MPM2H
Principles of Mathematics – Transfer Course
(Transfer from Grade 10 Applied Level to
Grade 10 Academic Level)
(Revised Nov. 2006)
MPM2H – Principles of Mathematics (Transfer Course)
Introduction
Grade 10 Mathematics Applied to Academic Transfer
Welcome to the Grade 10 Mathematics Applied to Academic Transfer Course, MPM2H.
This half-credit course is part of the new Ontario Secondary School curriculum.
This course will provide students who have successfully completed Foundations of
Mathematics, Grade 10, Applied with an opportunity to achieve the expectations not
covered in that course but included in Principles of Mathematics, Grade 10, Academic.
On successful completion of this transfer course, students will be able to proceed to
either the university preparation course or the university/college course in mathematics
in Grade 11.
This transfer course focuses on developing algebraic skills used in solving of quadratic
functions and in analytic geometry, and on investigating the trigonometry of acute
triangles.
Material
This course is self-contained and does not require a textbook. You will require lined
paper, graph paper, a ruler, a scientific calculator and a writing utensil.
Expectations
The overall expectations you will cover in the lesson are listed on the first page of each
lesson.
Lesson Description
Each lesson contains one or more concepts with each being followed by support
questions. At the end of the lesson the key questions covering all concepts in the
lesson are assigned and will be submitted for evaluation.
Evaluation
In each lesson, there are support questions and key questions. You will be evaluated
on your answers to the key questions in each of the ten lessons and the final exam.
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MPM2H – Principles of Mathematics (Transfer Course)
Introduction
Support Questions
These questions will help you understand the ideas and master the skills in each
lesson. They will also help you improve the way you communicate your ideas. The
support questions will prepare you for the key questions.
Write your answers to the support questions in your notebook. Do not submit these
answers for evaluation. You can check your answers against the suggested answers
that are given at the end of each unit.
Key Questions
The key questions evaluate your achievement of the expectations for the lesson. Your
answers will show how well you have understood the ideas and mastered the skills.
They will also show how well you communicate your ideas.
You must try all the key questions and complete most of them successfully in order to
pass each unit. Write your answers to the key questions on your own paper and
submit them for evaluation at the end of each unit. Make sure each lesson number
and question is clearly labelled on your submitted work.
Final Examination
In this course there is a final exam. The exam will incorporate the four learning
categories knowledge and understanding, application, communication, and thinking and
inquiry.
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MPM2H – Principles of Mathematics (Transfer Course)
Introduction
Unit One
Lesson One Concepts
¾
¾
¾
¾
collecting like terms
distributive law
expanding second degree polynomial expressions
simplifying second degree polynomial expressions
Lesson Two Concepts
¾ finding the greatest common factor
¾ factoring quadratic relations of the form ax 2 + bx + c where a = 1
¾ factoring quadratic relations of the form ax 2 + bx + c where a is an integer
and a ≠ 1
¾ factoring difference of squares trinomials
¾ factoring perfect square trinomials.
Lesson Three Concepts
¾ solving quadratic equations by factoring
¾ solving quadratic equations using the quadratic formula
Lesson Four Concepts
¾ calculating, identifying and graphing the x and y intercepts of a quadratic
equation
¾ calculating, identifying and graphing the vertex of a quadratic equation
¾ recognizing that a quadratic equation can be expressed in standard form and/or
factored form
¾ calculating the axis of symmetry
Lesson Five Concepts
¾ calculating the distance between two coordinates on a Cartesian plane
¾ calculating the coordinates of the midpoint given the end coordinates of a line
segment
¾ reviewing the calculation of slope using coordinates
¾ properties of various geometric shapes
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MPM2H – Principles of Mathematics (Transfer Course)
Introduction
Unit Two
Lesson Six Concepts
¾ determining the equation of a circle having centre (0, 0) and radius r, by applying
the distance formula
¾ identify the radius of a circle of centre (0, 0), given its equation
¾ write an equation of a circle given its radius
¾ sketch the graph of a circle given its equation
Lesson Seven Concepts
¾
¾
¾
¾
¾
¾
¾
recognizing and locating the circumcentre
recognizing and locating the centroid
finding the equation of a perpendicular bisector
finding the equation of the median lines of a triangle
calculating the point of intersection using algebra
calculating the coordinates of the midpoint of a line segment
calculating slope and perpendicular slopes
Lesson Eight Concepts
¾
¾
¾
¾
quadrilateral classifications
calculating slope
calculating lengths of segments
calculating perpendicular line segments
Lesson Nine Concepts
¾
¾
¾
¾
drawing a labelling triangle angles and sides
interpreting information and drawing diagrams based on that information
using the sine law to find an unknown side in a triangle
using the sine law to find an unknown side in a triangle
Lesson Ten Concepts
¾
¾
¾
¾
drawing a labelling triangle angles and sides
interpreting information and drawing diagrams based on that information
using the cosine law to find an unknown side in a triangle
using the cosine law to find an unknown side in a triangle
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Polynomials
Lesson 1
MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 1
Lesson One Concepts
¾
¾
¾
¾
collecting like terms
distributive law
expanding second degree polynomial expressions
simplifying second degree polynomial expressions
What is a Polynomial?
Polynomials are terms that have variables to whole number exponents.
5x 3
Is a polynomial
− 2 x −1
7
y2
Not a polynomial
Not a polynomial
The variable “x” has a whole number
exponents of 3
Has a negative exponent
Has the variable in the denominator
Below is an example of a second degree polynomial express:
The leading term has an
exponent of two making this a
second degree polynomial
The term without a
variable is the
constant term.
Expanding using the Distributive Law (Property)
Example #1
Simplify 3(x – 4)
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 1
Answer:
Example #2
Simplify 2y(3y + 2)
Answer:
Support Questions
Directions: Simplify each polynomial by using the distributive law.
1
a. 4(2w + 5)
d. –(3x + 4)
b. 5x(-3x – 7)
e. –6x(-x + 2)
c. 2y(-4y + 8)
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 1
Expanding and Simplifying Binomials using the Distributive
Law (Property)
Example #1
= 2x 2 − 6x + 4x − 12
= 2x 2 − 2x − 12
You can only add/subtract “like terms” if the terms
have the same variable(s) and each variable has
matching exponents. The only like terms in this
expression are –6x and 4x.
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 1
Example #2
Simplify (7x – 3)(x + 5)
Answer:
(7x − 3)(x + 5) = 7x(x) + 7x(5) − 3(x) − 3(5)
= 7x 2 + 35x − 3x − 15
= 7x 2 + 32x − 15
Support Questions
Directions: Expand and simplify each expression.
2.
a. (x + 3)(x - 4)
d. (3x + 1)(2x - 9)
g. (4n – 2) 2
b. (2y - 3)(7y - 1)
e. (7w - 3)(5w + 6)
h. (-7s –3)(s + 1)
c. (-t - 5)(t + 2)
f. (x - 5) 2
3.
a. 5(3y – 2)(y+7)
b. –2(y – 3)(5y – 6)
c. 6(4x + 3)(5x – 2)
4.
a. (x – 2y)(3x + 5y)
b. (-x + 3w)(w – 1)
c. –8(3w – 7y)(2w + y)
5.
a. –(x + 4) 2
b. 3x + (2x – 3)(x + 5)
c. -y(2y + 4x)(y - x)
6.
a. (2x + 3)(3x – 2) + x(3x –4)
c. (x + 4) 3
b. –2(3x – 2) 2 -5(x +2) 2
d. [(x + 2)(x – 3)] 2
e. (x – 1) 4
Key Question #1
Directions: For questions 1 – 5, expand and simplify each expression.
1.
a. (x - 6)(x - 7)
d. (3y + 1)(y - 4)
g. (-3n + 7) 2
b. (4y - 5)(8y + 3)
e. (6w - 4)(w + 9)
h. (-6t – 1)(t + 3)
c. (-x + 7)xt - 2)
f. (x - 9) 2
2.
a. 2(5y – 1)(y - 3)
b. –(y - 3)(2y – 5)
c. 3(2x - 5)(3x + 1)
3.
a. (x – y)(x + 2y)
b. (-w + 2x)(w - 8)
c. –8(3s – 4r)(9s + r)
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 1
Key Question #1 (continued)
4.
a. –(x - 3) 2
5.
a. (2x + 3)(x + 7) 2 + x(2x – 6)
c. (x - 2) 3
e. (x – 2) 4
6.
Write a simplified expression for the area of each object given below.
a.
b. 7x - (x – 4)(-x -3)
c. -d(2d - 1p)(-3d - p)
b. –(3y – 7) -5(y + 3) 2
d. [(x - 2)(x + 8)] 3
b.
7.
Given the expression (3s − 4)(4s + 5) , explain in words the steps taken to
completely simplify the expression.
8.
Directions: Fill in the missing terms (in your notebook)
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Factoring
Lesson 2
MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 2
Lesson Two Concepts
¾ finding the greatest common factor
¾ factoring quadratic relations of the form ax 2 + bx + c where a = 1
¾ factoring quadratic relations of the form ax 2 + bx + c where a is an integer
and a ≠ 1
¾ factoring difference of squares trinomials
¾ factoring perfect square trinomials
Finding the Greatest Common Factor
When factoring it is always important to factor out the greatest common factor first.
You are trying to factor out the largest number and largest degree of exponent for each
common variable from each term in the expression.
Example #1
Find the greatest common factor (GCF) for the following expression: 4x 3 − 8x 2 − 2x
Answer:
The largest number that divides evenly into each term is 2 and the largest
degree variable that divides into each term is x. The GCF is 2x
Example #2
Factor by factoring out the GCF: 12y 4 − 6y 3 − 24y 2
Answer:
The largest number that divides evenly into each term is 6 and the largest degree
variable that divides into each term is y 2 . The GCF is 6y 2
To factor by
taking out the
GCF you divide
each term by the
GCF.
12y 4 6y 3 24y 2
− 2 −
= 3y 2 − y − 4
2
2
6y
6y
6y
The factored form of 12y 4 − 6y 3 − 24y 2 is 6y 2 ( 3y 2 − y − 4 )
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 2
Example #3
Factor 9x +3
Answer: 3(x + 1)
3 is the GCF and is factored out of 9x and 3
Support Questions
Directions: factor completely
1.
a. 6c + 2
b. 100x - 20
d. 12x + 3x
2
2
e. 7r - 21r
c. 5g + 20
f. 16f + 12gf + 32
Factoring Quadratic Relations of the form ax + bx + c where
a=1
2
The following are examples of trinomials of the form ax 2 + bx + c where a = 1:
y 2 − 8y + 15
x 2 + 3x − 10
w 2 − 12w + 32
t 2 + 4t − 21
Example #1
Factor y 2 − 8y + 15
Answer:
Steps used to find the factors:
#1 list all the integers that multiply to equal + 15
15 , 1
-15 , -1
3,5
-3 , -5
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 2
#2 from the list find the pair that add to equal the value of the middle
term’s coefficient
Example #2
Factor x 2 + 3x - 10
Answer:
Support Questions
Directions: factor completely
2.
a. c 2 + 3c + 2
d. x 2 + 3x - 18
g. m 2 + 6m + 9
b. x 2 + 2x - 15
e. r 2 - 6r + 9
h. n 2 - 5n - 14
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c. g 2 + 5g + 4
f. f 2 - 2f - 8
i. b 2 - 8b + 15
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 2
Factoring Quadratic Relations of the form ax + bx + c where
a is an integer and a ≠ 1
2
The following are examples of trinomials of this form:
6y 2 − y - 12
5x 2 + 13x − 6
Example #1
Factor 6y 2 − y - 12
Answer:
Steps used to find the factors:
#1 list all the pairs of positive numbers that multiply to equal the product
of the first and last term’s coefficients. Only use the positive value of that
number. i.e. (6) x (–12) = -72 but use +72
6, 12;
24, 3;
9, 8;
72, 1;
36, 2;
18, 4
#2 from this list of numbers pick the pair that can give an answer that
equals the middle term’s coefficient when added or subtracted from each
other.
8 – 9 = -1
-1 is the value of the middle term.
#3 rewrite the expression using the pair chosen in step #2
6y 2 + 8y − 9y - 12
#4 factor out the GCF of the first two terms and factor out the GCF of the
last two terms
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 2
Example #2
Factor 5x 2 + 13x − 6
Answer:
30, 1;
15, 2
10 + 3 = 13
10, 3
6, 5
+13 is the value of the middle term
5x 2 + 10x + 3x − 6
5x(x + 2) + 3(x – 2) this one doesn’t work so try
15 – 2 = 13
+13 is the value of the middle term
5x 2 + 15x − 2x − 6
5x( x + 3) -2( x + 3)
Therefore 5x 2 + 13x − 6 = ( x + 3)(5x – 2)
Support Questions
Directions: factor completely
3.
a. 9c 2 + 12c + 4
d. 2x 2 + 3x + 1
g. 6m 2 + 5m + 1
b. 2x 2 + 5x + 2
e. 3r 2 - 11r - 4
h. 3n 2 - 5n - 2
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c. 10g 2 + 3g - 1
f. 4f 2 - 16f + 15
i. 10b 2 + b - 3
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 2
Factoring Quadratic Relations of the form a x − b ;
Difference of Squares
2
2
2
The following are examples of trinomials of this form:
9y 2 - 16
25x 2 − 1
Example #1
Both end terms must be
perfect squares and the two
terms must be separated by a
negative sign.
