part 1 set 2: straight lines we saw earlier that ax + cy = d was the equation for a straight line....

Part 1 Set 2: Straight Lines

We saw earlier that ax + cy = d was the equation for a straight line. What effect do the constants a, c, and d have on the line?

Straight Lines: standard form

ax + cy = dFirst of all, we can reduce the three constants, a, c,

and d, down to two by dividing every term by c: (a/c)x + y = d/c .

It is more convenient (as we’ll see) if we express the line as: y = (-a/c)x + (d/c), and replace the constant (-a/c) with m and the constant (d/c) with b to get:

y = mx + b .


y = mx + b

To see what the constant, b, tells us, consider the point where x = 0. In this case, y = b. Thus b is the value of y when x = 0. This is called the y intercept.


y = mx + b

To see what the constant, m, tells us, consider how y changes when x changes:

(y2 – y1) = ([mx2+b] – [mx1+b]) = m(x2 – x1)

or, m = (y2 – y1) / (x2 – x1) = Δy/Δx.

We call m the slope of the line.


What is the difference between the four colored straight lines below?



They all have the same slope, but their y intercepts are different!

The green line has thebiggest positive b,and the red line hasthe most negative b.



They all have the same y intercepts, but their slopes are different!

The blue line has thebiggest positive m,and the green line hasthe most negative m.

Straight Lines: parallel lines

How can we recognize parallel lines?

Straight Lines: parallel lines

How can we recognize parallel lines?

Parallel lines have the same slopebut different y intercepts.

Example: these two lines are parallel:

y = 2x + 6

y = 2x - 2

Straight Lines: perpendicular lines

1. How can we recognize perpendicular lines?

2. If the blue line is perpendicularto the green line, is it alsoperpendicular to the red linethat is parallel to the greenline?


The answer to the second question is YES. If the blue line is perpendicular to the green line, it is also perpendicular to the red line.

This means that the y interceptis not important when it comesto lines being perpendicular.

Straight Lines: perpendicular linesTo answer the first question, we note that the

green (and red) lines have a positive slope, while the perpendicularblue line has a negativeslope. It should be easyto see that perpendicular lineswill always have slopesthat have opposite signs.

Straight Lines: perpendicular linesBut not all lines with negative slopes will be

perpendicular to a line with a positive slope. The purple line added belowhas a negative slope, but it isnot perpendicular to thegreen and red lines.


We can see that as the slope of one line becomes steeper (such as the red linecompared to the green line)the slope of a line perpendicularto it must become flatter(such as the purple linecompared to theblue line).


It turns out that the condition for two lines to be perpendicular is that the product of the slopes must equal –1.

The slope of the red line is +2,while the slope of the perpendicular purpleline is –1/2.

The slope of the greenline is +1/3, whilethe slope of theperpendicular blueline is –3.

Straight Lines: scaling effectsScaling on the axes can affect the appearance of

straight lines. Below the perpendicular lines with equations y = (1/5)x + 1 and y = -5x +1 are plotted on two graphs with different scales on the x axes.

2 4-2 10 20-10

2 2

-2 -2

Straight Lines: scaling effects

Try this yourself on your calculator.

Choose two lines that are perpendicular.

Graph them.

Then change the scale for x and re-graph them.

Then change the scale for y and re-graph them.

Straight Lines: info needed

If we know the slope (m) and the y-intercept (b), we can write the equation for the line:

y = mx + b .

Are there other pieces of information that we can use to determine the equation for the line? What information would be sufficient to determine the equation?


If we are given two points, such as A at (-2,1) and B at (4,2), can we determine the equation of the straight line that connects them? In other words, how can we determine the slope,m, and the y intercept, b, so wecan determine theequation for this linein the form: y = mx + b?

