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Straight Lines II
MTH-4107-1
MTH-4107-1 C1-C4 StraightLines II_Layout 1 10-02-03 10:22 Page 1
STRAIGHT
LINES II
MTH-4107-1
Mathematics Coordinator: Jean-Paul Groleau
Author: Nicole Perreault
Content revision: Jean-Paul GroleauSuzie AsselinDaniel GélineauLouise Allard
Updates: Jean-Paul Groleau
Photocomposition and layout: Multitexte Plus
Desktop publishing for updated version: P.P.I. inc.
Cover page: Daniel Rémy
English version: Services à la commuauté anglophoneDirection de la production en langue anglaise
– Translation: William Gore– Revision: Lana Georgieff
Translation of updated sections: Claudia de Fulviis
Reprint: 2006
© Société de formation à distance des commissions scolaires du Québec
All rights for translation and adaptation, in whole or in part, reserved for all countries.Any reproduction, by mechanical or electronic means, including microreproduction, isforbidden without the written permission of a duly authorized representative of theSociété de formation à distance des commissions scolaires du Québec (SOFAD).
Legal Deposit — 2006
Bibliothèque et Archives nationales du Québec
Bibliothèque et Archives Canada
ISBN 2-89493-285-8
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TABLE OF CONTENTS
Introduction to the Program Flowchart ................................................... 0.4The Program Flowchart ............................................................................ 0.5How to Use This Guide ............................................................................. 0.6General Introduction................................................................................. 0.9Intermediate and Terminal Objectives of the Module ............................ 0.11Diagnostic Test on the Prerequisites ....................................................... 0.13Answer Key for the Diagnostic Test on the Prerequisites ...................... 0.19Analysis of the Diagnostic Test Results ................................................... 0.23Information for Distance Education Students......................................... 0.25
UNITS
1. Determining the Equation of a Line ........................................................ 1.12. Perpendicular and Parallel Lines............................................................. 2.13. Distance Between Two Points .................................................................. 3.14. Coordinates of a Point That Divides a Line Segment in a
Particular Ratio ......................................................................................... 4.1
5. Solving Analytical Geometry Problems ................................................... 5.1
Final Review .............................................................................................. 6.1Answer Key for the Final .......................................................................... 6.8Terminal Objectives .................................................................................. 6.12Self-Evaluation Test.................................................................................. 6.15Answer Key for the Self-Evaluation Test ................................................ 6.25Analysis of the Self-Evaluation Test Results .......................................... 6.31Final Evaluation........................................................................................ 6.32Answer Key for the Exercises ................................................................... 6.33Glossary ..................................................................................................... 6.81List of Symbols .......................................................................................... 6.86Bibliography .............................................................................................. 6.87
Review Activities ....................................................................................... 7.1
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INTRODUCTION TO THE PROGRAM FLOWCHART
Welcome to the World of Mathematics!
This mathematics program has been developed for the adult students of the
Adult Education Services of school boards and distance education. The learning
activities have been designed for individualized learning. If you encounter
difficulties, do not hesitate to consult your teacher or to telephone the resource
person assigned to you. The following flowchart shows where this module fits
into the overall program. It allows you to see how far you have progressed and
how much you still have to do to achieve your vocational goal. There are several
possible paths you can take, depending on your chosen goal.
The first path consists of modules MTH-3003-2 (MTH-314) and MTH-4104-2
(MTH-416), and leads to a Diploma of Vocational Studies (DVS).
The second path consists of modules MTH-4109-1 (MTH-426), MTH-4111-2
(MTH-436) and MTH-5104-1 (MTH-514), and leads to a Secondary School
Diploma (SSD), which allows you to enroll in certain Cegep-level programs that
do not call for a knowledge of advanced mathematics.
The third path consists of modules MTH-5109-1 (MTH-526) and MTH-5111-2
(MTH-536), and leads to Cegep programs that call for a solid knowledge of
mathematics in addition to other abilities.
If this is your first contact with this mathematics program, consult the flowchart
on the next page and then read the section “How to Use This Guide.” Otherwise,
go directly to the section entitled “General Introduction.” Enjoy your work!
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CEGEP
MTH-5110-1 Introduction to Vectors
MTH-5109-1 Geometry IV
MTH-5108-1 Trigonometric Functions and Equations
MTH-5107-1 Exponential and Logarithmic Functions and Equations
MTH-5106-1 Real Functions and Equations
MTH-5105-1 Conics
MTH-5104-1 Optimization II
MTH-5103-1 Probability II
MTH-5102-1 Statistics III
MTH-5101-1 Optimization I
MTH-4110-1 The Four Operations on Algebraic Fractions
MTH-4109-1 Sets, Relations and Functions
MTH-4108-1 Quadratic Functions
MTH-4107-1 Straight Lines II
MTH-4106-1 Factoring and Algebraic Functions
MTH-4105-1 Exponents and Radicals
MTH-4103-1 Trigonometry I
MTH-4102-1 Geometry III
MTH-536
MTH-526
MTH-514
MTH-436
MTH-426
MTH-416
MTH-314
MTH-216
MTH-116
MTH-3002-2 Geometry II
MTH-3001-2 The Four Operations on Polynomials
MTH-2008-2 Statistics and Probabilities I
MTH-2007-2 Geometry I
MTH-2006-2 Equations and Inequalities I
MTH-1007-2 Decimals and Percent
MTH-1006-2 The Four Operations on Fractions
MTH-1005-2 The Four Operations on Integers
MTH-5111-2 Complement and Synthesis II
MTH-4111-2 Complement and Synthesis I
MTH-4101-2 Equations and Inequalities II
MTH-3003-2 Straight Lines I
TradesDVS
MTH-5112-1 Logic
25 hours = 1 credit
50 hours = 2 credits
MTH-4104-2 Statistics II
THE PROGRAM FLOWCHART
You ar e here
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Hi! My name is Monica and I have beenasked to tell you about this math module.What’s your name?
