θ
θθ
θθ
θθ
θ
306 Section 2, Answer Key, Lesson 2 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 3 307
Module 3
)(
)( ( )
)( ( )
)(
)( ( )
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
.
308 Section 2, Answer Key, Lesson 3 Principles of Mathematics 11
Module 3
)(
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 3 309
Module 3
A
C
B
45°
15.2 m
19.8 m
Pitcher's Mound
Home
First Base
)( )(
)()( )( )(
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°
310 Section 2, Answer Key, Lesson 3 Principles of Mathematics 11
Module 3
)(
)(
)(
∆
∴
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°
Principles of Mathematics 11 Section 2, Answer Key, Lesson 3 311
Module 3
∆
∆
( )
( )
( )
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°
312 Section 2, Answer Key, Lesson 3 Principles of Mathematics 11
Module 3
∠
∆
∴
)()( )(
)(
)(
)(
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 4 313
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
314 Section 2, Answer Key, Lesson 4 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 4 315
Module 3
∠
∠ ∠ ∠∠
∆
∆
∆
( )
( )
( )
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
=
=
=
=
316 Section 2, Answer Key, Lesson 4 Principles of Mathematics 11
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 )
Principles of Mathematics 11 Section 2, Answer Key, Lesson 4 317
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 )
318 Section 2, Answer Key, Lesson 4 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 4 319
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
320 Section 2, Answer Key, Lesson 4 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 4 321
Module 3
∠
( )
( )
322 Section 2, Answer Key, Lesson 4 Principles of Mathematics 11
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)
Principles of Mathematics 11 Section 2, Answer Key, Review 323
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
324 Section 2, Answer Key, Review Principles of Mathematics 11
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
−
= + −
= + −
= −
= −
= −
= =
= − =
Principles of Mathematics 11 Section 2, Answer Key, Review 325
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
−
=
=
=
=
∠ =
326 Section 2, Answer Key, Review Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Review 327
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
328 Section 2, Answer Key, Review Principles of Mathematics 11
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.
Principles of Mathematics 11 Section 3, Answer Key, Lesson 1 329
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
330 Section 3, Answer Key, Lesson 1 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 3, Answer Key, Lesson 1 331
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
332 Section 3, Answer Key, Lesson 1 Principles of Mathematics 11
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
( ) ( )
( ) ( )
( )
( ) ( )
( ) ( )
( )
∴
Principles of Mathematics 11 Section 3, Answer Key, Lesson 1 333
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
334 Section 3, Answer Key, Lesson 1 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 3, Answer Key, Lesson 1 335
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
336 Section 3, Answer Key, Lesson 1 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 3, Answer Key, Lesson 2 337
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
338 Section 3, Answer Key, Lesson 2 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 3, Answer Key, Lesson 2 339
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
340 Section 3, Answer Key, Lesson 2 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 3, Answer Key, Lesson 2 341
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.
342 Section 3, Answer Key, Lesson 2 Principles of Mathematics 11
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).
Principles of Mathematics 11 Section 3, Answer Key, Lesson 2 343
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
,
( )
( )
( )
( )
( )
( ) ( )( )
)(
)(
)( )(
)(
)(
)(
)(
344 Section 3, Answer Key, Lesson 2 Principles of Mathematics 11
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.
Principles of Mathematics 11 Section 3, Answer Key, Lesson 3 345
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.
346 Section 3, Answer Key, Lesson 3 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 3, Answer Key, Lesson 3 347
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
=
=
=
348 Section 3, Answer Key, Lesson 3 Principles of Mathematics 11
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
= −
=− ..
Principles of Mathematics 11 Section 3, Answer Key, Lesson 3 349
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
350 Section 3, Answer Key, Lesson 3 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 3, Answer Key, Lesson 4 351
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.