A

B


A at (-2,1) and B at (4,2).From the definition of slope as Δ(y)/Δ(x), we have: m =

(2 – 1) / (4 – (-2)) = 1/6.Now if we use either point for x and y

in the basic equation y = mx+b, we can determine b.Using point A: y = mx + b becomes1 = (1/6)*-2) + b, so b = 1 + 1/3 = 4/3.Therefore we have: y = (1/6)x + 4/3

A

B


As you can see from the previous case, if we know one point and the slope, we can also determine the equation for the straight line in standard form: y = mx+b .

Review: what we need to know to determine the equation for a straight line is:1. slope and y intercept; or2. any two points; or3. slope and any one point.

Straight lines: symmetry

Do straight lines have any kind of symmetry? (through the origin, reflection through the vertical or reflection through the horizontal axes)?

Straight lines – symmetry

There is a sameness since straight lines are straight – the slope of the line doesn’t change.

However, there are no reflection symmetries and, unless b=0 there is no symmetry through the origin.

If b=0, we have the special case of our linear relation being a proportional relation and that does have symmetry through the origin.

Inverses

Whenever we do an operation or express a relation, is there an operation or a relation that will “undo” the initial operation or relation to get us back to where we started?

How about the operation of addition? Does it have an “inverse operation”?

How about the operation of multiplication? Does it have an “inverse operation”?

How about the operation of squaring? Does it have an “inverse operation”?

InversesFor addition: y = x + c. To get x back, we use

subtraction: x = y – c. Example: let c = 5. If x = 3, then y = 3 + 5 = 8. To get x back, x = y – c = 8 – 5 = 3.

For multiplication: y = c*x. To get x back, we use division: x = y/c.Example: let c = 5. If x = 3, then y = 3 * 5 = 15. To get x back, x = y/c = 15/5 = 3.

For squaring: y = x2. To get x back,we use the square root operation: x = √y.Example: If x=3, then y = 32 = 9.To get x back, x = ±√9 = ±3.

Inverses

Do inverses exist for relations, like our linear one: y = mx + b ?

If so, what would it look like?

Inversesy = mx + b

Example: suppose m = 4 and b = -8.If x = 3, then y = (4)*(3) + -8 = 4.How do we take y=4 and get back to x=3?

Let’s solve the above equation for x:x = (1/m)y – (b/m).

If we now substitute m = 4, b = -8, and y = 4, do we get back to x = 3?x = (1/4)(4) – (-8/4) = 1 + 2 = 3.

Inverses

Regular linear relation: y = mx + b

Inverse relation: x = (1/m)y – (b/m)

In general, note that y = mx + b = m[(1/m)y – (b/m)] + b = y.

InversesIf we look at the graphs of y vs. x, and its

inverse relation of x vs. y, we see (for our case of m=4 and b=-8): y = 4x + -8 x = (1/4)y – (-8/4)

y

x

x

y8 16-8-16

8

-8

8

-8

-16 -8 8 16

InversesTo get the x vs. y graph, you take the y vs. x

graph, rotate it 90 degrees (to bring the vertical y

to become horizontal), and then flip it (about the horizontal axis to bring the +x from the bottom to the top)

y = 4x + -8 x = (1/4)y – (-8/4)y

x

x

y8 16-8-16

8

-8

8

-8

-16 -8 8 16

InversesAnother way is to recognize that the point (x,y) on

the first graph is plotted (y,x) on the second graph. If we draw a line y=x, the inverse will graph as the reflection through this line

y = 4x + -8 x = (1/4)y – (-8/4)y

x

x

y8 16-8-16

8

-8

8

-8

-16 -8 8 16

Computer Homework

You should be able to do the third computer homework assignment on Linear Equations, Vol. 0, #2.

Straight Lines: from data

Sometimes we have a bunch of data points, and if we plot the data points, we see that the points seem to fit (more or less) a straight line.


We could try to draw a straight line through the data points and then use the two points method (with both points as close to our line as possible, or better yet, choose two points on the line) for determining that line. But is there a way to determine how well the data fit the line, and a way to get the line with the “best” fit?