I’m Andy.
Whether you areregistered at anadult education
center or atFormation àdistance, ...
You’ll see that with this method, math isa real breeze!
... you have probably taken aplacement test which tells youexactly which module youshould start with.
My results on the testindicate that I should beginwith this module.
Now, the module you have in yourhand is divided into threesections. The first section is...
... the entry activity, whichcontains the test on theprerequisites.
By carefully correcting this test using thecorresponding answer key, and record-ing your results on the analysis sheet ...
HOW TO USE THIS GUIDE
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?
The memo pad signals a brief reminder ofconcepts which you have already studied.
The calculator symbol reminds you thatyou will need to use your calculator.
The sheaf of wheat indicates a review designed toreinforce what you have just learned. A row ofsheaves near the end of the module indicates thefinal review, which helps you to interrelate all thelearning activities in the module.
The starting lineshows where thelearning activitiesbegin.
The little white question mark indicates the questionsfor which answers are given in the text.?
... you can tell if you’re well enoughprepared to do all the activities in themodule.
The boldface question markindicates practice exerciseswhich allow you to try out whatyou have just learned.
And if I’m not, if I need a littlereview before moving on, whathappens then?
In that case, before you start theactivities in the module, the resultsanalysis chart refers you to a reviewactivity near the end of the module.
In this way, I can be sure Ihave all the prerequisitesfor starting.
Exactly! The second sectioncontains the learning activities. It’sthe main part of the module.
Look closely at the box tothe right. It explains thesymbols used to identify thevarious activities.
The target precedes theobjective to be met.
I see!
?
START
Lastly, the finish line indicatesthat it is time to go on to the self-evaluationtest to verify how well you have understoodthe learning activities.
FINISH
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A “Did you know that...”?
Later ...
For example. words in bold-face italics appear in theglossary at the end of themodule...
G r e a t !
... statements in boxes are importantpoints to remember, like definitions, for-mulas and rules. I’m telling you, the for-mat makes everything much easier.
The third section contains the final re-view, which interrelates the differentparts of the module.
Yes, for example, short tidbitson the history of mathematicsand fun puzzles. They are in-teresting and relieve tension atthe same time.
No, it’s not part of the learn-ing activity. It’s just there togive you a breather.
There are also many fun thingsin this module. For example,when you see the drawing of asage, it introduces a “Did youknow that...”
Must I memorize what the sage says?
It’s the same for the “math whiz”pages, which are designed espe-cially for those who love math.
They are so stimulating thateven if you don’t have to dothem, you’ll still want to.
And the whole module hasbeen arranged to makelearning easier.
There is also a self-evaluationtest and answer key. They tellyou if you’re ready for the finalevaluation.
Thanks, Monica, you’ve been a bighelp.
I’m glad! Now,I’ve got to run.
See you!This is great! I never thought that I wouldlike mathematics as much as this!
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GENERAL INTRODUCTION
STRAIGHT LINES II: ONE STEP TOWARDS ANALYTICAL
GEOMETRY
This module deals with straight lines. The Cartesian plane, coordinates,
the abscissa, the origin, the slope and the equation of a line are concepts
emphasized throughout this learning guide. You should normally be familiar
with this subject matter, since it was the focus of a module that precedes this
course. In this module, you will expand your knowledge of straight lines.
This second course on straight lines is highly relevant, since the concepts you will
learn can be applied in fields such as physics, chemistry, economics and business
management. In fact, graphs can be used to represent a wide variety of
situations, and the coordinates of aligned or non-aligned points in a Cartesian
plane provide important information that makes it possible to interpret a given
phenomenon.
We will begin by studying different ways of finding the equation of a line when
we are given only two pieces of information. We can determine the equation of
a line given its slope and y-intercept, given its slope and a point on that line, or
given two points on that line. We can also determine the equation of a line given
a point on that line and the equation of another line parallel or perpendicular
to it. How do we do this? You will have no trouble answering this question after
reading the first two units.
We will then examine the concept of distance and learn how to find the distance
between two given points. The Pythagorean theorem will once again prove to
be essential in this case.
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We will also examine two new mathematical formulas that involve using the
coordinates of the endpoints of a line segment to determine the point that divides
that segment in a particular ratio. We will first look at the formula for finding
the point that divides a line segment in any given ratio and then use that formula
to derive the formula for locating the midpoint of a segment. These concepts are
extremely useful for finding the equation of the perpendicular bisector, median
and altitude of a triangle in analytical geometry. Analytical geometry is the
study of geometric figures in which algebraic reasoning is used and position is
represented by coordinates.