Factor 9y 2 - 16
Answer:
Example #2
Factor 25x 2 − 1
Answer:
25x 2 − 1 = (5x − 1)(5x + 1)
Support Questions
Directions: factor completely
4.
a. c 2 - 36
d. 16x 2 − 81
b. x 2 - 64
e. 50r 2 - 72
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c. 25g 2 - 49
f. 7f 2 - 28
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 2
Factoring Quadratic Relations of the form a x + 2abx + b ;
Perfect Square
2
2
2
The following are examples of trinomials of this form:
4y 2 - 12y + 9
25x 2 + 20x + 4
Example #1
Both end terms must
be perfect squares.
Factor 4y 2 - 12y + 9
Answer:
4y 2 - 12y + 9
Example #2
Factor 25x 2 + 20x + 4
Answer:
25x 2 + 20x + 4 = (5x + 2)(5x + 2) or (5x + 2) 2
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 2
Support Questions
Directions: factor completely
5.
a. 4c 2 - 20c + 25
d. 64x 2 + 16x + 1
b. 36x 2 + 60x + 25
e. 100r 2 - 180r + 81
c. 16g 2 - 24g + 9
f. 9r 2 - 12r + 4
Key Question #2
Directions: For questions 1 – 6, factor completely
1.
a. 8c + 4
d. 24x 2 + 16x
b. 75x - 25
e. 21r 2 - 35r
c. 6g + 30
f. 9f + 15gf + 18g
2.
a. c 2 + 4c + 4
d. x 2 - 7x - 18
b. x 2 − 2x - 8
e. r 2 + 10r + 9
c. g 2 + 8g + 12
f. x 2 - 10x + 24
3.
a. 3c 2 - c - 2
d. 2x 2 - x - 1
b. 2x 2 + 11x + 14
e. 3r 2 - 10r + 7
c. 4g 2 + 17g - 15
f. 6f 2 - f - 2
4.
a. c 2 - 49
d. 49x 2 − 81
b. x 2 - 100
e. 75r 2 - 108
c. 36g 2 - 25
f. 5f 2 - 20
5.
a. 9c 2 - 6c + 1
d. 49x 2 - 84x + 36
b. 36x 2 + 120x + 100
e. 4r 2 + 4r + 1
c. 25g 2 + 20g + 4
f. 9r 2 - 30r + 25
6.
a. 3c 2 + 6c
d. 9x 2 − 16
b. x 2 - 11x + 28
e. 3r 2 - 432
c. 10g 2 - 28g + 16
f. 16r 2 - 8r + 1
7.
Explain in the detail the steps need to factor 9x 2 − 6x − 24 .
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Solving Quadratic
Equations
Lesson 3
MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 3
Lesson Three Concepts
¾ solving quadratic equations by factoring
¾ solving quadratic equations using the quadratic formula
Solving Quadratic Equations with Factoring
Solving a quadratic equation is often called finding the zeros or roots and gives the
value(s) of x when y = 0.
Example #1
Solve: y = x 2 − 3x − 10
Answer:
Step #1 factor the quadratic equation
x 2 − 3x − 10 = (x – 5)(x + 2)
Step #2 set each factor that contains a variable equal to zero
x – 5 = 0 and x + 2 = 0
Step #3 solve for x for each equation in step #2
x–5=0
x=5
x+2=0
x = -2
Therefore the roots of x = 5 and x = -2
Example #2
Solve: y = 3n 2 − 6n
Answer:
Step #1 factor the quadratic equation
3n 2 − 6n = 3n(n – 2)
Step #2 set each factor that contains a variable equal to zero
3n = 0 and n – 2 = 0
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 3
Step #3 solve for x for each equation in step #2
3n = 0
n=0
n-2=0
n=2
Therefore the roots of n = 0 and n = 2
Example #3
A rock is thrown straight down from a building 95 metres high. The relation
h = −5t 2 − 25t + 95 is a model that gives the approximate height of the rock h, in metres,
at t seconds after it is thrown. How long does it take the rock to reach a ledge 20
metres from the base of the building?
Answer:
h = −5t 2 − 25t + 95
20 = − 5t 2 − 25t + 95
20 – 20 = − 5t 2 − 25t + 95 - 20
0 = − 5t 2 − 25t + 70
0 = -5( t 2 + 5t - 14 )
0 = -5(t + 7)(t – 2)
t – 2 = 0 and t + 7 = 0
t=2
t=-7
Therefore the roots of t = 2 and t = - 7
Since you cannot have negative time (–7 seconds) the rock reaches a
height of 25 metres at 2 seconds.
Support Questions
Directions: Factor each expression. Then find the roots of each equation.
1.
a. x 2 + 7x − 30 = 0
d. 4x 2 − 1 = 0
b. t 2 + 8t + 15 = 0
e. 9m 2 - 6m + 1 = 0
c. n 2 - 2n − 15 = 0
f. 5y 2 + 25y + 30 = 0
Directions: Solve each equation.
2.
a. 42 = y 2 - y
d. x 2 = 30 - 7x
b. t 2 - 4t = 21
e. 3 = 6m 2 - 7m
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c. n 2 = 2n + 48
f. 5y 2 = 3 - 2y
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 3
Support Questions (continued)
Directions: Determine all the values of n that satisfy the equation.
3.
a. y = n 2 − n − 30, when y = −24
c. y = n 2 − 7n + 12, when y = 2
b. y = n 2 − 3n − 28, when y = −10
d. y = n 2 − 8n + 16, when y = 4
4.
A person jumps of a 20 m high cliff into water below. The height of the person
above the ground at time t seconds is given by the equation h = −5t 2 + 30.
How long does it take the person to reach the half way mark to the water?
Solving Quadratic Equations with the Quadratic Formula
The quadratic formula can be used to solve any quadratic equation but is best used
when solving for quadratic equations that don’t factor. Factoring doesn’t always work
but the quadratic formula does.
Here is the quadratic formula
− b ± b − 4ac
x=
2a
2
2
The quadratic formula uses the “a”, “b” and “c” from ax + bx +c = 0, where a, b and c
are numbers.
It is important to understand that the quadratic formula only works if the quadratic
equation is set to or equal to zero.
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 3
Example #1
Solve: 0 = x 2 − 3x − 10
So the roots of the equation are x = 5 and x = -2
Example #2
Solve: 0 = 3x 2 − 5x + 2
Answer:
a=3
b = -5
c=2
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MPM2H – Principles of Mathematics (Transfer Course)
x=
− b ± b 2 − 4ac
2a
x=
− (-5) ± (-5) 2 − 4(3)(2)
2(3)
Unit 1 – Lesson 3
5 ± 25 − 24
2(3)
5 ±1
x=
6
2
x = 1 and x =
3
x=
Support Questions
Directions: Solve using the quadratic formula.
5.
a. 6x 2 - x - 15 = 0
d. 4x 2 − 10x + 25 = 0
b. t 2 − 16 = 0
e. 6m 2 + 32 = 80
c. n 2 + 9n + 20 = 0
f. (w − 5) 2 = 16
6.
a. The annual budget B, in $billions, for the NHL is approximated
by the quadratic relation B = −0.1492x 2 + 1.8058x + 9 , where x is
the number of years since 1988. What is the NHL’s budget in
2001?
b. When does the model indicate that the NHL’s budget was about
$8 billion?
7.
A cliff diver in Mexico dives from about 16 m above the water.
The diver’s height above the water h, in meters, after t seconds
is modelled by h = −4.9t 2 + 1.5t + 16 . How long is the diver in the
air?
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 3
Key Question #3
Directions: Factor each expression. Then find the roots of each equation.
1.
a. x 2 + 7x + 12 = 0
d. 25x 2 − 9 = 0
b. t 2 + 6t + 8 = 0
e. 16m 2 - 8m + 1 = 0
c. 2n 2 - 2n − 24 = 0
f. y 2 - 11y + 28 = 0
Directions: Solve each equation.
2.
a. 15 + y = 2y 2
d. x 2 = 2x + 48
b. t 2 + 5t = -6
e. 25 = 16m 2
c. 2n 2 = 3 - 5n
f. 2y 2 = −4 - 9y
Directions: Determine all the values of n that satisfy the equation.
3.
a. y = 6n 2 − n − 1, when y = 50
b. y = 2n 2 − 9n + 10, when y = 6
4.
A model rocket is shot straight up from the roof of a school. The height at any
time t is approximated by the model h = −5t 2 + 23t + 10 , where h is the height in
metres and t is the time in seconds. When does the rocket hit the ground?
Directions: Solve using the quadratic formula.
5.
a. 2x 2 + 2x - 9 = 0
d. 2x 2 + 5 = 167
b. 2t 2 − 12t + 18 = 0
e. m 2 − 8m − 24 = 0
c. 3n 2 − 4 = 0
f. 12x 2 − 40 = 17x
6.
a. The annual budget B, in $billions, for the NBA is approximated by
the quadratic relation B = −0.1592x 2 + 1.7058x + 8.5 , where x is the
number of years since 1989. What is the NBA’s budget in 2002?
b. When does the model indicate that the NBA’s budget was about
$8 billion?
7.
a. Find the roots of 2(x - 3) 2 -11 = 0 to two decimal places by isolating (x – 3) 2 ,
then taking the square root of both sides.
b. Solve the equation in (a) by expanding (x – 3) 2 , then using the quadratic
formula.
a. Which method involves the fewer steps?
8.
The altitude of a triangle is 3 m longer than its base. What are the dimensions of
the altitude and the base if the area of the triangle is 40 m 2 ?
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Sketching Quadratic
Equations
Lesson 4
MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 4
Lesson Four Concepts
¾ calculating, identifying and graphing the x and y intercepts of a quadratic
equation
¾ calculating, identifying and graphing the vertex of a quadratic equation
¾ recognizing that a quadratic equation can be expressed in standard form and/or
factored form
¾ calculating the axis of symmetry
Graphing a quadratic equation in factored form.
y = a(x − s)(x − t)
Below are four characteristics of a quadratic equation that will help produce a sketch of
a given quadratic equation.
1. If “a” is positive then the parabola produced by the quadratic equation opens up.
2. When the equation is in factored form the zeros can be determined to recognize
where/if the parabola crosses the x-axis.
3. Setting x = 0 and solving the quadratic equation for y determines the value where
the parabola crosses the y-axis.
4. Finding the equation of the axis of symmetry, which is midway between the zeros,
and substituting that value of x into the quadratic equation will determine the vertex
of the quadratic equation.
Example #1
Sketch y = (x − 3)(x + 5)
Answer:
1. since there is no value in front of the first set of parentheses then a = 1 and
the parabola opens up
2. solve for the zeros. x = 3 and x = -5 so the parabola crosses the x–axis at 3
and –5.
Copyright © 2005, Durham Continuing Education
Page 29 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 4
3. Set x = 0 and find the y-intercept
y = (x − 3)(x + 5)
y = (0 - 3)(0 + 5)
y = (-3)(5)
y = -15
4. The axis of symmetry is midway between the two zeros.
3 + −5
2
−2
x=
2
x = −1
x=
Copyright © 2005, Durham Continuing Education
Page 30 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 4
5. substitute the value of the axis of symmetry into the quadratic equation
to find the coordinates of the vertex.
y = (-1 − 3)(-1 + 5)
y = (-4)(4)
y = -16
The coordinates of the vertex are (-1, -16)
Copyright © 2005, Durham Continuing Education
Page 31 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 4
6. Once these points have all been graphed then connect the points with a
smooth curve producing a parabola to represent the quadratic equation
y = (x − 3)(x + 5) .
Example #2
Sketch y = - 12n 2 + 14n + 6
Answer:
1. - 12n 2 + 14n + 6 completely factored becomes –2( 2x –3 )(3x + 1)
2. since the value in front of the first set of parentheses then a = -2 and the
parabola opens down
3. solve for the zeros. x =
3
1
and x = −
so the parabola crosses the x – axis at
2
3
3
1
and − .
2
3
Copyright © 2005, Durham Continuing Education
Page 32 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 4
4. Set n = 0 and find the y-intercept
y = -12(0) 2 - 14(0) + 6
y=6
5. The axis of symmetry is midway between the two zeros.
x=
−
1 3
+
3 2
2
7
x= 6
2
7
x=
≈ 0.6
12
Copyright © 2005, Durham Continuing Education
Page 33 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 4
6. substitute the value of the axis of symmetry into the quadratic equation to find the
coordinates of the vertex.
2
⎛ 7 ⎞
⎛ 7 ⎞
y = -12⎜ ⎟ + 14⎜ ⎟ + 6
⎝ 12 ⎠
⎝ 12 ⎠
121
y=
12
⎛ 7 121 ⎞
The coordinates of the vertex are ⎜ ,
⎟
⎝ 12 12 ⎠
7. Once these points have all been graphed then connect the points with a smooth
curve producing a parabola to represent the quadratic equation
y = 12n 2 + 14n + 6 .
Copyright © 2005, Durham Continuing Education
Page 34 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 4
Support Questions
Directions: For each of the following quadratic equations:
a.
b.
c.
d.
e.
f.
