MTH-4107-1 S L II - SOFAD · · 2014-05-15MTH-4107-1 Straight Lines II MTH-4106-1 Factoring and...

Straight Lines II

MTH-4107-1

MTH-4107-1 C1-C4 StraightLines II_Layout 1 10-02-03 10:22

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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MTH-4107-1 Straight Lines II

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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-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

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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-


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-


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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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.

MTH-4107-1 S L II - SOFAD · · 2014-05-15MTH-4107-1 Straight Lines II MTH-4106-1 Factoring and...

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