In the last unit, you will be able to put what you have learned to the test by solving
problems involving several of the concepts you covered in the previous units
Without further ado, let’s enter the world of analytical geometry in which you
will discover a variety of new concepts.
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INTERMEDIATE AND TERMINAL OBJECTIVES OFTHE MODULE
Module MTH-4107-1 contains five units and requires 25 hours of study distrib-
uted as shown below. Each unit covers either an intermediate or a terminal
objective. The terminal objectives appear in boldface.
Objectives Number of Hours* % (Evaluation)
1 and 2 10 40%
3 3 15%
4 5 20%
5 6 25%
* One hour is allotted for the final evaluation.
1. Determining the Equation of a Line
Determine the equation of a line, given any one of the following:• the slope and the y-intercept of that line• the slope and a point on that line• two points on that line
2. Perpendicular and Parallel Lines
Determine the equation of a line, given either one of the following:• a point on that line and the equation of a line parallel to it,• a point on that line and the equation of a line perpendicular to it.The coefficients of the variables of these linear equations and thecoordinates of the points are rational numbers. The resultingequation must be of the form y = mx + b or the form Ax + By + C = 0.The steps in the solution must be shown.
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3. Distance Between Two Points
Determine the distance between two given points in the Cartesian
plane. The coordinates of these points and the distance are rational
numbers and the problems deal with everyday situations. The steps
in the solution must be shown and the distance must be stated in a
unit of measure.
4. Coordinates of a Point That Divides a Line Segment in a Particular
Ratio
Determine the point that divides a line segment in a particular ratio,
given the endpoints of the line segment. The coordinates of the
endpoints and the ratio are rational numbers and the problems deal
with everyday situations. The ratio in which the segment is divided
must be derived from the information given in the problem. The
steps in the solution must be shown.
5. Solving Analytical Geometry Problems
Solve problems that involve calculating the distance between two
points, determining the coordinates of a point that divides a line
segment and finding the equation of a line. The solutions may involve
all or some of these concepts.
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DIAGNOSTIC TEST ON THE PREREQUISITES
Instructions
1. Answer as many questions as you can.
2. You may use a calculator.
3. Write your answers on the test paper.
4. Don't waste any time. If you cannot answer a question, go on tothe next one immediately.
5. When you have answered as many questions as you can, correctyour answers using the answer key which follows the diagnostictest.
6. To be considered correct, your answers must be identical tothose in the key. In addition, the various steps in your solutionshould be equivalent to those shown in the answer key.
7. Transcribe your results onto the chart which follows the answerkey. This chart gives an analysis of the diagnostic test results.
8. Do only the review activities that apply to each of your incorrectanswers.
9. If all your answers are correct, you may begin working on thismodule.
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1. Plot the following points in the Cartesian plane below.
A(2, 4)
B 3, – 52
C(–2, 0)
D(0, 0)
E – 12 , –4
2. Graph the following equations in the Cartesian plane below.
a) 3y + 9x = –12 b) y + 6 = –3x
................................................... ........................................................
................................................... ........................................................
y
x
1
1
x
y
x
y
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y
x
1
1
c) What kind of lines did you draw in a) and b)? Explain your answer.
.......................................................................................................................
.......................................................................................................................
3. Find the slope of each of the following lines.
a) 2x – 3y – 2 = 0
.......................................................................................................................
.......................................................................................................................
b) y = – x4 + 13
.......................................................................................................................
.......................................................................................................................
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4. What is the slope of the line that passes through each of the following points?
a) (4, –2) and (–1, 0)
.......................................................................................................................
.......................................................................................................................
b) 12 , 1
3 and 0, 23
.......................................................................................................................
.......................................................................................................................
5. Solve the following equations, showing all the steps in your solution.
a) y – 22y + 4 = 3
2 b) 3x – 45 = 8 – 2x
3
................................................... ........................................................
................................................... ........................................................
................................................... ........................................................
................................................... ........................................................
................................................... ........................................................
................................................... ........................................................
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6. Calculate the length of the third side of the right triangles below.
a) b)C
A B
?
6
5
AC
B
?
2
9
................................................... ........................................................
................................................... ........................................................
................................................... ........................................................
................................................... ........................................................
................................................... ........................................................
7. The Fords are going away for the weekend. They will be travelling two
hundred and fifty kilometres due south and three hundred and eighty
kilometres east. They can cover four hundred and seventy-five kilometres on
one tank of gas and it costs them thirty-two dollars to fill up. Calculate the
straight-line distance (d) between their home and their destination.
................................................................
................................................................
................................................................
................................................................
................................................................
...........................................................................................................................
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ANSWER KEY FOR THE DIAGNOSTIC TESTON THE PREREQUISITES
1.
B 3, – 52
E – 12 , – 4
•
•
y
x
1
A(2, 4)
D(0, 0)C(–2, 0)1
•
• •
2. a) 3y + 9x = –12 b) y + 6 = –3x
3y = –9x – 12 y = –3x – 6
y = –3x – 4
y
a)b)
1
1 x
•
••
•
•
•
x –2 –1 0
y 2 –1 –4
x –2 –1 0
y 0 –3 –6
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c) They are parallel, distinct lines because they have the same slope
(m = –3) and different y-intercepts (–4 ≠ –6).