1.
state whether the quadratic relation opens up or down and why?
state the y-intercept for each equation
solve the quadratic equation
calculate then state the equation for the axis of symmetry
determine the coordinates of the vertex of each quadratic equation
sketch each quadratic equation using the y-intercept, the zeros, and the
vertex of each.
a. y = 3(x + 2)(x − 7)
d. y = −2x 2 + 8x
b. y = (2x + 1)(x − 4)
e. y = 5x 2 - 10x − 15
c. y = 3x 2 + 5x − 2
f. y = 24x 2 - 30x + 9
Key Question #4
Directions: For each of the following quadratic equations in question one:
a.
b.
c.
d.
e.
f.
1.
State whether the quadratic relation opens up or down and why?
State the y-intercept for each equation.
Solve the quadratic equation.
Calculate then state the equation for the axis of symmetry.
Determine the coordinates of the vertex of each quadratic equation.
Sketch each quadratic equation using the y-intercept, the zeros, and the
vertex of each.
a. y = x 2 - x
d. y = x - 5x − 14
2
2.
b. y = (x + 3)(x − 2)
e. y = 2x 2 + x − 3
c. y = -3(2x - 1)(x + 3)
f. y = -3x 2 - 3x + 6
Describe and sketch the transformation of relations when the quadratic equation
y = x 2 + 4x + 3 is changed to y = -2x 2 - 8x − 6.
Copyright © 2005, Durham Continuing Education
Page 35 of 110
Midpoint and Length
of Line Segments
Lesson 5
MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 5
Lesson Five Concepts
¾ calculating the distance between two coordinates on a Cartesian plane
¾ calculating the coordinates of the midpoint given the end coordinates of a line
segment
Length of a Line Segment (distance between coordinates)
The distance formula is a version of the Pythagorean Theorem.
To find the length of a side in a right angle triangle the Pythagorean Theorem is used.
The Pythagorean Theorem states: a 2 + b 2 = c 2 , where a, b and c are lengths of sides in
a right angle triangle.
This formula can be simplified to c = a 2 + b 2
Example #1
Find the distance between the two points (-3, 2) and (1, 4).
Answer:
A lines could be drawn creating a right angle triangle.
Then the length of each side could be determined
Copyright © 2005, Durham Continuing Education
Page 37 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 5
Then Pythagorean Theorem could be applied to find the distance between
the two points.
c = 42 + 22
c = 16 + 4
c = 20
c ≈ 4.47
This entire process can be simplified by using the following formula to
calculate the distance between two points.
Again using the coordinates given in this example
d = (x − x )2 + (y − y )2
2
1
2
1
d=
[( −3) − (1)]2 + [(2) − (4)]2
d = ( −4) 2 + ( −2) 2
d = 16 + 4
d = 20
d ≈ 4.47
Copyright © 2005, Durham Continuing Education
Page 38 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 5
Example #2
Find the distance between the two points (5, 3) and (-2, 9).
Answer:
d = (x − x )2 + (y − y )2
2
1
2
1
d=
[(5) − (-2)]2 + [(3) − (9)]2
d = (7)2 + ( −6) 2
d = 49 + 36
d = 85
d ≈ 9.22
Support Questions
1.
Calculate the distance between each pair of the coordinates given.
a. (4, -2); (-1, -6)
d. (7, 8); (-5, 5)
b. (-2, 6); (9, 4)
e. (1, 6); (4, 10)
c. (11, 3); (5, 6)
f. (-7, -3); (-5, -4)
2.
A quadrilateral has vertices at A(-5, 1), B(3, 3); C(4, -1); D(-4, -3). Find the slope
and length of each side and then use this information to identify the type of
quadrilateral described by the given vertices.
3.
What is the distance by air from Toronto to Hamilton? One unit represents 50
km.
Copyright © 2005, Durham Continuing Education
Page 39 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 5
The Midpoint between Two Coordinates
When finding middle value between two numbers the two numbers are added together
and divided by two. The mean average is found.
To find the midpoint between two coordinates this same principle is applied.
To find the midpoint between two coordinates the following formula is used:
⎛x +x
y +y ⎞
Midpoint = ⎜⎜ 1 2 , 1 2 ⎟⎟
2
2 ⎟⎠
⎜
⎝
Example #1
Find the coordinates of the midpoint between (-3, 2) and (1, 4).
Answer:
⎛ x + x 2 y1 + y 2 ⎞
⎟
Midpoint = ⎜ 1
,
⎜
⎟
2
2
⎝
⎠
⎛ (-3) + (1) (2) + (4) ⎞
,
=⎜
⎟
2
2 ⎠
⎝
⎛ −2 6⎞
=⎜
, ⎟
⎝ 2 2⎠
= ( −1,3)
Example #2
Find the coordinates of the midpoint between (7, 2) and (-8, 9).
Answer:
⎛ x + x 2 y1 + y 2 ⎞
⎟
Midpoint = ⎜ 1
,
⎜
⎟
2
2
⎝
⎠
⎛ (7) + (-8) (2) + (-9) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛ −1 - 7 ⎞
=⎜ , ⎟
⎝ 2 2 ⎠
= ( −0.5, - 3.5)
Copyright © 2005, Durham Continuing Education
Page 40 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 5
Support Questions
4.
Calculate the midpoint between each pair of the coordinates given.
a. (4, -2); (-1, -6)
d. (7, 8); (-5, 5)
b. (-2, 6); (9, 4)
e. (1, 6); (4, 10)
c. (11, 3); (5, 6)
f. (-7, -3); (-5, -4)
5.
A rectangle is defined by the vertices A(10, 0), B(-8, 6), C(-12, -6), and D(6, -12).
Show that the diagonals bisect each other.
6.
Show that the mid-segments of a square with the vertices E(2, -12), F(-10, -8),
G(-6, 4), and H(6, 0) form a square.
Key Question #5
1.
Calculate the distance between each pair of the coordinates given.
a. (5, 2); (-3, -6)
d. (6, 7); (-6, 4)
b. (-1, 9); (6, 4)
e. (-2, -7); (3, -9)
c. (10, -2); (4, 5)
f. (-8, -2); (-6, -3)
2.
A quadrilateral has vertices at A(-3, 2), B(2, 4); C(6, -1); D(1, -3). Find the slope
and length of each side then use this information to identify the type of
quadrilateral described by the given vertices.
3.
What is the distance by air from Mississauga to Markham? One unit represents
50 km.
Copyright © 2005, Durham Continuing Education
Page 41 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 1 – Lesson 5
Key Question #5 (continued)
4.
A builder needs to connect a partially built house to a
temporary power supply. On the plan, the coordinates of
the house are (50, 112) and the power supply is at (145,
83). What is the least amount of cable needed? Why is it
likely that the builder
will use more cable than this?
5.
A fence is needed behind a barn to hold horses. The plans
for the pasture show an irregular pentagonal area with
vertices at (2, 0), (1, 6), (8, 9), (10, 7), and (6, 0). What
amount of fencing to keep the horses in the pasture? Each
unit represents 20 m.
6.
Calculate the midpoint between each pair of the coordinates given.
a. (5, 2); (-3, -6)
d. (6, 7); (-6, 4)
b. (-1, 9); (6, 4)
e. (-2, -7); (3, -9)
c. (10, -2); (4, 5)
f. (-8, -2); (-6, -3)
7.
A rectangle is defined by the vertices A(-6, 5), B(12, -1), C(8, -13), and D(-10, -7).
Prove that the diagonals bisect each other.
8.
Show that the mid-segments of a rhombus with the vertices E(-5, 2), F(-1, 3),
G(-2, -1), and H(-6, -2) form a rectangle.
9.
What characteristics are needed too prove the classification of a shape when just
given the coordinates of it vertices. Explain with words and an example.
Copyright © 2005, Durham Continuing Education
Page 42 of 110
Equations of Circles
Lesson 6
MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 6
Lesson Six Concepts
¾ determining the equation of a circle having centre (0, 0) and radius r, by applying
the distance formula
¾ identify the radius of a circle of centre (0, 0), given its equation
¾ write an equation of a circle given its radius
¾ sketch the graph of a circle given its equation
The Circle and its Equation
A circle is a set of all points in a plane that are the same distance from a fixed point, the
centre. The distance from any point on the circle to the centre is called the radius “r”.
If the centre of the circle is at the origin of the x-y Cartesian plane and the radius is r
units then r = x 2 + y 2 , which is equal to x 2 + y 2 = r 2 .
x 2 + y 2 = r 2 is the equation of a circle with centre (0, 0) and radius r.
Example #1
Write the equation of a circle with its centre (0,0) and a radius of 3 units.
Answer:
x2 + y2 = r 2
x 2 + y 2 = (3) 2
x2 + y2 = 9
Copyright © 2005, Durham Continuing Education
Page 44 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 6
Example #2
Sketch the graph given by the equation x 2 + y 2 = 25 and what are its x and y intercepts?
Answer:
Example #3
What is the radius of a circle that has the equation x 2 + y 2 = 121 and what would be this
equations x and y intercepts if graphed?
Answer:
x 2 + y 2 = 121
x 2 + y 2 = (11) 2
∴ r = 11
The x intercepts are ± 11 and the y intercepts are also ± 11.
Support Questions
1.
Write the equation of the circle with its centre at (0, 0) and given the radius, r.
a. r = 4
d. r = 15
b. r = 10
1
e. r =
3
Copyright © 2005, Durham Continuing Education
c. r = 0.5
Page 45 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 6
Support Questions (continued)
2.
Graph each circle and state the x- and y-intercepts.
a. x 2 + y 2 = 36
1
d. x 2 + y 2 =
16
3.
b. x 2 + y 2 = 0.25
c. x 2 + y 2 = 1
Find the radius of a circle with centre (0, 0) and passing through
a. (-3 ,4)
d. (6, -8)
b. (5, 0)
c. (0, -4)
4.
What are the equations for each in question 3.
5.
Determine if the point is on, inside or outside the circle given by x 2 + y 2 = 100 .
Explain your reasoning for each.
a. (6, 8)
d. (10,1)
b. (5, 9)
c. (7, 7)
Key Question #6
1.
Write the equation of the circle with its centre at (0, 0) and given the radius, r.
a. r = 2
d. r = 13
2.
c. r = 1
Graph each circle and state the x- and y-intercepts.
a. x 2 + y 2 = 2.25
25
d. x 2 + y 2 =
4
3.
b. r = 12
1
e. r =
2
b. x 2 + y 2 = 14.44
c. x 2 + y 2 = 64
Find the radius of a circle with centre (0, 0) and passing through
a. (2.5 ,3)
d. (-8,-6 )
b. (0, -7)
Copyright © 2005, Durham Continuing Education
c. (14, 0)
Page 46 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 6
Key Question #6 (continued)
4.
What are the equations for each in question 3.
5.
Determine if the point is on, inside or outside the circle given by x 2 + y 2 = 49 .
Explain your reasoning for each.
a. (2, 6)
d. (6, 5)
b. (-3, 5)
c. (-4, 5)
6.
Points (a, -5) and (-9, b) are on the circle x 2 + y 2 = 125 . What are the values of
“a” and “b”?
7.
Two satellites are orbiting Earth. The path of
one has an equation of x 2 + y 2 = 4 840 000 .
The orbit of the other is 250 km farther from the
centre of the earth. In one orbit, how much
farther does the second satellite travel than the
first?
8.
Give two examples of real life situations that can be modelled by a circle. Include
an equation.
Copyright © 2005, Durham Continuing Education
Page 47 of 110
Triangle Properties
Lesson 7
MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 7
Lesson Seven Concepts
¾
¾
¾
¾
¾
¾
¾
recognizing and locating the circumcentre
recognizing and locating the centroid
finding the equation of a perpendicular bisector
finding the equation of the median lines of a triangle
calculating the point of intersection using algebra
calculating the coordinates of the midpoint of a line segment
calculating slope and perpendicular slopes
The Centroid of a Triangle
The centroid of a triangle can be determined by finding the equations of two of the
medians lines, then finding the point of intersection of those two lines.
Then centroid is the centre of mass or “balance point” of the triangle.
Example #1
A triangle has the vertices A(-4, 5), B(3, 3) and C(-1, 2). Draw this triangle, calculate
the midpoints of each line segment and show the point of intersection of the three
median lines.
Answer:
⎡ ( −4) + (3) (5) + (3) ⎤
Midpoint AB = ⎢
,
2
2 ⎥⎦
⎣
= ( −0.5, 4)
⎡ (3) + (-1) (3) + (2) ⎤
Midpoint BC = ⎢
,
2
2 ⎥⎦
⎣
= (1, 2.5)
⎡ ( −1) + (-4) (2) + (5) ⎤
Midpoint CA = ⎢
,
2
2 ⎥⎦
⎣
= ( −2.5, 3.5)
Copyright © 2005, Durham Continuing Education
Page 49 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 7
Support Questions
1.
A triangle has the vertices A(7, 1), B(-3, 6) and C(5, -3). Draw this triangle,
calculate the midpoints of each line segment and show the point of intersection of
the three median lines.
2.
A triangle has the vertices A(-6, 8), B(3, 5) and C(-1, -4). Draw this triangle,
calculate the midpoints of each line segment and show the point of intersection of
the three median lines.
3.
A triangle has the vertices A(6, -3), B(3, 4) and C(-4, 2). Draw this triangle,
calculate the midpoints of each line segment and show the point of intersection of
the three median lines.
