Hawkes Learning Systems: College Algebra

HAWKES LEARNING SYSTEMS

math courseware specialists

Copyright © 2011 Hawkes Learning Systems. All rights reserved.

Hawkes Learning Systems:College Algebra3.3: Forms of Linear Equations




Objectives

o Understand the meaning of and to be able to calculate the slope of a line.

o Be able to write the equation of a line in slope-intercept form.

o Be able to write the equation of a line in point-slope form.




The Slope of a Line

o There are several ways to characterize a given line in the Cartesian plane.

o We have already used one way repeatedly: plotting two distinct points in the Cartesian plane to determine a unique line.

o Another approach is to identify just one point on the line and to indicate how “steeply” the line is rising or falling as we scan the plane from left to right. A single number is sufficient to convey this notion of “steepness”.




The Slope of a Line

Let stand for a given line in the Cartesian plane, and let and be the coordinates of any two distinct points on . The slope, , of the line, is the ratio

which, can be described in words as “change in over change in ” or “rise over run.”

L 1 1,x y 2 2,x y

LL

2 1

2 1

y ymx x

yx

m




The Slope of a Line

Rise and Run Between Two Pointsy

x

2 1Rise y y

2 1Run x x 2 1,x y

2 2,x y

1 1,x y

As drawn above, the ratio is positive, and we say that the line has a positive slope. If the rise and run have opposite signs, the slope of the line would be negative and the line under consideration would be falling from the upper left to the lower right.

2 1

2 1

y yx x




The Slope of a Line

Caution!

It doesn’t matter how you assign the labels

and to the two points you are using to calculate

slope, but it is important that you are consistent as you

apply the formula. That is, don’t change the order in

which you are subtracting as you determine the

numerator and denominator in the formula .

1 1,x y

2 2,x y

2 1

2 1

y yx x




Slopes of Horizontal Lines

Horizontal lines all have slopes of 0, and horizontal lines are the only lines with slope equal to 0. The equation of a horizontal line can be written in the form , where is a constant.y c c

y

x

y c




Slopes of Vertical Lines

Vertical lines all have undefined slopes, and vertical lines are the only lines for which the slope is undefined. The equation of a vertical line can be written in the form where is a constant. x c c

y

x

x c




Example 1: Finding Slope Using Two Points

Determine the slopes of the line passing through the following points.

8,1 and 2,33 12 8

m

210

m

15

m

2 1

2 1

y ymx x





Determine the slopes of the line passing through the following points.

5,4 and 8,4 4 48 5

m

03

m

0m

Note: The two points lie on a horizontal line.

2 1

2 1

y ymx x





Determine the slope of the line passing through the following pair of points.

2 5 4, and ,1

3 4 3

514

4 23 3

m

9 34 2

m

278

m




Finding the Slope of a Line

o We already know how to identify any number of ordered pairs that lie on a line, given the equation for the line. Identifying just two such ordered pairs allows us to calculate the slope of a line defined by an equation.

o In the next example, we will first find two points on the line. Then, we will use these points to determine the slope.




Example 4: Finding the Slope of a Line

Determine the slope of the line defined by the following equation.

2 4 16x y

2 4 0 16x

8x

-intercept: 8,0x

2 0 4 16y

4y

-intercept: 0,4y

Solution: First, find two points on the line.




Example 4: Finding the Slope of a Line (Cont.)

Next, use these points to determine the slope.

4 00 8

m

48

m

12

m

2 4 16x y

-intercept: 8,0x

-intercept: 0,4y





Determine the slope of the line defined by the following equation. 3 4 7x y

First point on the line: 1,1 7-intercept: 0,4

y 71

41 0

m

34

1m

34

m





Determine the slope of the line defined by the following equation. 5x

First point: 5,2 Second point: 5,8

60

Slope is undefined.

As soon as we realize that the line defined by the equation is vertical, we can state that the slope is undefined.