3. a) m = – AB
= –2–3 = 2
3 (Equation of the form Ax + By + C = 0)
b) m = coefficient of x = – 14 (Equation of the form y = mx + b)
4. a) m = y2 – y1x2 – x1
= 0 – (–2)–1 – 4 = 2
–5 = – 25
b) m =
y2 – y1x 2 – x 1
=23 – 1
30 – 1
2
=13
– 12
= 13 × – 2
1 = – 23
5. a) y – 22y + 4 = 3
2 b) 3x – 45 = 8 – 2x
3
2(y – 2) = 3(2y + 4) 3(3x – 4) = 5(8 – 2x)
2y – 4 = 6y + 12 9x – 12 = 40 – 10x
2y – 6y = 12 + 4 9x + 10x = 40 + 12
–4y = 16 19x = 52
y = 16–4 x = 52
19 or 2 1419
y = –4
6. a) a2 = b2 + c2 b) c2 = a2 – b2
a2 = 52 + 62 c2 = 92 – 22
a2 = 25 + 36 c2 = 81 – 4
a2 = 61 c2 = 77
a = 7.81 c = 8.77
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7.
d2 = 2502 + 3802
d2 = 62 500 + 144 400
d2 = 206 900
d = 454.86 (to the nearest hundredth)
Their home and destination are 454.9 kilometres apart.
d
Destination380
250
Home
N
S
EW
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ANALYSIS OF THE DIAGNOSTICTEST RESULTS
QuestionsAnswers Review Before going on
Correct Incorrect Section Page to unit
1. 7.2 7.19 Unit 12. a) 7.2 7.19 Unit 1
b) 7.2 7.19 Unit 1c) 7.2 7.19 Unit 1
3. a) 7.2 7.19 Unit 1b) 7.2 7.19 Unit 1
4. a) 7.2 7.19 Unit 1b) 7.2 7.19 Unit 1
5. a) 7.3 7.32 Unit 1b) 7.3 7.32 Unit 1
6. a) 7.1 7.4 Unit 3b) 7.1 7.4 Unit 3
7. 7.1 7.4 Unit 3
• If all your answers are correct, you may begin working on this module.
• For each incorrect answer, find the related section listed in the “Review”
column. Complete this section before beginning the unit listed in the right-
hand column under the heading “Before Going on to Unit”.
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INFORMATION FOR DISTANCEEDUCATION STUDENTS
You now have the learning material for MTH-4107-1 together with the home-
work assignments. Enclosed with this material is a letter of introduction from
your tutor indicating the various ways in which you can communicate with him
or her (e.g. by letter or telephone) as well as the times when he or she is available.
Your tutor will correct your work and help you with your studies. Do not hesitate
to make use of his or her services if you have any questions.
DEVELOPING EFFECTIVE STUDY HABITS
Distance education is a process which offers considerable flexibility, but which
also requires active involvement on your part. It demands regular study and
sustained effort. Efficient study habits will simplify your task. To ensure
effective and continuous progress in your studies, it is strongly recommended
that you:
• draw up a study timetable that takes your working habits into account and
is compatible with your leisure time and other activities;
• develop a habit of regular and concentrated study.
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The following guidelines concerning the theory, examples, exercises and assign-
ments are designed to help you succeed in this mathematics course.
Theory
To make sure you thoroughly grasp the theoretical concepts:
1. Read the lesson carefully and underline the important points.
2. Memorize the definitions, formulas and procedures used to solve a given
problem, since this will make the lesson much easier to understand.
3. At the end of an assignment, make a note of any points that you do not
understand. Your tutor will then be able to give you pertinent explanations.
4. Try to continue studying even if you run into a particular problem. However,
if a major difficulty hinders your learning, ask for explanations before
sending in your assignment. Contact your tutor, using the procedure
outlined in his or her letter of introduction.
Examples
The examples given throughout the course are an application of the theory you
are studying. They illustrate the steps involved in doing the exercises. Carefully
study the solutions given in the examples and redo them yourself before starting
the exercises.
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Exercises
The exercises in each unit are generally modelled on the examples provided.
Here are a few suggestions to help you complete these exercises.
1. Write up your solutions, using the examples in the unit as models. It is
important not to refer to the answer key found on the coloured pages at the
end of the module until you have completed the exercises.
2. Compare your solutions with those in the answer key only after having done
all the exercises. Careful! Examine the steps in your solution carefully even
if your answers are correct.
3. If you find a mistake in your answer or your solution, review the concepts that
you did not understand, as well as the pertinent examples. Then, redo the
exercise.
4. Make sure you have successfully completed all the exercises in a unit before
moving on to the next one.
Homework Assignments
Module MTH-4107-1 contains three assignments. The first page of each
assignment indicates the units to which the questions refer. The assignments
are designed to evaluate how well you have understood the material studied.
They also provide a means of communicating with your tutor.
When you have understood the material and have successfully done the perti-
nent exercises, do the corresponding assignment immediately. Here are a few
suggestions.
1. Do a rough draft first and then, if necessary, revise your solutions before
submitting a clean copy of your answer.
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2. Copy out your final answers or solutions in the blank spaces of the document
to be sent to your tutor. It is preferable to use a pencil.
3. Include a clear and detailed solution with the answer if the problem involves
several steps.
4. Mail only one homework assignment at a time. After correcting the assign-
ment, your tutor will return it to you.