The Circumcentre of a Triangle
Then circumcentre is the centre of a circle that passes through all three vertices of a
triangle.
Copyright © 2005, Durham Continuing Education
Page 50 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 7
The circumcentre of a triangle is determined by finding the equations of the
perpendicular bisectors of two sides, then finding the point of intersection of those two
lines.
Example #1
A triangle has the vertices A(-4, 5), B(3, 3) and C(-1, 2). Calculate the circumcentre
of ΔABC .
Answer:
Slope of AB =
y1 − y 2
x1 − x 2
=
(5) − (3)
( −4) − (3)
⎡ ( −4) + (3) (5) + (3) ⎤
Midpoint AB = ⎢
,
2
2 ⎥⎦
⎣
= ( −0.5, 4)
2
−7
y − y2
Slope of BC = 1
x1 − x 2
=
=
(3) − (2)
(3) − (-1)
1
=
4
⎡ (3) + (-1) (3) + (2) ⎤
Midpoint BC = ⎢
,
2
2 ⎥⎦
⎣
= (1, 2.5)
The slope of the line segment ⊥ AB is
7
2
4
The slope of the line segment ⊥ BC is = 4
1
This symbol means
perpendicular.
Finding “b” in y = mx + b for both perpendicular equations.
Copyright © 2005, Durham Continuing Education
Page 51 of 110
MPM2H – Principles of Mathematics (Transfer Course)
" b" for line ⊥ to AB
Unit 2 – Lesson 7
" b" for line ⊥ to BC
y = mx + b
y = mx + b
7
y = 4x + b
y = x+b
2
(2.5) = 4(1) + b
7
2.5 = 4 + b
(4) = ( −0.5) + b
2
2.5 − 4 = b
7
- 1.5 = b
4=− +b
4
7
4+ =b
4
3
5 =b
4
Since “b” and “m” is now known for both perpendicular bisectors the two
equations are:
y=
7
23
x+
and
2
4
y = 4x −
3
2
Finding the point of intersection using the perpendicular equations.
Set the first y equal to the other and solve for x.
7
23
3
= 4x −
x+
2
4
2
⎛7⎞
⎛ 23 ⎞
⎛3⎞
4⎜ ⎟ x + 4⎜ ⎟ = 4(4x ) − 4⎜ ⎟
⎝2⎠
⎝ 4 ⎠
⎝2⎠
⎛7⎞
⎛ 23 ⎞
⎛3⎞
4/ 2 ⎜ ⎟ x + 4/ 1 ⎜ ⎟ = 4(4x ) − 4/ 2 ⎜ ⎟
⎝ 2/ ⎠
⎝ 4/ ⎠
⎝ 2/ ⎠
14x + 23 = 16x − 6
− 2x = −29
29
x=
2
Substitute x into either equation to find y.
Copyright © 2005, Durham Continuing Education
Page 52 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 7
3
2
⎛ 29 ⎞ 3
y = 4⎜ ⎟ −
⎝ 2 ⎠ 2
116 3
−
y=
2
2
113
y=
2
y=4 −
So the point of intersection of the two perpendicular bisectors
⎛ 29 113 ⎞
is ⎜ ,
⎟ and this is the circumcentre of ΔABC .
⎝ 2 2 ⎠
Support Questions
4.
Find the perpendicular slope of the line segment containing each of the given two
coordinates.
a. A(3, 5), B(4, 9)
5.
c. E(7, -2), F(3, 1)
Solve each of the following algebraic equations.
a.
6.
b. C(-2, 3), D(4, 7)
1
x + 12 = 2x − 7
3
b.
1 2
3
x - = 5x −
2 3
4
c. 4x + 9 = 3x + 2
Find the b in y = mx + b if the slope is 5 and its line contains
a. (-3 ,4)
d. (-1, -2)
b. (5, 2)
e. (0, 6)
c. (7, -4)
7.
A triangle has the vertices A(5, 1), B(-2, 0) and C(4, 8). Calculate the
circumcentre of ΔABC .
8.
A triangle has the vertices D(1, 8), E(9, 2) and F(8, 9). Calculate the
circumcentre of ΔDEF .
Copyright © 2005, Durham Continuing Education
Page 53 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 7
Key Question #7
1.
A triangle has the vertices A(-3, 2), B(-5, -6) and C(5, 0). Draw this triangle,
calculate the midpoints of each line segment and show the point of intersection of
the three median lines.
2.
A triangle has the vertices A(-2, 12), B(-10, 4) and C(2, 8). Draw this triangle,
calculate the midpoints of each line segment and show the point of intersection of
the three median lines.
3.
Find the perpendicular slope of the line segment containing each of the given two
coordinates.
a. A(4, 3), B(-3, 0)
4.
c. E(8, -2), F(7, -5)
Solve each of the following algebraic equations.
a. 7x − 13 = 4x − 7
5.
b. C(6, 3), D(4, -1)
b.
1 2
3
x - = -2x +
3
5
2
c.
1
x - 2 = 5x + 7
4
Find the b in y = mx + b if the slope is -2 and its line contains
a. (-3 ,4)
d. (-1, -2)
b. (5, 2)
e. (0, 6)
c. (7, -4)
6.
A triangle has the vertices A(-2, 7), B(-4, 2) and C(6, -2). Calculate the
circumcentre of ΔABC .
7.
A triangle has the vertices D(10, 10), E(-7, 3) and F(0, -14). Calculate the
circumcentre of ΔDEF .
9.
If you know the vertex coordinates of a triangle, how can
you find the coordinates of the centroid and
circumcentre of the triangle?
10.
Three houses in a small town are going to share a
single light standard. What coordinates should the light
standard be constructed so that each of the homes is
equally illuminated? The coordinates of the house are
X(-16, 32), Y(22, -24), and Z(56, 8)
Copyright © 2005, Durham Continuing Education
Page 54 of 110
Quadrilateral
Characteristics
Lesson 8
MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 8
Lesson Eight Concepts
¾
¾
¾
¾
quadrilateral classifications
calculating slope
calculating lengths of segments
calculating perpendicular line segments
Types of Quadrilaterals
A parallelogram has:
•
•
•
opposite sides the same length
adjacent sides different lengths
both pairs of opposite sides are parallel
•
•
•
opposite sides the same length
adjacent sides different lengths
adjacent sides are perpendicular to each other
•
•
all sides the same length
both pairs of opposite sides are parallel
•
•
all sides the same length
adjacent sides are perpendicular to each other
A rectangle has:
A rhombus has:
A square has:
When given the coordinates of the vertices of the quadrilateral, the slope and side
lengths can be calculated.
With the slopes and lengths of all sides of a quadrilateral, the quadrilateral can be
classified.
Copyright © 2005, Durham Continuing Education
Page 56 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 8
Example #1
Show that ABCD, where A(-2, 3), B(2, 1), C(0, -3) and D(-4, -1) is a square.
Answer:
For a square, need numerical proof that:
•
•
all sides the same length
adjacent sides are perpendicular to each other
Proving that the adjacent sides are perpendicular
(y 1 − y 2 )
(x 1 − x 2 )
(y 1 − y 2 )
(y − y 2 )
mBC
(x 1 − x 2 ) mDA = 1
(x 1 − x 2 )
(1 − 3)
(-1 − −3)
=
=
( −3 − 1)
(-1 − 3)
=
=
(2 − −2)
( −4 − 0)
(0 − 2)
(-4 − −2)
( −2)
(2)
=
=
( −4)
( −4)
=
=
(4)
(-4)
( −2)
(-2)
1
1
=−
=−
=2
=2
2
2
So adjacent sides AB ⊥ BC and BC is ⊥ CD .
Since all sides are the same length and the adjacent sides are
perpendicular then ABCD is a square.
m AB =
(y − y 2 )
= 1
(x 1 − x 2 )
Copyright © 2005, Durham Continuing Education
m CD =
Page 57 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 8
Support Questions
1.
Show that the line passing through (1, 4) and (5, 5) is parallel to the line passing
through (3, -4) and (7, -3).
2.
Show that the line passing through (-1, 7) and (3, 5) is perpendicular to the line
passing through (-4, 1) and (-1, 7).
3.
Show that the quadrilateral with vertices W(-2, 3), X(-2, -2), Y(2,1), Z(2, 6) is a
rhombus.
4.
Describe the type of quadrilateral described by each set of vertices. Give
numerical proof by calculating lengths and slopes for you answer.
a. A(-5, 2), B(-1, 3), C(-2, -1), D(-6, -2)
b. E(-5, 1), F(3, 3), G(4, -1), H(-4, -3)
c. X(-5, -4), Y(-5, 1), Z(7, 4), W(7, -1)
Key Question #8
1.
Show that the line passing through (-4, 3) and (0, 5) is parallel to the line passing
through (1, 1) and (5, 3).
2.
Show that the line passing through (-1, 7) and (2, 1) is perpendicular to the line
passing through (-4, 3) and (0, 5).
3.
Show that the quadrilateral with vertices W(0, -5), X(-9, 2), Y(-5,8), Z(4, 2) is
not a rectangle.
4.
Describe the type of quadrilateral described by each set of vertices. Give
numerical proof by calculating lengths and slopes for you answer.
a. A(3, -4), B(9, 2), C(14, -3), D(8, -9)
b. E(-1, 3), F(6, 4), G(4, -1), H(-3, -2)
5.
A person is marking the corners of a piece of land on which a park
will be made. If the corners have the coordinates (-1, -1), (-4, 1),
(-1, 3), and (2, 1). What shape is the lot? Give numerical proof by
calculating lengths and slopes for you answer.
6.
Give the coordinates of a parallelogram and then prove numerically that your
vertices create a parallelogram.
Copyright © 2005, Durham Continuing Education
Page 58 of 110
Law of Sines
Lesson 9
MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 9
Lesson Nine Concepts
¾
¾
¾
¾
¾
drawing a labelling triangle angles and sides
interpreting information and drawing diagrams based on that information
using the sine law to find an unknown side in a triangle
using the sine law to find an unknown side in a triangle
drawing diagrams and solving questions, using the Sine Law, involving the angle
of elevation and depression
Trigonometry: The Sine Law
The sine law is the relationship between the ratios of the sines of the angles of a
triangle and the lengths of the opposite sides.
For the triangle given below:
The Sine Law can be written as:
Sin A Sin B Sin C
a
b
c
=
=
or
=
=
a
b
c
Sin A Sin B Sin C
Recognizing when to use the Sine Law.
You need:
•
•
an angle and the value of its side opposite
a second angle and the need to find its unknown opposite side
OR
•
an angle and the value of its side opposite
Copyright © 2005, Durham Continuing Education
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MPM2H – Principles of Mathematics (Transfer Course)
•
Unit 2 – Lesson 9
a second side and the need to find its unknown opposite angle
Example #1
x Sin24 D = 8 Sin39 D
D
x Sin24
8 Sin39 D
=
Sin24 D
Sin24 D
8 Sin39 D
Sin24 D
x ≈ 12.4
x=
Copyright © 2005, Durham Continuing Education
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MPM2H – Principles of Mathematics (Transfer Course)
Copyright © 2005, Durham Continuing Education
Unit 2 – Lesson 9
Page 62 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 9
Support Questions
1.
Given each of the ratios below, state whether the unknown angle will be larger or
smaller than the one given.
a.
2.
Sin 67 Sin X
=
13
7.2
b.
Sin Y Sin 81
=
7.6
8.4
c.
Sin 37 Sin A
=
4
8
Given each of the ratios below, state whether the unknown angle will be larger or
smaller than the one given.
a.
Sin 55 Sin 71
=
13
x
b.
Sin 16 Sin 77
=
12.5
y
c.
Sin 83 Sin 41
=
a
7.2
3.
Find the length of the indicated side.
4.
Find the measure of angle A.
5.
In Δ PQR, find the value of q, if ∠R = 73°, ∠Q = 32°, and r = 23 cm.
6.
In Δ ABC, find the value of c, if ∠A = 52°, ∠C = 47°, and a = 12 m.
7.
In Δ TUV, find the value of ∠U, if ∠T = 37°, u = 16 cm, and t = 22 cm.
8.
In Δ XYZ, find the value of ∠Z, if ∠X = 72°, x = 41 m, and z = 11 m.
9.
A radio tower is supported by two wires on opposite sides. The wires form an
angle of 60° at the top of the post. On the ground, the ends of the wire are 17 m
apart, and one wire is at a 45° angle to the ground. How long will the wires be?
Copyright © 2005, Durham Continuing Education
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 9
Key Question #9
1.
Given each of the ratios below, state whether the unknown angle will be larger or
smaller than the one given.
a.
2.
Sin B Sin 35
=
10
5.1
b.
Sin Y Sin 64
=
9.1
8.4
c.
Sin 37 Sin A
=
6.1
3.8
Given each of the ratios below, state whether the unknown angle will be larger or
smaller than the one given.
a.
Sin 13 Sin 52
=
8
x
b.
Sin 54 Sin 23
=
12.5
y
c.
Sin 17 Sin 76
=
a
7.2
3.
Find the length of the indicated side.
4.
Find the measure of angle C.
5.
In Δ TUV, find the value of t, if ∠T = 61°, ∠U = 24°, and u = 17 cm.
6.
In Δ EFG, find the value of e, if ∠F = 52°, ∠G = 47°, and f = 40 m.