352 Section 3, Answer Key, Lesson 4 Principles of Mathematics 11
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
+ + =+ + =
= −= −= += − +=
∴
∴
)(
)(
Principles of Mathematics 11 Section 3, Answer Key, Lesson 4 353
Module 3
53
23
1
x yy
x y
+ =
+ =
d)
Equation 1Equation 2
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
,
( )
354 Section 3, Answer Key, Lesson 4 Principles of Mathematics 11
Module 3
e)
Equation 1Equation 2
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
( )
( ) ( ) ( )
∴
∴
Principles of Mathematics 11 Section 3, Answer Key, Lesson 4 355
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 1Equation 2
Equation 1 x 2Equation 2Subtract Equation 2 from(Equation 1)x 2
10 6 6
10 6 00 6
x y
x y
+ =+ =
=
∴
∴
∴
( ) ( ) ( )
356 Section 3, Answer Key, Lesson 4 Principles of Mathematics 11
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
− =− = −− =− = −
==
− = −− = −− = −− = −
=
Equation 1 x 2Equation 2 x 3
Subtract
Substitute x = 3 into Equation 2
∴
( )
( ) ( ) ( ) ( )
( )
Principles of Mathematics 11 Section 3, Answer Key, Lesson 4 357
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)
Equation 1Equation 2
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.
∴
∴
∴
( ) ( ) ( ) ( )
( ) ( )
358 Section 3, Answer Key, Lesson 4 Principles of Mathematics 11
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
Substitute x = 0 into Equation 1
Replace y in Equation 2
Substitute x = –2 into Equation 1
∴
( )
( )
( ) ( )
( )
Principles of Mathematics 11 Section 3, Answer Key, Lesson 4 359
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
+ =− − =
+ =− − =
− == −
+ =+ − =
==
Equation 1 x 3Equation 2 x 4
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
==
∴
( )( )( ) ( )
( )
( )( )
( )( )
∴
360 Section 3, Answer Key, Lesson 4 Principles of Mathematics 11
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
+ =− =
Equation 1Equation 2
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
− =− + =
Equation 1Equation 2
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}.
Principles of Mathematics 11 Section 3, Answer Key, Lesson 4 361
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
Equation 1 x 7Equation 2
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.
Equation 1Equation 2
( )
( )
( )
( ) ( )
362 Section 3, Answer Key, Lesson 4 Principles of Mathematics 11
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
+ − =
+ =− =+ =− =
= −− =
− − =+ =
==
Equation 1 x 3Equation 2
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
Substitute x = 6 into Equation 1
∴
( )( )
( )( )
( )
Lesson 5Answer Key
1. Substitute the point into the equation to test whether it is asolution.
Principles of Mathematics 11 Section 3, Answer Key, Lesson 5 363
a)
(1, –2, –3) is asolution
(0, –2, –4) 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
(2, –3, –1) is 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
364 Section 3, Answer Key, Lesson 5 Principles of Mathematics 11
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):
( )( )
( )
( )
( )
( )( )
Principles of Mathematics 11 Section 3, Answer Key, Lesson 5 365
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
∴
∴
( )
( )
( )
( ) ( )
( )
366 Section 3, Answer Key, Lesson 5 Principles of Mathematics 11
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
− =− − =
− =− − =
− == −
∴
( )
( )
Principles of Mathematics 11 Section 3, Answer Key, Lesson 5 367
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
− + + ==
∴
∴
( )
( )( )
368 Section 3, Answer Key, Lesson 5 Principles of Mathematics 11
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
+ + − =
+ + + − + = + +
+ + − =
Principles of Mathematics 11 Section 3, Answer Key, Review 369
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 )
370 Section 3, Answer Key, Review Principles of Mathematics 11
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
,
,
Principles of Mathematics 11 Section 3, Answer Key, Review 371
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
372 Section 3, Answer Key, Review Principles of Mathematics 11
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
Principles of Mathematics 11 Section 3, Answer Key, Review 373
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
+ + = −+ + = −
= −= −
Equation 1Equation 2
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
− ===
374 Section 3, Answer Key, Review Principles of Mathematics 11
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
Principles of Mathematics 11 Section 3, Answer Key, Review 375
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
+ − = −− + = −
+ − = −
376 Section 3, Answer Key, Review Principles of Mathematics 11
Module 3
− −119
23
19
, ,
Module 3
Lesson 1
Answer Key