In the section “Student’s Questions”, write any questions which you may wish to
have answered by your tutor. He or she will give you advice and guide you in your
studies, if necessary.
In this course
Homework Assignment 1 is based on units 1 and 2.
Homework Assignment 2 is based on units 3 to 5.
Homework Assignment 3 is based on units 1 to 5.
CERTIFICATION
When you have completed all the work, and provided you have maintained an
average of at least 60%, you will be eligible to write the examination for this
course.
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UNIT 1
DETERMINING THE EQUATIONOF A LINE
1.1 SETTING THE CONTEXT
Temperature Scales and Linear Equations
Steve is starting to feel the heat! He has to hand in his math
homework tomorrow and doesn’t really know how to solve the
three problems he has been assigned even though his teacher,
Mr. Mercurio, clearly explained the concepts required to do this
work. Steve feels a bit lost in a world where terms like slope,
abscissa, ordinate, and intercept points are just words on a
page.
The three problems are outlined on the next page.
-30-25-20-15-10-505
1015202530°C
START
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3. When placed in melting ice, a
thermometer with a Celsius
scale reads 0°, while a ther-
mometer with a Fahrenheit
scale reads 32°. When these
thermometers are immersed
in boiling water, they read
100°C and 212°F respec-
tively. Find the equation of
the line that shows how tem-
perature readings in degrees
Fahrenheit vary as a function
of temperature readings in
degrees Celsius.
1. Find the equation of the line
that shows how temperature
readings in degrees Kelvin
vary as a function of tem-
perature readings in degrees
Celsius. The slope of this line
is 1 and its y-intercept is 273.
2. What is the equation of the
line that shows how tem-
perature readings in degrees
Réaumur vary as a function
of temperature readings in
degrees Celsius? The slope of
this line is 45 . When placed
in boiling water, a thermom-
eter with a Celsius scale
reads 100°, whereas a ther-
mometer with a Réaumur
scale reads 80°.
To achieve the objective of this unit, you should be able to find the
equation of a line given its slope and y-intercept, given the slope and the
coordinates of a point on that line, or given the coordinates of two points
on that line.
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Let’s return to Steve’s first problem. The slope and the y-intercept are given.
The Cartesian plane in Figure 1.1 illustrates this situation.
400
300
200
100
(0, 273) m = 1
(y-intercept)
Origin
6 12 18 24 x C
K°
°
y
Fig. 1.1 Temperature readings in degrees Kelvin as a functionof temperature readings in degrees Celsius
• The y-intercept is the second coordinate (or ordinate) of the
point where the line crosses the y-axis. This point is usually
expressed as follows: (0, b).
• The x-intercept is the first coordinate (or abscissa) of the
point where the line crosses the x-axis. This point is usually
expressed as follows: (a, 0).
• The slope (or rate of change) of a line is the change in the
ordinates (or y-coordinates) divided by the change in the
abscissas (or x-coordinates). This change is represented by
the mathematical equation m = y2 – y1x2 – x1
.
We already know that there are two ways of writing the equation of a line.
y = mx + b or Ax + By + C = 0
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In the first equation, m represents the slope of the line and b represents the
y-intercept, namely the y-coordinate of the point where the line crosses the y-
axis.
If we are given the slope and the y-intercept of a line, as is the case in Steve’s first
problem, we can easily find the equation of that line.
1. Let m = the value of the slope and b = the value of the y-intercept in the
equation y = mx + b.
Since m = 1 and b = 273,
y = 1x + 273 or y = x + 273.
2. Check the resulting equation by replacing each variable with the correspond-
ing coordinate of the point (0, 273).
y = x + 273
273 = 0 + 273
273 = 273
N.B. All the points on the line satisfy this equation. You can check this for
yourself, using any point on the line in Figure 1.1.
In any equation of the form y = mx + b, m is the slope and b is the
y-intercept.
The slope of a line is the coefficient of x and the y-intercept is the
constant term when the y variable is isolated.
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In any equation of the form Ax + By + C = 0, known as the general
form of the equation of a line, the slope is equal to – AB
if
B ≠ 0 and the y-intercept is equal to – CB
if B ≠ 0.
Example 1
Find the equation of a line that crosses the y-axis at the point (0, –2.5) and
that has a slope of –3.
If m = –3 and b = –2.5,
then y = –3x – 2.5.
Check: –2.5 = –3(0) – 2.5
–2.5 = –2.5
• Lines rising to the right have a posi-
tive slope (m > 0).
• Lines falling to the right have a
negative slope (m < 0).
• The slope of a horizontal line is zero
(m = 0).
• The slope of a vertical line is unde-
fined.
Fig. 1.2 Position of a
line depending on its
slope
That was fairly easy! Now let’s move on to Steve’s second problem, which
involves finding the equation of a line, given its slope and a point on that line.
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The problem states that when a thermometer with a Celsius scale reads 100, the
one with a Réaumur scale reads 80. This corresponds to the point (100, 80) in the
Cartesian plane.
The point (80, 100) is not represented in this situation, because the problem calls
for the equation that shows how temperature readings in degrees Réaumur vary
as a function of temperature readings in degrees Celsius. This means that the
Réaumur readings are on the y-axis and the Celsius readings are on the x-axis
and not vice-versa. Compare figures 1.3 and 1.4.