7.
In Δ ABC, find the value of ∠A, if ∠B = 37°, a = 10 cm, and b = 17 cm.
8.
In Δ QRS, find the value of ∠Q, if ∠S = 72°, q = 26 m, and s = 35 m.
9.
An architect designs a house that is 10 m wide. The rafters holding
up the roof are equal length and meet at an angle of 65°. The rafters
extend 0.4 m beyond the supporting wall. How long are the rafters?
10.
Does the sine law apply to right triangles? Explain your answer.
Copyright © 2005, Durham Continuing Education
Page 64 of 110
Law of Cosine
Lesson 10
MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 10
Lesson Ten Concepts
¾
¾
¾
¾
drawing a labelling triangle angles and sides
interpreting information and drawing diagrams based on that information
using the cosine law to find an unknown side in a triangle
using the cosine law to find an unknown side in a triangle
Trigonometry: The Cosine Law
The cosine law is used to find the third side of a triangle when two sides and a
contained angle are known or to find an angle measure when the length of three sides
are known.
The contained angle in a triangle is the angle between the two given sides of the
triangle. In the example given below C is the contained angle between the sides CA
and CB.
The Cosine Law states that for a ΔABC :
c 2 = a2 + b2 − 2ab CosC
b2 = a2 + c 2 − 2ac CosB
a2 = b2 + c 2 − 2bc CosA
Example #1
Find the missing side shown in the diagram below using the cosine law.
Copyright © 2005, Durham Continuing Education
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 10
Answer:
a 2 = b 2 + c 2 − 2bc CosA
a 2 = (16) 2 + (11) 2 − 2(16)(11) CosA
Example #1 cont.
So the missing length is 23.3 units long.
Example #2
Find the angle A using the cosine law.
Answer:
First using algebra manipulate the formula to isolate Cos A.
Copyright © 2005, Durham Continuing Education
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 10
a 2 = b 2 + c 2 − 2bc CosA
a 2 - b 2 − c 2 = −2bc CosA
a 2 - b 2 − c 2 − 2bc CosA
=
− 2bc
− 2bc
a2 - b2 − c 2
= CosA
− 2bc
CosA =
c 2 + b2 - a2
2bc
Example #2 cont.
Copyright © 2005, Durham Continuing Education
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MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 10
Support Questions
1.
Find the length of the indicated side.
2.
Find the measure of angle X.
3.
In Δ PQR, find the value of r, if p = 13 cm, q = 8 cm and ∠R = 63°.
4.
In Δ ABC, find the value of a, if b = 17 cm, c = 28 cm and ∠A = 105°.
5.
In Δ TUV, find the value of ∠U, if v = 10 cm, u = 16 cm, and t = 22 cm.
6.
In Δ XYZ, find the value of ∠Z, if x = 41 m, y = 32 m, and z = 28 m.
7.
The bases on a softball diamond are 15 m apart. A player picks up
a fair ground ball 3 m from third base, along the line from second to
third base. How far must he throw it to first base?
8.
Find the measure, to the nearest degree, of the middle angle in a
triangle that has side lengths of 8 cm, 19 cm, and 21 cm.
9.
A plane leaves Hamilton and flies due east for 75 km. At the same time, a
second plane flies in a direction 40° southeast for 120 km. How far apart are the
planes when they reach their destinations?
Copyright © 2005, Durham Continuing Education
Page 69 of 110
MPM2H – Principles of Mathematics (Transfer Course)
Unit 2 – Lesson 10
Key Question #10
1.
Find the length of the indicated side.
2.
Find the measure of angle C.
3.
In Δ EFG, find the value of f, if e = 7.1 cm, g = 8.9 cm and ∠E = 71°.
4.
In Δ LMO, find the value of m, if o = 34 cm, l = 62 cm and ∠M = 98°.
5.
In Δ BCD, find the value of ∠B, if b = 46 cm, c = 70 cm, and d = 58 cm.
6.
In Δ ABC, find the value of ∠C, if a = 5.2 m, b = 3.8 m, and c = 6.7 m.
7.
Two paths diverge at a 48° angle. Two mountain bike riders take
separate routes at 8 km/hr and 12 km/hr. How far apart are they after
2 hours? Include a diagram.
8.
Draw a diagram to solve this problem: Oshawa is 10 km due east of
Ajax. Uxbridge is 18 km NW of Oshawa. How far is it from Ajax to Uxbridge?
Explain whether you have enough information to solve this problem.
10.
A golfer hits a tee shot on a 325 m long straight
golf hole. The ball is hooked (hit at an angle)
18° to the left. The ball lands 185 m from the
tee. How far is the ball from the hole?
11.
Given the points A(0, 0), B(3, 1) and C(1, 4),
what is the measure of ∠ ABC?
Copyright © 2005, Durham Continuing Education
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MPM2H – Principles of Mathematics (Transfer Course)
Support Question Answers
Support Question Answers
Lesson One
1.
a. 4(2w + 5) = 4(2w) + 4(5)
= 8w + 20
b. 5x(-3x - 7) = 5x(-3x) + 5x(-7)
= -15x 2 -35x
c. 2y(-4y + 8) = 2y(-4y) + 2y(8)
= -8y 2 +16y
d. -(3x + 4) = -(3x) - (4)
= -3x – 4
e. -6x(-x +2) = -6x(-x) -6x(2)
= 6x 2 -12x
(x + 3)(x − 4) = x(x) + x(-4) + 3(x) + 3(-4)
2.
a.
= x 2 - 4x + 3x − 12
= x 2 - x − 12
(2y − 3)(7y − 1) = 2y(7y) + 2y(-1) − 3(7y) − 3(-1)
b.
= 14y 2 - 2y - 21y + 3
= 14y 2 - 23y + 3
(-t − 5)(t + 2) = -t(t) - t(2) − 5(t) − 5(2)
c.
= -t 2 - 2t - 5t - 10
= -t 2 - 7t - 10
(3x + 1)(2x − 9) = 3x(2x) + 3x(-9) + 1(2x) + 1(-9)
d.
= 6x 2 - 27x + 2x - 9
= 6x 2 - 25x - 9
(7w − 3)(5w + 6) = 7w(5w) + 7w(6) − 3(5w) − 3(6)
e.
= 35w 2 + 42w - 15w - 18
= 35w 2 + 27w - 18
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MPM2H – Principles of Mathematics (Transfer Course)
f.
Support Question Answers
(x − 5) 2 = (x − 5)(x − 5)
= x(x) + x(-5) − 5(x) − 5(-5)
= x 2 - 5x - 5x + 25
= x 2 - 10x + 25
g.
(4n − 2) 2 = (4n − 2)(4n − 2)
= 4n(4n) + 4n(-2) − 2(4n) − 2(-2)
= 16n 2 - 8n - 8n + 4
= 16n 2 - 16n + 4
(-7s − 3)(s + 1) = −7s(s) − 7s(1) − 3(s) − 3(1)
h.
= -7s 2 - 7s - 3s - 3
= -7s 2 - 10s - 3
5(3y − 2)(y + 7) = 5[3y(y) + 3y(7) − 2(y) − 2(7)]
[
= 5[3y
= 5 3y 2 + 21y - 2y - 14
3.
a.
+ 19y - 14
2
]
]
= 5(3y ) + 5(19y) + 5( −14)
2
= 15y 2 + 95y − 70
- 2(y − 3)(5y − 6) = -2[y(5y) + y(-6) − 3(5y) − 3(-6)]
[
= -2[5y
= -2 5y 2 - 6y - 15y + 18
b.
2
- 21y + 18
]
]
= -2(5y 2 ) - 2(-21y) - 2(18)
= -10y 2 + 42y − 36
6(4x + 3)(5x − 2) = 6[4x(5x) + 4x(-2) + 3(5x) + 3(-2)]
[
= 6[20x
= 6 20x 2 - 8x + 15x - 6
c.
2
+ 7x - 6
]
]
= 6(20x 2 ) + 6(7x) + 6(-6)
= 120x 2 + 42x − 36
(x − 2y)(3x + 5y) = x(3x) + x(5y) − 2y(3x) − 2y(5y)
4.
a.
= 3x 2 + 5xy − 6xy − 10y 2
= 3x 2 − xy − 10y 2
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MPM2H – Principles of Mathematics (Transfer Course)
Support Question Answers
(-x + 3w)(w − 1) = − x(w) − x(-1) + 3w(w) + 3w(-1)
b.
= -xw + x − 3w 2 − 3w
= −3w 2 − xw − 3w + x
- 8(3w - 7y)(2w + y) = -8[3w(2w) + 3w(y) - 2y(2w) - 2y(y)]
[
= -8[6w
= -8 6w 2 + 3wy − 4wy - 2y 2
c.
2
- wy - 2y 2
]
]
= -8(6w ) - 8(wy) - 8(-2y 2 )
2
= -48x 2 - 8wy + 16y 2
- (x + 4)(x + 4) = -[x(x) + x(4) + 4(x) + 4(4)]
[
= -[x
= - x 2 + 4x + 4x + 16
5.
a.
2
+ 8x + 16
]
]
= -(x 2 ) - (8x) - (16)
= -x 2 - 8x + 16
3x + (2x − 3)(x + 5) = 3x + 2x(x) + 2x(5) − 3(x) − 3(5)
b.
= 3x + 2x 2 + 10x − 3x − 15
= 2x 2 + 10x − 15
- y(2y + 4x)(y − x) = -y[2y(y) + 2y(-x) + 4x(y) + 4x(-x)]
[
= -y[2y
= -y 2y 2 - 2xy + 4xy - 4x 2
c.
2
+ 2xy - 4x
2
]
]
= -y(2y ) - y(2xy) - y(-4x 2 )
2
= -y 3 - 2xy 2 + 4x 2 y
(2x + 3)(3x - 2) + x(3x - 4) = 2x(3x) + 2x(-2) + 3(3x) + 3(-2) + x(3x) + x(-4)
6.
a.
= 6x 2 - 4x + 9x - 6 + 3x 2 − 4x
= 9x 2 + x - 6
- 2(3x - 2) 2 − 5(x + 2) 2 = −2(3x − 2)(3x − 2) − 5(x + 2)(x + 2)
b.
[
] [
= −2 9x 2 − 12x + 4) − 5 x 2 + 4x + 4
]
= −18x + 24x − 8 − 5x − 20x − 20
2
2
= −23x 2 + 4x − 28
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MPM2H – Principles of Mathematics (Transfer Course)
Support Question Answers
(x - 4) 3 = (x + 4)(x + 4)(x + 4)
= (x 2 + 8x + 16)(x + 4)
= x(x 2 ) + x(8x) + x(16) + 4(x 2 ) + 4(8x) + 4(16)
c.
= x 3 + 8x 2 + 16x + 4x 2 + 32x + 64
= x 3 + 12x 2 + 48x + 64
[(x + 2)(x - 3)]2 = (x + 2)(x − 3)(x + 2)(x - 3)
= (x 2 − x − 6)(x 2 − x − 6)
d.
= x 4 − x 3 − 6x 2 − x 3 + x 2 + 6x − 6x 2 + 6x + 36
= x 4 − 2x 3 − 11x 2 + 12x + 36
(x − 1) 4 = (x - 1)(x − 1)(x - 1)(x − 1)
= (x 2 − 2x + 1)(x 2 − 2x + 1)
e.
= x 4 − 2x 3 + x 2 − 2x 3 + 4x 2 - 2x + x 2 - 2x + 1
= x 4 − 4x 3 + 6x 2 - 4x + 1
Lesson Two
1.
a. 2(3c + 1)
d. 3x(4x + 1)
b. 20(5x – 1)
e. 7r(r – 3)
2.
a. product 2 and 1 add to 3: (c + 2)(x + 1)
b. product 5 and -3 add to 2: (x + 5)(x - 3)
c. product 4 and 1 add to 5: (x + 4)(x + 1)
d. product 6 and -3 add to 3: (x + 6)(x - 3)
e. product -3 and -3 add to -6: (x - 3)(x - 3) = (x –3) 2
f. product -4 and 2 add to -2: (x - 4)(x + 2)
g. product 3 and 3 add to 6: (x + 3)(x + 3) = (x + 3) 2
h. product -7 and 2 add to -5: (x - 7)(x + 2)
i. product -5 and -3 add to -8: (x - 5)(x - 3)
3.
a. 9 x 4 = 36
c. 5(g + 4)
f. 4(4f +3gf +8g)
6 x 6 = 36 and 6 + 6 = 12
9c 2 + 12c + 4 = 9c 2 + 6c + 6c + 4
= 3c(c + 2) + 2(3c + 2)
= (3c + 2)(3c + 2)
= (3c + 2) 2
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MPM2H – Principles of Mathematics (Transfer Course)
b. 2 x 2 = 4
Support Question Answers
4 x 1 = 4 and 4 + 1 = 5
2x 2 + 5x + 2 = 2x 2 + 4x + 1x + 2
= 2x(x + 2) + 1(x + 2)
= (2x + 1)(x + 2)
c. 10 x 1 = 10
5 x 2 = 10 and 5 - 2 = 3
10g 2 + 3g - 1 = 10g 2 + 5g - 2g - 1
= 5g(2g + 1) - 1(2g + 1)
= (5g − 1)(2g + 1)
d. 2 x 1 = 2 and 2 + 1 = 3
2x 2 + 3x + 1 = 2x 2 + 2x + 1x + 1
= 2x(x + 1) + 1(x + 1)
= (2x + 1)(x + 1)
e. 3 x 4 = 12
12 x 1 = 12 and 1 - 12 = -11
3r 2 - 11r - 4 = 3r 2 − 12r + 1r − 4
= 3r(r - 4) + 1(r - 4)
= (3r + 1)(r - 4)
f. 4 x 15 = 60
6 x 10 = 60 and -6 -10 = -16
4f 2 - 16f + 15 = 4f 2 - 6f - 10f + 15
= 2f(2f - 3) - 5(2f - 3)
= (2f - 5)(2f - 3)
g. 6 x 1 = 6 2 x 3 = 6 and 2 + 3 = 5
6m 2 + 5m + 1 = 6m 2 + 2m + 3m + 1
= 2m(3m + 1) + 1(3m + 1)
= (2m + 1)(3m + 1)
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MPM2H – Principles of Mathematics (Transfer Course)
h. 3 x 2 = 6
Support Question Answers
and -3 -2 = -5
3n 2 − 5n − 2 = 3n 2 - 3n - 2n − 2
= 3n(n - 1) − 2(n - 1)
= (3n - 2)(n - 1)
i. 10 x 3 = 30
6 x 5 = 30 and 6 - 5 = 1
10b 2 + b - 3 = 10b 2 + 6b - 5b - 3
= 2b(5b + 3) − 1(5b + 3)
= (2b − 1)(5b + 3)
4.
a.
c 2 = c and
36 = 6 ; (c + 6)(c – 6)
b.
x 2 = x and
64 = 8 ; (x + 8)(x – 8)
c.