Principles of Mathematics 11 Section 4, Answer Key, Lesson 1 377
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
378 Section 4, Answer Key, Lesson 1 Principles of Mathematics 11
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
( )
Principles of Mathematics 11 Section 4, Answer Key, Lesson 1 379
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
380 Section 4, Answer Key, Lesson 1 Principles of Mathematics 11
Module 3
Substitute y = 10 into Eqn (5):
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
( )
÷
Principles of Mathematics 11 Section 4, Answer Key, Lesson 1 381
Module 3
Substitute y = 10 into Eqn (4):
Substitute x = 10, y = 10 into Eqn (1):
Ordered triple: (10, 10, 5)
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
( )
382 Section 4, Answer Key, Lesson 1 Principles of Mathematics 11
Module 3
Substitute z = 25 into Eqn (3):
Substitute x = 11, z = 25 into Eqn (1):
Ordered triple: (11, 16, 25)
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
Principles of Mathematics 11 Section 4, Answer Key, Lesson 2 383
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
384 Section 4, Answer Key, Lesson 2 Principles of Mathematics 11
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
( )
( )
( )( )
( )( )
( )( )
Principles of Mathematics 11 Section 4, Answer Key, Lesson 2 385
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
386 Section 4, Answer Key, Lesson 2 Principles of Mathematics 11
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
, ,( )
( )
( )
Principles of Mathematics 11 Section 4, Answer Key, Lesson 2 387
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
.
( )
( ) ( )
( ) ( )
( )
388 Section 4, Answer Key, Lesson 2 Principles of Mathematics 11
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
( )( )
( )( ) ( )( )
Principles of Mathematics 11 Section 4, Answer Key, Lesson 2 389
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
( )( )
( )( )
390 Section 4, Answer Key, Lesson 2 Principles of Mathematics 11
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
( )
∴ ∴
( )
( )( )
( )( )
Principles of Mathematics 11 Section 4, Answer Key, Lesson 2 391
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
392 Section 4, Answer Key, Lesson 2 Principles of Mathematics 11
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):
( )
( )( )
Principles of Mathematics 11 Section 4, Answer Key, Lesson 2 393
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
7. Let x = one numbery = second number
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
==
( )
( )( )
( )
( )( )
( )
394 Section 4, Answer Key, Lesson 2 Principles of Mathematics 11
Module 3
8. Let x = one numbery = second number
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
( )( )
( )
Principles of Mathematics 11 Section 4, Answer Key, Lesson 2 395
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
396 Section 4, Answer Key, Lesson 2 Principles of Mathematics 11
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.
Principles of Mathematics 11 Section 4, Answer Key, Lesson 3 397
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:
398 Section 4, Answer Key, Lesson 3 Principles of Mathematics 11
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.
Principles of Mathematics 11 Section 4, Answer Key, Lesson 3 399
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
Rewrite in the form y = mx + b.
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.
Algebraically, this would be written as
400 Section 4, Answer Key, Lesson 3 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 4, Answer Key, Lesson 3 401
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{ }( ) { }( )
402 Section 4, Answer Key, Lesson 3 Principles of Mathematics 11
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
Algebraically, this would be written as
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( ){ } ( ){ }
Principles of Mathematics 11 Section 4, Answer Key, Lesson 3 403
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.
Algebraically, this would be written as
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
Rewrite in the form y = mx + b.
(1)
(2)
( ) { }( ){ }
404 Section 4, Answer Key, Lesson 3 Principles of Mathematics 11
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
Rewrite in the form y = mx + b.
(1)
(2)
(3)
≤
≥
Principles of Mathematics 11 Section 4, Answer Key, Lesson 3 405
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).
Rewrite in the form y = mx + b.
406 Section 4, Answer Key, Lesson 3 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 4, Answer Key, Lesson 4 407
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
408 Section 4, Answer Key, Lesson 4 Principles of Mathematics 11
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.