°R
80
20
20 100 °C
(100, 80)•
°C
100
20
20 80 °R
(80, 100)•
Fig. 1.3 Temperature readingsin degrees Réaumur as a func-tion of temperature readings indegrees Celsius
We already know that the slope is the vertical change (change in y or rise)
divided by the horizontal change (change in x or run). In other words,
m = y2 – y1x2 – x1
. In this case, the slope is 45 , which means that the slope between the
point (100, 80) and any other point (x, y) on the line will also be 45 . Therefore, we
can say:
45 = y – 80
x – 100
Fig. 1.4 Temperature change indegrees Celsius in relation totemperature change in degreesRéaumur
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To find the equation of the line, you need only solve this equation expressed as
a proportion.
Solving an equation expressed as a proportion involves applying
the fundamental property of proportions, which states that the
product of the extremes is equal to the product of the means.
Hence,
4x2 = 6
74x × 7 = 2 × 6
28x = 12
x = 1228
x = 37
Therefore:
45 = y – 80
x – 1005(y – 80) = 4(x – 100)
5y – 400 = 4x – 400
5y = 4x – 400 + 400
5y = 4x + 0
y = 45 x
N.B. In this case, b = 0. This means that the line passes through the origin,
namely the point (0, 0).
After finding the equation of the line, you have to check it by replacing each
variable with the corresponding coordinate of the point given in the statement
of the problem, i.e., (100, 80).
y = 45 x
80 = 45 × 100
80 = 80
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Hence, we can say that the equation represents a line that passes through the
point (100, 80) and has a slope of 45 . By examining the figure below, you will see
that any point on the line satisfies this equation.
°C
°R
(100, 80)
m = 4/5
(0, 0) °C
°R
(100, 80)
m = 4/5
(0, 0)•
•
Fig.1.5 Temperature readings in degrees Réaumur as a
function of temperature readings in degrees Celsius
To find the equation of a line when its slope m and the coordinates (x1, y1) of a point
on that line are known, simply use the slope formula by letting (x, y) represent
any point on that line. Hence, m = y – y1x – x1
.
Example 2
What is the equation of the line that has a slope of – 56 and that passes
through the point 23 , – 1
2 ?
1. Apply the formula m = y – y1x – x1
where (x1, y1) are the coordinates of the
given point.
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– 56 =
y – – 12
x – 23
– 56 =
y + 12
x – 23
6 y + 12 = –5 x – 2
3
6y + 3 = –5x + 103
6y = –5x + 103 – 3
6y = –5x + 13
y = – 56 x + 1
18
2. Check the equation by replacing each variable with the corresponding
coordinate of the given point.
y = – 56 x + 1
18
– 12 = – 5
623 + 1
18
– 12 = – 10
18 + 118
– 12 = – 9
18 – 12 = – 1
2
Steve is now more confident and doesn’t find these problems so difficult after all.
By doing the following exercises, you will see whether you too can find the
equation of a line using certain information. Before going on to the exercise,
however, let’s look at a bit of history.
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Did you know that...
... temperature scales are almost always named after the
scientist who devised them? Swedish astronomer and
physicist Anders Celsius (1701-1744) came up with the idea
of dividing a thermometer into 100 equal units.
The Fahrenheit scale was created by German physicist Gabriel Daniel
Fahrenheit (1686-1736). He perfected the thermometer by using mercury
rather than alcohol and devised a new scale. He also invented a hydrom-
eter, an instrument used to measure the specific gravity of liquids.
The Kelvin scale, also known as the absolute scale, was devised and first
used by William Thomson, alias Lord Kelvin (1824-1907). This British
physicist made significant contributions to the study of thermodynamics
and invented a galvanometer (a device that measures small electric
currents) as well as an electrometer (an instrument used to measure
differences in electric potential).
French physicist and naturalist René Antoine Ferchault de Réaumur
(1683-1757) also invented a thermometer that bears his name, and his
work on steel and tin production made him “the father of metallurgy.” He
also did a great deal of research on invertebrates.
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Exercise 1.1
1. Find the equation of each line below, given its y-intercept and slope. Check
your answer.
a) The y-intercept is 5 and the slope is 2.
b) The y-intercept is –3 and the slope is 12 .
c) The y-intercept is –5.35 and the slope is –0.54.
2. What is the equation of the line that satisfies each set of conditions stated
below? Check your answer.
a) The line passes through (2, 5) and m = – 12 .
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b) The line passes through (–2, –1) and m = 2.
c) The line passes through 13 , 1
4 and m = – 16 .
d) The line passes through (0, 3) and m = –8.
e) The line passes through (–2.4, 6.3) and m = 5.6.
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Steve is now set to tackle the third problem in his homework assignment, which
involves finding the equation of the line that shows how temperature readings
in degrees Fahrenheit vary as a function of temperature readings in degrees
Celsius.
? Along which axis of the Cartesian plane should the Fahrenheit readings be
scaled?
...........................................................................................................................
? Along which axis of the Cartesian plane will the Celsius readings be scaled?
...........................................................................................................................
Since the problem states that the line should show how temperature readings in
degrees Fahrenheit vary as a function of temperature readings in degrees
Celsius, the Fahrenheit readings should be scaled along the y-axis and the
Celsius readings along the x-axis.