25g 2 = 5g and
d.
16x 2 = 4x and
e. 2(25r -36);
2
f. 7(f - 4);
2
5.
49 = 7 ; (5g - 7)(5g + 7)
81 = 9 ; (4x – 9)(4x + 9)
25r = 5r and
2
f = f and
36 = 6 ; 2(5r + 6)(5r - 6)
4 = 2 ; 7(f + 2)(f - 2)
2
a.
4c 2 = 2c and
25 = 5 ; (2c + 5)(2c + 5) = (2c + 5) 2
b.
36x 2 = 6x and
25 = 5 ; (6x + 5)(6x + 5) = (6x + 5) 2
c.
16g 2 = 4g and
9 = 3 ; (4g - 3)(4g - 3) = (4g - 3) 2
d.
64x 2 = 8x and
1 = 1 ; (8x + 1)(8x + 1) = (8x + 1) 2
e.
100r 2 = 10r and
f.
9r 2 = 3r and
81 = 9 ; (10r - 9)(10r - 9) = (10r - 9) 2
4 = 2 ; (3r - 2)(3r - 2) = (3r - 2) 2
Lesson Three
x 2 + 7x − 30 = 0
1.
a.
b.
(x + 10)(x − 3) = 0
x + 10 = 0
and
x = −10 and
x −3 = 0
x=3
t 2 + 8t + 15 = 0
(t + 5)(t + 3) = 0
t+5 =0
and
t+3 =0
t = −5 and
t = −3
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Support Question Answers
n 2 - 2n − 15 = 0
c.
(n - 5)(n + 3) = 0
n-5 = 0
and n + 3 = 0
n = +5 and
n = −3
4x 2 − 1 = 0
(2x + 1)(2x − 1) = 0
d.
e.
2x + 1 = 0
2x = −1
1
x=−
2
2x − 1 = 0
2x = 1
1
and
x=
2
and
and
9m 2 − 6m + 1 = 0
(3x − 1)(3x − 1) = 0
3x − 1 = 0
3x = 1
1
x=
3
5y 2 + 25y + 30 = 0
f.
5(y 2 + 5y + 6)
5(y + 3)(y + 2) = 0
y+3 =0
and
y+2=0
y = −3
and
y = -2
y 2 - y − 42 = 0
(y + 6)(y − 7) = 0
y+6=0
and
y = −6 and
y −7 = 0
y =7
42 = y 2 - y
2.
a.
t 2 - 4t = 21
b.
t 2 - 4t − 21 = 0
(t + 3)(t − 7) = 0
t+3 =0
and
t = −3 and
t−7 = 0
t=7
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n 2 = 2n + 48
c.
n 2 - 2n − 48 = 0
(n + 6)(n − 8) = 0
n+6 = 0
and
n = −6 and
n−8 = 0
n=8
x 2 = 30 - 7x
d.
x 2 + 7x − 30 = 0
(x + 10)(x − 3) = 0
x + 10 = 0
and
x = −10 and
x −3 = 0
x=3
3 = 6m 2 - 7m
e.
f.
6m 2 - 7m − 3 = 0
(3m + 1)(2m − 3) = 0
3m + 1 = 0
and 2m − 3 = 0
3m = −1 and
2m = 3
1
3
m = − and
m=
3
2
5y 2 + 2y - 3 = 0
(5y - 3)(y + 1) = 0
5y − 3 = 0
and
5y = −3 and
3
y = − and
5
y +1= 0
y = -1
- 24 = n 2 - n - 30
3.
a.
n2 − n − 6 = 0
(n - 3)(n + 2) = 0
n−3 = 0
and n + 2 = 0
n = 3 and
n = -2
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- 10 = n 2 - 3n - 28
b.
n 2 − 3n − 18 = 0
(n - 6)(n + 3) = 0
n−6 = 0
and n + 3 = 0
n = 6 and
n = -3
2 = n 2 - 7n + 12
c.
n 2 − 7n + 10 = 0
(n - 5)(n + 2) = 0
n−5 = 0
and n − 2 = 0
n = 5 and
n=2
4 = n 2 - 8n + 16
d.
n 2 − 8n + 12 = 0
(n - 6)(n - 2) = 0
n−6 = 0
and n - 2 = 0
n = 6 and
n=2
10 = -5t 2 + 30
- 5t 2 + 20 = 0
4.
- 5(t 2 − 4) = 0
- 5(t - 2)(t + 2) = 0
t−2=0
and t + 2 = 0
t=2
and
t = -2
Cannot have negative time so it would take 2 seconds to reach the water.
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5.
a.
a=6
b = -1
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c = -15
x=
− b ± b 2 − 4ac
2a
x=
− (-1) ± (-1)2 − 4(6)(-15 )
2(6)
1 ± 1 + 360
2(6)
1 ± 19
x=
12
5
3
x = and x = −
3
2
x=
b.
a=1
x=
b=0
c = -16
− b ± b 2 − 4ac
2a
− (0) ± (0) 2 − 4(1)(-16)
x=
2(1)
0 ± 64
2(1)
±8
x=
2
x = −4 and x = 4
x=
c.
a=1
b=9
c = 20
− b ± b 2 − 4ac
x=
2a
− (9) ± (9) 2 − 4(1)(20)
x=
2(1)
-9± 1
2(1)
− 9 ±1
x=
2
x = −5 and x = -4
x=
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d.
a=4
b = -20
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c = 25
x=
− b ± b 2 − 4ac
2a
x=
− (-20 ) ± (-20) 2 − 4( 4 )(25 )
2( 4 )
20 ± 0
2(4)
20
x=
8
5
x=
2
x=
e.
a=6
b=0
c = 48
− b ± b 2 − 4ac
x=
2a
x=
− (0) ± (0) 2 − 4(6)(-48 )
2(6)
0 ± 1152
(12)
x ≈ ±2.82
x=
(w − 5) 2 = 16
f.
w 2 − 10w + 25 = 16
w 2 − 10w + 9 = 0
a=1
b = -10
c=9
x=
− b ± b 2 − 4ac
2a
x=
− (-10) ± (-10) 2 − 4(1)(9)
2(1)
10 ± 100 - 36
2(1)
10 ± 8
x=
2
x = 9 and x = 1
x=
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6.
a = -0.1492
b = 1.8058
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c=9
x=
− b ± b 2 − 4ac
2a
x=
− (1.8058 ) ± (1.8058 ) 2 − 4(-0.1492 )(9)
2(-0.1492 )
x=
- 1.8058 ± 8.6321
- 0.2984
x ≈ 15.9 and x ≈ -3.79
16 + 1988 = 2004
7.
a = -0.1492
b = 1.8058
c=9
t=
− (1.5 ) ± (1.5 ) 2 − 4(-4.9)(16)
2(-4.9)
t=
- 1.5 ± 315.85
- 9.8
t ≈ −1.66 and t ≈ 1.97 The diver is in the air approximat ely 2 seconds
Lesson Four
1.
a.
i.
ii.
up because “a” is positive (3)
Set x = 0 and find the y-intercept
y = 3(x + 2)(x - 7)
y = 3(0 + 2)(0 - 7)
y = (3)(2)(-7)
iii.
iv.
y = −42
x = -2 and x = 7
-2+7
2
5
x=
2
x = 2.5
x=
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v.
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y = 3(2.5 + 2)(2.5 - 7)
y = −60.75
The vertex is (2.5,-60.75)
vi.
b.
i.
ii.
up because “a” is positive (1)
Set x = 0 and find the y-intercept
y = (2x + 1)(x - 4)
y = (2(0) + 1)(0 - 4)
y = (1)(-4)
iii.
iv.
y = −4
1
x = - and x = 4
2
- 0.5 + 4
2
3.5
x=
2
x = 1.75
x=
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v.
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y = (2(1.75) + 1)(1.75 - 4)
y = −10.125
The vertex is (1.75,-10. 125)
vi.
c.
i.
iii.
up because “a” is positive (3)
Set x = 0 and find the y-intercept
y = 3x 2 + 5x − 2
iii.
y = 3(0) 2 + 5(0) - 2
y = −2
(3x - 1)(x + 2)
1
x=
and x = -2
3
iv.
1
-2
x= 3
2
5
x=−
6
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v.
y = 3x 2 + 5x − 2
5
5
y = 3(- ) 2 + 5(- ) - 2
6
6
49
y=−
12
5 49
so the vertex is ( − ,−
)
6 12
vi.
d.
i.
ii.
down because “a” is negative (-2)
Set x = 0 and find the y-intercept
y = −2x 2 + 8x
ii.
y = -2(0) 2 + 8(0)
y=0
-2x(x - 4)
x = 0 and x = 4
iv.
0+4
2
x=2
x=
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v.
y = -2x 2 + 8x
y = -2(2) 2 + 8(2)
y=8
so the vertex is (2, 8)
vi.
e.
i.
ii.
up because “a” is positive (24)
Set x = 0 and find the y-intercept
y = 5x 2 - 10x - 15
iii.
y = 5(0) 2 - 10(0) - 15
y = −15
5(x – 3)(x + 1)
x = 3 and x = -1
iv.
3 -1
2
x =1
x=
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v.
y = 5x 2 - 10x - 15
y = 5(1) 2 - 10(1) - 15
y = -20
so the vertex is (1, -20)
vi.
f.
i.
iv.
up because “a” is positive (3)
Set x = 0 and find the y-intercept
y = 24x 2 - 30x + 9
iii.
iv.
y = 24(0) 2 - 30(0) + 9
y=9
(2x - 1)(4x - 3)
1
3
x=
and x =
2
4
0.5 + 0.75
2
x = 0.625
x=
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v.
y = 24x 2 - 30x + 9
y = 24(0.625) 2 - 30(0.625) + 9
y = −0.375
so the vertex is (0.625,-0.375)
vi.
Lesson Five
1.
a.
b.
d = (x − x )2 + (y − y )2
2
1
2
1
d = (x − x )2 + (y − y )2
2
1
2
1
d=
[(4) − (-2)]2 + [(-1) − (-6)]2
d=
[(-2) − (9)]2 + [(6) − (4)]2
d = (6)2 + (5)2
d = (-11)2 + (2)2
d = 36 + 25
d = 121 + 4
d = 61
d ≈ 7.81
d = 125
d ≈ 11.18
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c.
d.
d = (x − x )2 + (y − y )2
2
1
2
1
d=
[(11) − (5)]2 + [(3) − (6)]2
d=
[(7) − (-5)]2 + [(8) − (5)]2
d = (12) 2 + (3)2
d = 36 + 9
d = 144 + 9
d = 45
d ≈ 6.7
d = 153
d ≈ 12.37
e.
f.
d = (x − x )2 + (y − y )2
2
1
2
1
2.
d = (x − x )2 + (y − y )2
2
1
2
1
d = (6)2 + (-3)2
d=
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[(1) − (4)]2 + [(6) − (10)]2
d = (x − x )2 + (y − y )2
2
1
2
1
d=
[(-7) − (-5)]2 + [(-3) − (-4)]2
d = (-3)2 + (-4)2
d = (-2)2 + (1)2
d = 9 + 16
d = 4 +1
d = 25
d=5
d= 5
d ≈ 2.24
(y 2 − y 1 ) (3) − (1) 2 1
=
= =
(x 2 − x 1 ) (3) − (-5) 8 4
(y − y 1 ) (-1) − (3) - 4
= -4
BC slope = 2
=
=
(x 2 − x 1 ) (4) − (3)
1
(y − y 1 ) (-3) − (-1) - 2 1
=
=
CD slope = 2
=
(x 2 − x 1 ) (-4) − (4) - 8 4
(y − y 1 ) (-3) − (1) - 4
=
= −4
DA slope = 2
=
(x 2 − x 1 ) (-4) − (-5) 1
AB slope =
d
AB
d=
= (x − x )2 + (y − y )2
2
1
2
1
d
= (x − x )2 + (y − y )2
2
1
2
1
BC
[(-5) − (3)]2 + [(1) − (3)]2
d=
[(3) − (4)]2 + [(3) − (-1)]2
d = (-8)2 + (-2)2
d = (-1)2 + (4)2
d = 64 + 4
d = 1 + 16
d = 68
d = 17
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d
CD
d=
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= (x − x )2 + (y − y )2
2
1
2
1
d
= (x − x )2 + (y − y )2
2
1
2
1
DA
[(4) − (-4)]2 + [(-1) − (-3)]2
d=
[(-4) − (-5)]2 + [(-3) − (1)]2
d = (8)2 + (2)2
d = (1)2 + (-4)2
d = 64 + 4
d = 1 + 16
d = 68
d = 17
Since opposite sides are equal length and slopes are perpendicular then the
quadrilateral is a rectangle.