( )
( )
( )
( )
( ) ( )
Principles of Mathematics 11 Section 4, Answer Key, Lesson 4 409
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
( )( )
( )( ) ( )
( )( ) ( )
( )
( )( ) ( )
( )
( ) ( ) {=
≤
410 Section 4, Answer Key, Lesson 4 Principles of Mathematics 11
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
( ) {=+( )
Principles of Mathematics 11 Section 4, Answer Key, Lesson 4 411
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
+ − +signs of factors sign of produc
( ) ( + ) {=
( )( ) ( )
( ){ } { } ∞ −52
,
( )( )
( )( )( )
( ) ( )( )
∞
t
– +signs of factors sign of produc
( ) {=−( )
( )( )( )
( )( )( )
t
+ –signs of factors sign of produc
( ) {=−( )
∞
( )( )( )
( )( )( )
t
+ − +signs of factors sign of produc
( ) ( + ) {=
( )
412 Section 4, Answer Key, Lesson 4 Principles of Mathematics 11
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
( ) {=( – )
Principles of Mathematics 11 Section 4, Answer Key, Lesson 4 413
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
( ) ( ) {=expression
+ + +signs of factors sign of
( ) ( ) {=expression
≠
( )
( )( )
414 Section 4, Answer Key, Lesson 4 Principles of Mathematics 11
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
–{
{
Principles of Mathematics 11 Section 4, Answer Key, Lesson 4 415
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
∞
416 Section 4, Answer Key, Lesson 4 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 4, Answer Key, Lesson 4 417
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
( )( )
( )( )
( )( )
( )( )
( )
( )
( )
( )
( )( )( )( )
( )( )( )( )
( )
( )
( )( )( )( )
( )( )( )( )
418 Section 4, Answer Key, Lesson 4 Principles of Mathematics 11
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
− − = − = + − − +
= − + − − − − − +
∞
∞
∞
( )( )
( )( )
( )( )
( )( )
( )
( )
( )( )
( )( )
∞
( )
( )
( )( )( )
( )( )( )( )
( )
( )
( )
( )( )( )
( )( )( )( )
( )
Principles of Mathematics 11 Section 4, Answer Key, Lesson 4 419
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
∞
( )( )
( )( )
( )( )
( )( )
( )
( )
( )( )
( )( )
( )( )
( )( )( )
( )
420 Section 4, Answer Key, Lesson 4 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 4, Answer Key, Lesson 4 421
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
422 Section 4, Answer Key, Lesson 4 Principles of Mathematics 11
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 )
Principles of Mathematics 11 Section 4, Answer Key, Lesson 5 423
∴
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
424 Section 4, Answer Key, Lesson 5 Principles of Mathematics 11
(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.∆
Principles of Mathematics 11 Section 4, Answer Key, Lesson 5 425
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
⊥ ⋅ = −
= −− −
=− −
−=
−= −
.
426 Section 4, Answer Key, Lesson 5 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 4, Answer Key, Lesson 5 427
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
− =+ =
==
− =− =− = −
=
428 Section 4, Answer Key, Lesson 5 Principles of Mathematics 11
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 )
Principles of Mathematics 11 Section 4, Answer Key, Lesson 5 429
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
430 Section 4, Answer Key, Lesson 5 Principles of Mathematics 11
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
+ =+ =
Principles of Mathematics 11 Section 4, Answer Key, Review 431
Module 3
( )
( )
( )
( )
432 Section 4, Answer Key, Review Principles of Mathematics 11
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.
Equation 1Equation 2
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( )
( )
∴
( )
Principles of Mathematics 11 Section 4, Answer Key, Review 433
Module 3
d) Solve for y in Equation 1 and substitute that value intoEquation 2.
Equation 1Equation 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
434 Section 4, Answer Key, Review Principles of Mathematics 11
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
Principles of Mathematics 11 Section 4, Answer Key, Review 435
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
436 Section 4, Answer Key, Review Principles of Mathematics 11
Module 3
−
− −
( )
( )
( )