The problem also states that when placed in melting ice, the Fahrenheit
thermometer reads 32° and the Celsius thermometer reads 0°. This means that
one of the points on the graph is (0, 32). In boiling water, these thermometers
read 212 degrees Fahrenheit and 100 degrees Celsius respectively. These
measurements correspond to the point (100, 212).
Steve seems to remember Mr. Mercurio saying that they should draw the line
representing a given situation because a diagram often makes it easier to solve
the problem.
? Plot the points (0, 32) and (100, 212) in the following Cartesian plane and
draw the line that passes through these two points.
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°F
°C
Fig. 1.6 Cartesian plane
We need to find the equation of the line that passes through these two points. The
figure below shows the line representing the situation described in the problem.
°F
°C
(100, 212)
10010 20
204060
(0, 32)
80100120140160180200220
30 40 50 60 70 80 90
•
•
Fig. 1.7 Temperature readings in degrees Fahrenheit as a
function of temperature readings in degrees Celsius
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Find the equation of a line, given two points on that line.
1. Calculate the slope of this line, using the coordinates of the two points.
(x1, y1) = (0, 32) and (x2, y2) = (100, 212).
m = y2 – y1x2 – x1
m = 212 – 32100 – 0
m = 180100
m = 95
N.B. For the sake of convenience, the points (0, 32) and (100, 212) have been
designated as (x1, y1) and (x2, y2) respectively so that the difference between
the y-coordinates and the difference between the x-coordinates would be
positive. Nevertheless, we would get the same result if we designated the
points (100, 212) and (0, 32) as (x1, y1) and (x2, y2) respectively.
2. Use the value of the slope and the coordinates of one of the two points to find
the equation.
m = y – y1x – x1
95 = y – 32
x – 05(y – 32) = 9 (x – 0)
5y – 160 = 9x
5y = 9x + 160
y = 95 x + 160
5
y = 95 x + 32
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N.B. We can also use the point (100, 212) to find the equation. The calcula-
tions are as follows:
95 = y – 212
x – 100
5(y – 212) = 9(x – 100)
5y – 1 060 = 9x – 900
5y = 9x + 160
y = 95 x + 160
5
y = 95 x + 32
N.B. Because (0, 32) is the y-intercept, we can also find the equation of the line
by letting m = 95 and b = 32. Hence,
y = mx + b
y = 95 x + 32
3. Check the resulting equation by replacing the variables with the coordinates
of either of the given points.
y = 95 x + 32
32 = 95 (0) + 32
32 = 32
N.B. Of course, you can check your answer using the point (100, 212).
Example 3
What is the equation of the line that passes through the points (2.53, –3.02)
and (2.45, 0)?
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1. Calculate the slope of the line passing through these two points.
m = y2 – y1x2 – x1
m = 0 – (–3.02)2.45 – 2.53
m = 3.02– 0.08
m = –37.75
2. Use the value of the slope and the coordinates of one of the points to find
the equation.
m = y – y1x – x1
–37.75 = y – 0x – 2.45
y = –37.75(x – 2.45)
y = –37.75x + 92.49
N.B. We get 92.49 by rounding off to the nearest hundredth.
3. Check the resulting equation by replacing the variables with the coordi-
nates of either of the given points.
y = –37.75x + 92.49
0 = –37.75(2.45) + 92.49
0 = –92.49 + 92.49
0 = 0
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Steve has finished his homework, realizing that there was no great mystery to
this after all! Once you understand the terms related to Cartesian planes, all you
have to do is solve equations expressed as proportions. Now let’s do a few
exercises to see if you can manage as well as Steve did.
Exercise 1.2
Determine the equation of the line that passes through each pair of points given
below. Show all the steps in your solution and check your answer.
N.B. Your results must be rounded to the nearest hundredth, when necessary.
1. (1, –3) and (2, 2)
2. (2.3, –5,1) and (3.2, 2.7)
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3. 12 , –1 and – 3
2 , 2
4. (0, 4) and 12 , 5
2
5. (2.1, –3.4) and (–1.3, –2.7)
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6. (0, 3.25) and (5.14, 0)
7. 0, 13 and 1
8 , 0
8. (0, 0) and (–3.2, –1.3)
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When trying to find the equation of a line, you may come across two special
situations. The following two examples illustrate these cases.
Example 4
What is the equation of the line that passes through the points (–3, 3) and (4, 3)?
1. First, calculate the slope.
m = 3 – 34 – (–3)
= 04 + 3 = 0
7 = 0
Since its slope is equal to 0, this line is horizontal.
2. Find its equation.
0 = y – 3x + 3
y – 3 = 0(x + 3)
y – 3 = 0
y = 3
Equations of the form y = b, namely
those which do not have an x vari-
able, are represented by a horizontal
line parallel to the x-axis.
Fig. 1.8 Horizontal line
The second part of this example is easier to do.
Since the distance between the x-axis and each point on a horizontal line is
the same, all the points have the same y-coordinate.
If b = 3 and y = b , then y = 3.
(–3, 3) (4, 3)
10
1
m = 0y
x
• •
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Example 5
What is the equation of the line that passes through the points (1, 4) and (1, –2)?
1. First, calculate the slope.
m = –2 – 41 – 1 = –6
0 , which is undefined.
Since its slope is undefined, this line is vertical.