3.
Make Toronto (0,0) then the coordinates of Hamilton are (-2.5, 3)
d = (x − x )2 + (y − y )2
2
1
2
1
d=
[(0) − (-2.5)]2 + [(0) − (3)]2
d = (2.5) 2 + (3) 2
d = 6.25 + 9
d = 15.25
d ≈ 3.9 x 50 = 195 km
4.
a.
b.
⎛ x + x 2 y1 + y 2 ⎞
⎟
Midpoint = ⎜ 1
,
⎜
⎟
2
2
⎝
⎠
⎛ (4) + (-1) (-2) + (-6) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛ x + x 2 y1 + y 2 ⎞
⎟
Midpoint = ⎜ 1
,
⎜
⎟
2
2
⎝
⎠
⎛ (-2) + (9) (6) + (4) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛3 -8⎞
=⎜ , ⎟
⎝2 2 ⎠
3
= ( , - 4)
2
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⎛ 7 10 ⎞
=⎜ , ⎟
⎝2 2 ⎠
7
= ( , 5)
2
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c.
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d.
⎛ x + x 2 y1 + y 2 ⎞
⎟
Midpoint = ⎜ 1
,
⎜
⎟
2
2
⎝
⎠
⎛ (11) + (5) (3) + (6) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛ 16 11 ⎞
=⎜ , ⎟
⎝ 2 2⎠
11
= (8, )
2
e.
⎛ x + x 2 y1 + y 2 ⎞
⎟
Midpoint = ⎜ 1
,
⎜
⎟
2
2
⎝
⎠
⎛ (7) + (-5) (8) + (5) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛ 2 13 ⎞
=⎜ , ⎟
⎝2 2 ⎠
13
= (1, )
2
f.
⎛ x + x 2 y1 + y 2 ⎞
⎟
Midpoint = ⎜ 1
,
⎜
⎟
2
2
⎝
⎠
⎛ (1) + (4) (6) + (10) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛ x + x 2 y1 + y 2 ⎞
⎟
Midpoint = ⎜ 1
,
⎜
⎟
2
2
⎝
⎠
⎛ (-7) + (-5) (-3) + (-4) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛ 5 16 ⎞
=⎜ , ⎟
⎝2 2 ⎠
5
= ( , 8)
2
5.
Midpoint
⎛ x + x 2 y1 + y 2 ⎞
⎟
=⎜ 1
,
⎟
AC ⎜
2
2
⎝
⎠
⎛ (-3) + (6) (2) + (-1) ⎞
,
=⎜
⎟
2
2
⎝
⎠
⎛3 1⎞
=⎜ , ⎟
⎝2 2⎠
⎛ - 12 - 7 ⎞
=⎜
, ⎟
⎝ 2 2 ⎠
7
= (-6, - )
2
Midpoint
⎛ x + x 2 y1 + y 2 ⎞
⎟
=⎜ 1
,
⎟
BD ⎜
2
2
⎝
⎠
⎛ (2) + (1) (4) + (-3) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛ 3 1⎞
=⎜ , ⎟
⎝2 2⎠
Both AC and BD have the same midpoint coordinates so the diagonals bisect.
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6.
Midpoint
⎛ x + x 2 y1 + y 2 ⎞
⎟
=⎜ 1
,
⎟
EF ⎜
2
2
⎝
⎠
⎛ (2) + (-10) (-12) + (-8) ⎞
=⎜
,
⎟
2
2
⎝
⎠
Midpoint
⎛ x + x 2 y1 + y 2 ⎞
⎟
=⎜ 1
,
⎟
FG ⎜
2
2
⎝
⎠
⎛ (-10) + (-6) (-8) + (4) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛ - 8 - 20 ⎞
=⎜ ,
⎟
⎝ 2 2 ⎠
= A (- 4, - 10 )
Midpoint
⎛ - 16 - 4 ⎞
=⎜
, ⎟
⎝ 2 2 ⎠
= B( −8,−2)
⎛ x + x 2 y1 + y 2 ⎞
⎟
=⎜ 1
,
⎟
GH ⎜
2
2
⎝
⎠
⎛ (-6) + (6) (4) + (0) ⎞
=⎜
,
⎟
2
2
⎝
⎠
Midpoint
⎛ x + x 2 y1 + y 2 ⎞
⎟
=⎜ 1
,
⎟
HE ⎜
2
2
⎝
⎠
⎛ (6) + (2) (0) + (-12) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛0 4⎞
=⎜ , ⎟
⎝2 2⎠
= C(0, 2)
d
AB
d=
⎛ 8 - 12 ⎞
=⎜ ,
⎟
⎝2 2 ⎠
= D(4,−6)
= (x − x )2 + (y − y )2
2
1
2
1
d
= (x − x )2 + (y − y )2
2
1
2
1
BC
[(-8) − (-4)]2 + [(-2) − (-10)]2
d=
[(-8) − (0)]2 + [(-2) − (2)]2
d = (-4)2 + (8)2
d = (-8)2 + (-4)2
d = 16 + 64
d = 64 + 16
d = 80
d = 80
d
CD
d=
= (x − x )2 + (y − y )2
2
1
2
1
d
= (x − x )2 + (y − y )2
2
1
2
1
DA
[(0) − (4)]2 + [(2) − (-6)]2
d=
[(4) − (-4)]2 + [(-6) − (-10)]2
d = (-4)2 + (8)2
d = (8)2 + (4)2
d = 16 + 64
d = 64 + 16
d = 80
d = 80
(y 2 − y 1 ) (-2) − (-10) 8
=
= −2
=
(x 2 − x 1 ) (-8) − (-4) - 4
(y − y 1 ) (2) − (-2) 4 1
= =
BC slope = 2
=
(x 2 − x 1 ) (0) − (-8) 8 2
All sides are equal and right angle makes the midpoints form a square.
AB slope =
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Lesson Six
1.
a.
b.
c.
x2 + y2 = r 2
x2 + y2 = r 2
x2 + y2 = r 2
x 2 + y 2 = (4) 2
x 2 + y 2 = (10) 2
x 2 + y 2 = (0.5) 2
x 2 + y 2 = 16
x 2 + y 2 = 100
x 2 + y 2 = 0.25
d.
e.
x2 + y2 = r 2
x 2 + y 2 = (15) 2
x 2 + y 2 = 225
2.
x2 + y2 = r2
1
x 2 + y 2 = ( )2
3
1
x2 + y2 =
9
a.
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b.
c.
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d.
3.
a.
b.
x 2 + y2 = r2
x 2 + y2 = r2
(-3) 2 + (4) 2 = r 2
c.
(5) 2 + (0) 2 = r 2
x 2 + y2 = r2
(0) 2 + (-4) 2 = r 2
9 + 16 = r 2
25 + 0 = r 2
0 + 16 = r 2
25 = r 2
25 = r 2
16 = r 2
5=r
5=r
4=r
d.
x2 + y2 = r2
(6) 2 + (8) 2 = r 2
36 + 64 = r 2
100 = r 2
10 = r
4.
a.
b.
c.
x2 + y2 = r 2
x2 + y2 = r2
x2 + y2 = r 2
x 2 + y 2 = (5) 2
x 2 + y 2 = (5) 2
x 2 + y 2 = (4) 2
x 2 + y 2 = 25
x 2 + y 2 = 25
x 2 + y 2 = 16
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d.
x2 + y2 = r2
x 2 + y 2 = (10) 2
x 2 + y 2 = 100
5.
a.
b.
x + y = 100
2
x 2 + y 2 = 100
2
(6) 2 + (8) 2 = 100
36 + 64 = 100
100 = 100
lies on the circle
c.
(5) 2 + (9) 2 = 100
25 + 81 = 100
106 > 100
lies outside the circle
d.
x + y = 100
2
2
(7)2 + (7)2 = 100
49 + 49 = 100
98 < 100
lies inside the circle
x 2 + y 2 = 100
(10) 2 + (1) 2 = 100
100 + 1 = 100
101 > 100
lies outside the circle
Lesson Seven
1.
Midpoint
⎛ x + x 2 y1 + y 2 ⎞
⎟
=⎜ 1
,
⎟
AB ⎜
2
2
⎝
⎠
⎛ (7) + (-3) (1) + (6) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛4 7⎞
=⎜ , ⎟
⎝ 2 2⎠
7
= (2, )
2
Copyright © 2005, Durham Continuing Education
Midpoint
⎛ x + x 2 y1 + y 2 ⎞
⎟
=⎜ 1
,
⎟
BC ⎜
2
2
⎝
⎠
⎛ (-3) + (5) (6) + (-3) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛2 3⎞
=⎜ , ⎟
⎝2 2⎠
3
= (1, )
2
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Midpoint
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⎛ x + x 2 y1 + y 2 ⎞
⎟
=⎜ 1
,
⎟
CA ⎜
2
2
⎝
⎠
⎛ (5) + (7) (-3) + (1) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛ 12 - 2 ⎞
=⎜ , ⎟
⎝ 2 2 ⎠
= (6, - 1)
2.
⎛ x + x 2 y1 + y 2 ⎞
⎟
=⎜ 1
Midpoint
,
⎟
AB ⎜
2
2
⎝
⎠
⎛ (-6) + (3) (8) + (5) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛ - 3 13 ⎞
=⎜ , ⎟
⎝ 2 2 ⎠
Copyright © 2005, Durham Continuing Education
Midpoint
⎛ x + x 2 y1 + y 2 ⎞
⎟
,
=⎜ 1
⎟
BC ⎜
2
2
⎠
⎝
⎛ (3) + (1) (5) + (-4) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛ 4 1⎞
=⎜ , ⎟
⎝2 2⎠
1
= (2, )
2
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Midpoint
Support Question Answers
⎛ x + x 2 y1 + y 2 ⎞
⎟
,
=⎜ 1
⎟
CA ⎜
2
2
⎠
⎝
⎛ (-1) + (-6) (-4) + (8) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛-7 4⎞
=⎜ , ⎟
⎝ 2 2⎠
7
= (- , 2)
2
3.
⎛ x + x 2 y1 + y 2 ⎞
⎟
Midpoint
,
=⎜ 1
⎟
AB ⎜
2
2
⎠
⎝
⎛ (6) + (3) (-3) + (4) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛ 9 1⎞
=⎜ , ⎟
⎝2 2⎠
Copyright © 2005, Durham Continuing Education
Midpoint
⎛ x + x 2 y1 + y 2 ⎞
⎟
,
=⎜ 1
⎟
BC ⎜
2
2
⎠
⎝
⎛ (3) + (-4) (4) + (2) ⎞
=⎜
,
⎟
2
2
⎝
⎠
⎛ -1 6 ⎞
=⎜ , ⎟
⎝ 2 2⎠
1
= (- , 3)
2
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MPM2H – Principles of Mathematics (Transfer Course)
Midpoint
Support Question Answers
⎛ x + x 2 y1 + y 2 ⎞
⎟
,
=⎜ 1
⎟
CA ⎜
2
2
⎝
⎠
⎛ (-4) + (6) (2) + (-3) ⎞
,
=⎜
⎟
2
2
⎝
⎠
⎛ - 2 - 1⎞
=⎜ , ⎟
⎝ 2 2⎠
1⎞
⎛
= ⎜ - 1, - ⎟
2⎠
⎝
4.
a.
m AB =
1
(5) − (9) − 4
= 4 ∴ the ⊥ is =
4
(3) − (4) − 1
m CD =
(3) − (7) − 3 1
= ∴ the ⊥ is - 2
=
(-2) − (4) − 6 2
mEF =
4
(-2) − (1) − 3
=
∴ the ⊥ is
3
(7) − (3)
4
b.
c.
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MPM2H – Principles of Mathematics (Transfer Course)
5
a.
b.
1
x + 12 = 2x − 7
3
(3)(
6.
3
1 2
x - = 5x −
4
2 3
3
2
1x
(12)( ) − (12)( ) = (12)(5x) − (12)( )
4
3
2
6x - 8 = 60x - 9
1 = 54x
1
=x
54
1x
) + (3)(12) = (3)(2x) − (3)(7)
3
x + 36 = 6x - 21
57 = 5x
57
=x
5
c.