8 25 5

2 1

2 1

y ymx x




Slope-Intercept Form of a Line

If the equation of a non-vertical line in and is solved for , the result is an equation of the form

The constant is the slope of the line, and the line crosses the -axis at ; that is, the -intercept of the line is . If the variable does not appear in the equation, the slope is 0 and the equation is simply of the form .

xy

y

.y mx b

my b y

0,b x

y b




Slope-Intercept Form of a Line

y mx b y

x

intercept, y b 1 1,x y

2 2,x y

12x x

12y y

1

2 1

2m yx

yx

The constant is the slope of the line, and the line crosses the y-axis at ; that is, the y-intercept of the line is .

mb

0,b




Example 7: Graphing With Slope-Intercept Form

Use the slope-intercept form of the line to graph the equation .4 3 6x y

4 3 6x y

3 4 6y x

4 23

y x




Example 8: Graphing With Slope-Intercept Form

Find the equation of the line that passes through the point and has a slope of . Then graph. 0,3 3

5

3 35

y x

In Slope Intercept Form:

y mx b 35

m

3b




Point-Slope Form of a Line

Given an ordered pair and a real number an equation for the line passing through the point with slope is

Note that , , and are all constants, and that and are variables. Note also that since the line, by definition, has slope , vertical lines cannot be described in this form.

1 1,x y m,

1 1,x y m

m 1x 1y xy

m

1 1 .y y m x x




Example 9: Finding Slope-Intercept Form

Find the equation, in slope-intercept form, of the line that passes through the point with slope . 4, 1 2

4, 1 slope: 21x 1y m

1 1my xy x

1 2 4y x

1 2 8y x

2 9y x




Example 10: Finding Slope-Intercept Form

Find the equation, in slope-intercept form, of the line that passes through the two points and . 3,5 2,3

5 33 2

m

2m

5 2 3y x

5 2 9y x

2 4y x




Linear Regression

o In many practical applications, the goal is to take a given number of points and find the equation whose graph comes closest to fitting those points.

o For example, the next slide contains a plot of the growth of a new financial advising firm, with the vertical axis representing the number of clients acquired and the horizontal axis representing the number of weeks after the firm’s opening. The actual data points are listed in the accompanying table.

1 1 2 2 ,, , , ,..., n nx y x y x yy mx b




Linear Regression

0 5 10 15 20 2505

10152025303540

Weeks after opening

Num

ber o

f clie

ntsWeeks after

opening Number of clients

1 14

3 18

6 21

10 25

14 28

18 30

22 35

x y

The owner of the business would like to be able to make a projection about the number of customers in the near future, based on the assumption that the growth can be approximately modeled by a straight line.




Linear Regression

A crude method would be to “eyeball” a straight line- that is, to literally draw a straight line on the above graph coming as close as possible to the given data points.

0 5 10 15 20 2505

10152025303540

Weeks after opening

Num

ber o

f clie

nts




Linear Regression

o The “eyeball” method’s disadvantage is imprecision. o This flaw can be avoided by using linear regression, a

method that results in the slop-intercept form for the line whose graph minimizes the deviations between the line and the actual data points.




Linear Regression

Step 1: Calculate and . These are the averages (or means) of the -values and -values, respectively, of all the given data points. For our example,

x yx y

1 3 6 10 14 18 22 10.577

x

and

14 18 21 25 28 30 35 24.437

.y




Linear Regression (Cont.)

Step 2: Calculate and values. For each -value, there is a corresponding , which

represents the difference between the -value and . The same is true for .

xV yV xxVx x

1 -9.57

3 -7.57

6 -4.57

10 -0.57

14 3.43

18 7.43

22 11.43

x14 -10.43

18 -6.43

21 -3.43

25 0.57

28 3.57

30 5.57

35 10.57

x x x V y y y y VyV





Step 3: Calculate and indicates that corresponding and from the tables on the last slide should be multiplied together, and their resulting products added.

x y .x x x y 'sx 'sy

-9.57 -10.43 99.82-7.57 -6.43 48.68-4.57 -3.43 15.68-0.57 0.57 -0.323.43 3.57 12.257.43 5.57 41.39

11.43 10.57 120.82

x y x y

338.32x y

-9.57 -9.57 91.58-7.57 -7.57 57.30-4.57 -4.57 20.88-0.57 -0.57 0.323.43 3.43 11.767.43 7.43 55.20

11.43 11.43 130.64

x x x x

367.68x x





Step 4: Calculate slope and -intercept for the linear regression “best-fit” line. The slope for the linear regression line is

and the y-intercept is

So the equation for the linear regression line is

m by

338.32367.68

x ymx x

24.43 0.92 10.57 14.71.b y mx

0.92 71.14.y x




Linear Regression

This is the graph of the line we have just found and the original data points.

Documents

Hawkes Learning Systems: College Algebra