Equations of the form x = a, namely
those which do not have a y variable,
are represented by a vertical line par-
allel to the y-axis.
Fig. 1.9 Vertical line
2. Find its equation.
Since the distance between the y-axis and each point on a vertical line is the
same, all the points have the same x-coordinate.
If a = 1 and x = a, then x = 1.
By simply looking at the coordinates, find the equation of the line that passes
through each of the following pairs of points.
? a) (–2, 5) and –2, 34 : ................... b) (0.25, 3.75) and (4, 3.75): ..............
m: un-defined
0
(1, 4)
(1, –2)
1
1 x
y •
•
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If your equations are x = –2 and y = 3.75, you have used the quickest way of
finding the solution. In the first set of ordered pairs, the x-coordinate (–2)
remains the same, whereas in the second set of points, the y-coordinate (3.75)
remains the same.
You don’t have to do any calculations in these situations. You need only
remember that x = a represents a vertical line and that y = b corresponds to a
horizontal line.
Before going on to the practice exercises, let’s summarize the steps involved in
finding the equation of a line when different types of information are given.
Procedure for finding the equation of a line, given its
slope and its y-intercept:
1. Let m = the value of the slope and b = the value of the
y-intercept in the equation y = mx + b;
2. Check the resulting equation by replacing each variable
with the corresponding coordinate of the point (0, b).
Procedure for finding the equation of a line, given its
slope and a point on that line:
1. Apply the formula m = y – y1x – x1
where (x1, y1) are the
coordinates of the given point;
2. Check the resulting equation by replacing each variable
with the corresponding coordinate of the given point.
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Procedure for finding the equation of a line, given two
points on that line:
1. Calculate the slope of the line by substituting the coordi-
nates of the two given points for the variables in the
formula m = y2 – y1x2 – x1
;
2. Find the equation of the line by substituting the value of
the slope and the coordinates of one of the given points in
the formula m = y – y1x – x1
;
3. Check the resulting equation by replacing the variables
with the coordinates of either of the given points.
We’ve now come to the end of this unit. The practice exercises will test your
understanding of the concepts described so far. If you have trouble with these
exercises, reread the relevant explanations. You have to understand these
concepts to be able to do the rest of the work in this course. For now though, let’s
take a look at the world of science.
Did you know that…
…American scientists have discovered a relationship be-
tween air temperature and the number of times a cricket
chirps per minute? This means that you could use a cricket
as a thermometer!
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If a cricket chirps 40 times per minute
at 50°F and 140 times per minute at
75°F, find the equation that American
scientists devised to show how the tem-
perature in degrees Fahrenheit is re-
lated to the number of times a cricket
chirps per minute. Using this equation,
determine the temperature (first in de-
grees Fahrenheit, then in degrees Cel-
sius) at which a cricket stops chirping.
N.B. The solution is in the answer key for the exercises.
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?1.2 PRACTICE EXERCISES
For each of the following problems, find the equation of the line that satisfies the
given conditions. Show all the steps in your solution and check your answers.
N.B. The results must be rounded off to the nearest hundredth, when necessary.
1. The line has a slope of 3 and passes through the point (–2, 7).
2. The line has a slope of –2 and passes through the point (2, –5).
3. The line has a slope of – 43 and passes through the point (0, 4).
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4. The line passes through the points (–2, 6) and (3, –5).
5. The line passes through the points (2.3, –1.5) and (1.7, 2.8).
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6. The line has a slope of – 23 and its y-intercept is 4.
7. The line has a slope of 13 and its x-intercept is –3.
8. The y-intercept is –2.5 and the x-intercept is –5.3.
9. The line passes through the points (–3, 2) and (4, 2).
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10. The y-intercept is – 12 and the line passes through the point (3, –2).
11. The line has a slope of – 23 and passes through the origin.
12. The line has a slope of –2.3 and passes through the point (2.5, 0.4).
13. The line passes through the points – 52 , 7 and – 5
2 , 0 .
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14. The line has a slope of 0 and passes through the point – 12 , – 2
3 .
15. The height of the column of mercury in a thermometer is zero at 0°C and
11 cm when the temperature rises to 33°C. Find the equation of the line for
each case below.
a) The height of the column of mercury is scaled along the y-axis and the
temperature in °C is scaled along the x-axis.
b) The temperature in °C is scaled along the y-axis and the height of the
column of mercury is scaled along the x-axis.
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1.3 REVIEW EXERCISES
1. Complete the following sentences by writing in the correct missing terms or
expressions.
The slope of a line is the ....................................... of x when the y variable is
isolated.
The y-intercept of a line is the ................................. ............................. when
the y variable is isolated.
2. Find the equation of each line, given the information in the following table.
m x-intercept b P1(x1, y1) P2(x2, y2) Equation
of the line
6 –3 a)
–3 4 b)
(2.3, –3.6) (4.2, 3.7) c)
– 34
25 d)
– 32
13 e)
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1.4 THE MATH WHIZ PAGE
Intercept Points
If a and b are respectively the intercept points of any given line, then
that line passes through the points (a, 0) and (0, b). We already know
that the slope determined using the coordinates of two given points is
equal to the slope between any point on the line and one of the two
given points. This relationship is expressed as follows:
y2 – y1x2 – x1
= y – y1x – x1
Using this equation, find the equation of a line, given its intercept
points.