Note: not all data will fit a linear model all that well. We’ll develop other possible models as we go through the course.

Straight Lines: from dataOne way to determine how well the line fits the data

is to measure how far (in the y direction) each data point is removed from our line. This “distance” for each point is called the residual. For points above the line the residual will be positive, and for those below the line the residual will be negative. It would seem a good idea to adjust our line so that the sum of the residuals will be zero.


However, there may be lots of lines we can draw that will have the sum of the residuals be zero. How can we choose among these lines?


One of the “best fit” lines is called the least squares regression line. In this method we minimize the sum of the squares of the residuals. Your calculator and some spreadsheets can perform this process which is called linear regression. A measure of how well this “best” line fits the data is the correlation coefficient, r. The value of r will always be between –1 and 1, and a perfect fit will have |r| = 1, so the closer r is to zero the worse the fit.

Note: This method assumes each data point is equally as good as any other. In MATH 201 and MATH 309 we discuss this topic in much greater detail.

Straight Lines: from dataInstructions for calculator: (type from Home screen)

{1,2,3,4,5}→L1 (enter) (where 1,2,3,4,5 are the x values; → is found near top of catalog menu, use alpha key to type letter L)

{2,4,6,8,10.3 }→L2 (enter) (where 2,4,6,8,10.3 are the y values)

LinReg L1,L2 (enter) (use alpha key to type letters)

ShowStat (enter) (use alpha key to type letters)

Regeq(x) →y1(x) NewPlot 1,1,L1,L2 (enter) (use alpha key to type letters)

(press the diamond then graph key)

Regular Homework Set #3(continued on next page)

1. Plot the data on the right (I suggest x y that you use an excel spreadsheet 1 20 to create and print out the graph, 2 32 and use your eye and a ruler to 3 39

draw the best line and then 4 49 determine the equation of your line. 5 61

2. Use your calculator to perform the 6 68 the linear regression, and compare its 7 77 equation with yours by calculating each of its residuals, each of the squares of its residuals, the sum of its residuals, and the sum of the squares of its residuals.

Regular Collected Homework(continued from previous page)

3. Put the equation 6x + 4y = 11 in the standard linear form: y = mx + b

4. What is the equation of the line (in standard form) that is perpendicular to the line of problem 3?

5. What is the equation (in standard form) for the line that goes through the points (-2,7) and (8,-3)?

Straight Lines: systemsIf we have a linear equation, for each value of x

there is a value of y. Example:y = 5x – 7. The following points satisfy this equation: (0, -7); (2, 3); (10, 43); as well as an infinite number of other points. Any point “on the line” satisfies this equation.

Suppose, though, that we need to find a point or points that satisfy two linear equations at the same time. Can we do this? How many points can we expect to find that do satisfy both conditions (equations) at the same time (simultaneously)?

Straight Lines: systemsWe can “see” this question if we use graphs.

Let’s start out with an example. Is there a point or points that will satisfy bothy = 3x – 7 and y = -2x +4 ?

As we can see, there is onepoint of intersection whichis on both lines, and so is asolution of both equations.

Straight Lines: systemsy = 3x – 7 and y = -2x +4

You can use your graphing calculator to locate (reasonably closely) the point of intersection. Do this now.

By looking at the graph,can you see any casesin which there is nosolution, or any caseswhere there would bemore than one solution?

Straight Lines: systemsIf the lines are parallel, we know that parallel

lines never meet, so there would be zero points of intersection and hence zero solutions. The condition forparallel lines is that theslopes of the two linesare the same, but theyhave different y intercepts.

Straight Lines: systems

However, if the two lines are really the same line (same slope, same y intercept), then we don’t really have two separate equations, we only have one. So in this case, there would be an infinite number of solutions (just like one equation).