Support Question Answers
4x + 9 = 3x + 2
x = -7
a.
b.
c.
y = 5x + b
y = 5x + b
y = 5x + b
(4) = 5(-3) + b
4 = -15 + b
19 = b
(2) = 5(5) + b
2 = 25 + b
- 23 = b
(-4) = 5(7) + b
- 4 = 35 + b
- 39 = b
d.
e.
y = 5x + b
(-2) = 5(-1) + b
- 2 = -5 + b
3=b
y = 5x + b
(6) = 5(0) + b
6 = 0+b
6=b
Copyright © 2005, Durham Continuing Education
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MPM2H – Principles of Mathematics (Transfer Course)
7.
Slope of AB =
y1 − y 2
x1 − x 2
=
(1) − (0)
(5) − (-2)
Support Question Answers
⎡ (5) + (-2) (1) + (0) ⎤
,
Midpoint AB = ⎢
2 ⎥⎦
2
⎣
= (2.5, 0.5)
1
7
y1 − y 2
Slope of BC =
x1 − x 2
=
=
(0) − (8)
(-2) − (4)
⎡ (-2) + (4) (0) + (8) ⎤
,
Midpoint BC = ⎢
2 ⎥⎦
2
⎣
= (1, 4)
-8
-6
4
=
3
The slope of the line segment ⊥ AB is - 7
=
The slope of the line segment ⊥ BC is -
3
4
" b" for line ⊥ to BC
" b" for line ⊥ to AB
y = mx + b
y = −7x + b
y = mx + b
3
x+b
4
3
(4) = - (1) + b
4
4 = −0.75 + b
y=-
(0.5) = -7(1.5) + b
0.5 = −10.5 + b
11 = b
4.75 = b
y = −7x + 11
and
y = -0.75x + 4.75
− 7x + 11 = −0.75x + 4.75
6.25 = 6.25x
1= x
y = -7(1)+11
y = -7+11
y=4
So the point of intersection of the two perpendicular bisectors is (1, 4) and
this is the circumcentre of ΔABC .
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MPM2H – Principles of Mathematics (Transfer Course)
8.
Slope of DE =
y1 − y 2
x1 − x 2
=
(8) − (2)
(1) − (9)
Support Question Answers
⎡ (1) + (9) (8) + (2) ⎤
,
Midpoint DE = ⎢
2 ⎥⎦
⎣ 2
= (5, 5)
6
3
=−
-8
4
y − y2
Slope of EF = 1
x1 − x 2
=
=
(2) − (9)
(9) − (8)
-7
=
1
= −7
⎡ (9) + (8) (2) + (9) ⎤
,
Midpoint EF = ⎢
2 ⎥⎦
⎣ 2
= (8.5, 5.5)
The slope of the line segment ⊥ DE is
4
3
The slope of the line segment ⊥ EF is
1
7
" b" for line ⊥ to DE
y = mx + b
y=
" b" for line ⊥ to EF
y = mx + b
1
y = x+b
7
11 1 17
( ) = ( )+b
7 2
2
11 17
=
+b
2 14
30
=b
7
4
x +b
3
4
(5) + b
3
20
5=
+b
3
5
- =b
3
(5) =
y=
5
4
x3
3
and
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y=
30
1
x+
7
7
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MPM2H – Principles of Mathematics (Transfer Course)
Support Question Answers
30
5 1
4
x- = x+
7
3 7
3
30
1x
5
4x
(21)( ) - (21)( ) = (21)( ) + (21)( )
7
7
3
3
28x - 35 = 3x + 90
25x = 125
125
x=
25
x=5
4
5
x3
3
4
5
y = (5) 3
3
y=5
y=
So the point of intersection of the two perpendicular bisectors is (5, 5) and
this is the circumcentre of ΔABC .
Lesson Eight
1.
(-4) − (-3) − 1 1
(4) - (5) - 1 1
=
= and =
=
(3) − (7)
(1) - (5) - 4 4
−4 4
∴ the the lines containing these segments are parallel.
m=
2.
(7) − (5)
2
1
(1) - (7) - 6
=
= − and =
=2
(-1) − (3) − 4
2
(-4) - (-1) - 3
∴ the the lines containing these segments are perpendicu lar.
m=
3.
d
WX
d=
= (x − x )2 + (y − y )2
2
1
2
1
[(-2) − (-2)]2 + [(3) − (-2)]2
d
XY
d=
= (x − x )2 + (y − y )2
2
1
2
1
[(-2) − (2)]2 + [(-2) − (1)]2
d = (0)2 + (5)2
d = (-4)2 + (-3)2
d = 25
d = 16 + 9
d = 25
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MPM2H – Principles of Mathematics (Transfer Course)
d
YZ
d=
= (x − x )2 + (y − y )2
2
1
2
1
d
[(2) − (2)]2 + [(1) − (6)]2
d=
Support Question Answers
= (x − x )2 + (y − y )2
2
1
2
1
ZW
[(2) − (-2)]2 + [(6) − (3)]2
d = (0)2 + (-5)2
d = (4)2 + (3)2
d = 25
d = 16 + 9
d = 25
(y 2
(x 2
(y
XY slope = 2
(x 2
WX slope =
− y 1 ) (3) − (-2) 5
=
= = undefined
− x 1 ) (-2) − (-2) 0
− y 1 ) (-2) − (1) - 3 3
=
=
=
− x 1 ) (-2) − (2) - 4 4
All sides are equal do not contain a right angle making the vertices form a
rhombus.
4.
a.
d
= (x − x )2 + (y − y )2
2
1
2
1
AB
d=
[(-5) − (-1)]2 + [(2) − (3)]2
d
= (x − x )2 + (y − y )2
2
1
2
1
BC
d=
[(-1) − (-2)]2 + [(3) − (-1)]2
d = (-4)2 + (-1)2
d = (1)2 + (4)2
d = 16 + 1
d = 1 + 16
d = 17
d = 17
d
= (x − x )2 + (y − y )2
2
1
2
1
CD
d=
[(-2) − (-6)]2 + [(-1) − (-2)]2
d
= (x − x )2 + (y − y )2
2
1
2
1
DA
d=
[(-6) − (-5)]2 + [(-2) − (2)]2
d = (4)2 + (1)2
d = (-1)2 + (-4)2
d = 16 + 1
d = 1 + 16
d = 17
d = 17
(y 2 − y 1 ) (-4) − (2) - 6
=
=
=1
(x 2 − x 1 ) (3) − (9) - 6
(y − y 1 ) (2) − (-3)
5
=
=
= −1
CD slope = 2
(x 2 − x 1 ) (9) − (14) - 5
AB slope =
All sides are equal and contain a right angle making the vertices form a square.
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MPM2H – Principles of Mathematics (Transfer Course)
Support Question Answers
b.
d
= (x − x )2 + (y − y )2
2
1
2
1
EF
d=
[(-5) − (3)]2 + [(1) − (3)]2
d
= (x − x )2 + (y − y )2
2
1
2
1
FG
d=
[(3) − (4)]2 + [(3) − (-1)]2
d = (-8)2 + (-2)2
d = (-1)2 + (4)2
d = 64 + 4
d = 1 + 16
d = 68
d = 17
d
GH
d=
= (x − x )2 + (y − y )2
2
1
2
1
d
= (x − x )2 + (y − y )2
2
1
2
1
HE
[(4) − (-4)]2 + [(-1) − (-3)]2
d=
[(-4) − (-5)]2 + [(-3) − (1)]2
d = (8)2 + (2)2
d = (1)2 + (-4)2
d = 64 + 4
d = 1 + 16
d = 68
d = 17
(y 2 − y 1 ) (1) − (3) - 2 1
=
=
=
(x 2 − x 1 ) (-5) − (3) - 8 4
(y − y 1 ) (3) − (-1) 4
=
=
= −4
FG slope = 2
(x 2 − x 1 ) (3) − (4) - 1
EF slope =
Opposite sides are equal and contain a right angle making the vertices form a
rectangle.
c.
d
XY
d=
= (x − x )2 + (y − y )2
2
1
2
1
d
[(-5) − (-5)]2 + [(-4) − (1)]2
d=
YZ
= (x − x )2 + (y − y )2
2
1
2
1
[(-5) − (7)]2 + [(1) − (4)]2
d = (0)2 + (-5)2
d = (-12)2 + (-3)2
d = 25
d = 144 + 9
d = 153
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MPM2H – Principles of Mathematics (Transfer Course)
d
ZW
d=
= (x − x )2 + (y − y )2
2
1
2
1
[(7) − (7)]2 + [(4) − (-1)]2
d
W
d=
Support Question Answers
X = (x − x )2 + (y − y )2
2
1
2
1
[(7) − (-5)]2 + [(-1) − (-4)]2
d = (0)2 + (5)2
d = (12)2 + (3)2
d = 25
d = 144 + 9
d = 153
(y 2 − y 1 ) (-4) − (1) - 5
=
=
= undefined
(x 2 − x 1 ) (-5) − (-5) 0
(y − y 1 ) (1) − (4)
-3 1
YZ slope = 2
=
=
=
(x 2 − x 1 ) (-5) − (7) - 12 4
Opposite sides are equal do not contain a right angle making the vertices form a
parallelogram.
XY slope =
Lesson Nine
1.
a. 7.2 < 13 so ∠X is smaller than 67
c. 8 > 4 so ∠A is larger than 37
2.
a. 71° > 55° so x is larger than 13
c. 83° > 41° so a is larger than 7.2
b. 7.6 < 8.4 so ∠Y is smaller than 81
b. 77° > 16° so y is larger than 12.5
3.
Sin54° Sin27°
=
x
8
8 Sin54° = x Sin27°
8 Sin54° x Sin27°
=
Sin27°
Sin27°
8 Sin54°
=x
Sin27°
14.3 ≈ x
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MPM2H – Principles of Mathematics (Transfer Course)
4.
Support Question Answers
SinA Sin26°
=
6.7
3.2
6.7 Sin26° = 3.2 SinA
6.7 Sin26° 3.2 SinA
=
3.2
3.2
6.7 Sin26°
= Sin A
3.2
0.9178 ≈ Sin A
sin -1 0.9178 ≈ ∠A
66.6° ≈ ∠A
5.
Sin32° Sin73°
=
q
23
23 Sin32° = q Sin73°
23 Sin32° q Sin73°
=
Sin73°
Sin73°
23 Sin32°
=q
Sin73°
12.7 ≈ q
6.
Sin47° Sin52°
=
c
12
12 Sin47° = c Sin52°
12 Sin47° c Sin52°
=
Sin52°
Sin52°
12 Sin47°
=c
Sin52°
11.1 ≈ c
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MPM2H – Principles of Mathematics (Transfer Course)
Support Question Answers
7.
SinU Sin37°
=
16
22
16 Sin37° = 22 SinU
16 Sin37° 22 SinU
=
22
22
16 Sin37°
= Sin U
22
0.4377 ≈ Sin U
sin -1 0.4377 ≈ ∠U
26° ≈ ∠U
8.
SinZ Sin72°
=
11
41
11 Sin72° = 41 SinZ
11 Sin72° 41 SinZ
=
41
41
11 Sin72°
= Sin Z
41
0.1795 ≈ Sin Z
sin -1 0.1795 ≈ ∠Z
10.3° ≈ ∠Z
9.
Sin60° Sin45°
=
15
x
15 Sin45° = x Sin60°
15 Sin45° x Sin60°
=
Sin60°
Sin60°
15 Sin45°
=x
Sin60°
12.2 ≈ x
Sin60° Sin75°
=
15
y
15 Sin75° = y Sin60°
15 Sin75° y Sin60°
=
Sin60°
Sin60°
15 Sin75°
=y
Sin60°
16.7 ≈ y
∴the lengths of the wires are 12.2 m and 16.7 m
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MPM2H – Principles of Mathematics (Transfer Course)
Support Question Answers
Lesson Ten
1.
x 2 = 30 2 + 50 2 − 2(30)(50)Cos100°
x 2 ≈ 3920.9
x ≈ 62.6
2.
35 2 + 17 2 − 28 2
2(35)(17)
CosX ≈ 0.6134
CosX =
∠X ≈ Cos −1 0.6134
∠X ≈ 52.2°
3.
r 2 = 13 2 + 8 2 − 2(13)(8)Cos63°
r 2 ≈ 138.57
r ≈ 11.8
4.
a 2 = 17 2 + 28 2 − 2(17)(28)Cos105°
a 2 ≈ 1319.4
a ≈ 36.3°
5.
10 2 + 22 2 − 16 2
2(10)(22)
CosU ≈ 0.7455
CosU =
∠U ≈ Cos −1 0.7455
∠U ≈ 41.8°
6.
412 + 32 2 − 28 2
2(41)(32)
CosZ ≈ 0.732
CosZ =
∠Z ≈ Cos −1 0.732
∠Z ≈ 42.9°
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MPM2H – Principles of Mathematics (Transfer Course)
Support Question Answers
7.
x 2 = 12 2 + 15 2 − 2(12)(15)Cos90°
x 2 ≈ 369
x ≈ 19.2
8.
8 2 + 212 − 19 2
CosX =
2(8)(21)
CosX ≈ 0.4286
∠X ≈ Cos −1 0.4286
∠X ≈ 64.6°
9.
x 2 = 75 2 + 120 2 − 2(75)(120)Cos45°
x 2 ≈ 7297.1
x ≈ 85.4
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