As long as we have two equations that are not parallel and not just different forms of the same equation, we can expect that there will be one answer. We can graphically get an approximate answer. Can we calculate the answer? We’ll now look at two methods.


Simultaneous EquationsMethod of Substitution

In this method, we use one equation to express one of the unknowns, say y, in terms of the other unknown, x. We then substitute this expression for y in the other equation. On the next slide we show an example.


Example: 3x + 4y = 14 and 2x – 7y = -5First we solve one equation for one of the

variables. Let’s choose the first equation and solve for y: y = (-3/4)x + 14/4 .

Now we substitute this expression for y into the second equation: 2x – 7((-3/4)x + 14/4)= -5

Multiplying it out gives: 2x + (21/4)x – 98/4 = -5Combining terms gives: (29/4)x = 78/4 , or x = 78/29 ≈ 2.690. We now substitute this into the y

equation above to get y = (-3/4)(78/29) + 14/4 ≈ 1.483 . Our solution: x ≈ 2.690, y ≈ 1.483 .

Check: 3(2.690) + 4(1.483) ≈ 14 and 2(2.690) – 7(1.483) ≈ -5 .


Simultaneous Equations

Method of EliminationIn this method, we multiply through each equation

by a factor that will make the coefficient of one of the variables to be 1. We then subtract the two equations which results in one of the variables being eliminated.

We demonstrate this method next through an example.


Example: 3x + 4y = 14 and 2x – 7y = -5First we will eliminate x from each equation, so we

multiply the first equation by 1/3 and the second equation by ½:

x + (4/3)y = 14/3 and x – (7/2)y = -5/2 Now we subtract the second equation from the first:

(4/3)y – (-7/2)y = (14/3) – (-5/2) .Combining terms: (29/6)y = 43/6, or

y = 43/29 ≈ 1.483.Now we substitute this value of y in either equation

and solve for x: x – (7/2)y = -5/2 , or x ≈ -(5/2) + (7/2)(1.483) = 2.690.


If we have linear equations, either method will work not only for two equations in two unknowns, but will work for n equations in n unknowns. You simply repeat the process to reduce the unknowns down until you get to the point where you have one unknown and can solve one equation for that one unknown.

There are geometric concepts we can employ, but we’ll leave those for a future course.

Composite function

Sometimes we have one quantity that depends on another, and that second quantity depends on a third. So the first quantity actually does depend on the third. We can express that with something called a composite function. There is a notation for this: (f ◦ g)(x) = f(g(x)).

Composite function

An example would be that pressure of a gas in a container depends on temperature: P(T) = nRT/V (ideal gas law). Now let’s suppose that the temperature depends on the time of day according to the following formula: T(t) = To + bt. For this case:

(f ◦ g)(x) = f(g(x)) becomes (P ◦ T)(t) = P(T(t)) = nR(To + bt)/V.

Computer Homework

You should be able to do the fourth Computer Homework assignment on Simultaneous Equations, Vol. 0, #3.

Regular Homework Set #4(continued on the next page)

1. Choose two lines that are perpendicular. Graph them on your calculator, and make a drawing of the graph from the calculator. Estimate the coordinates of the point where they intersect.

2. Change the scale for x and re-graph the two equations from problem 1, and make a drawing of the graph from the calculator.

3. Choose one line. Determine its inverse. Graph both the original line and its inverse relation on your calculator and make a drawing of the graphs.

4. Solve for the values of x and y that satisfy both of these equations: 3x – 7y = 13, and y = 3x + 10.

Regular Homework Set #4(continued from previous page)

5. Graph the two lines from problem 4 on your calculator, and make a drawing of the graph from your calculator.

6. Solve the following three equations for the three unknowns, x, y, and z: x + 2y + 3z = 4; -2x + 3y + 4z = -7; 7x – 3y – 2z = -13.

WebCT Test #1

You should be able to do the first WebCT test now.

part 1 set 2: straight lines we saw earlier that ax + cy = d was the equation for a straight line....

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