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Lecture Comments (7)

1 answer

Last reply by: sherman boey
Tue Aug 12, 2014 11:13 AM

Post by sherman boey on August 12, 2014

example 5 is wrong.. y =8 not -8

1 answer

Last reply by: Professor Eric Smith
Sun Jul 6, 2014 2:18 PM

Post by patrick guerin on June 24, 2014

Thank you for the lecture!

0 answers

Post by Farhat Muruwat on March 21, 2014

can someone please fix the answers to the questions on the practice questions.

0 answers

Post by Professor Eric Smith on December 10, 2013

The key you want to recognize for many of these problems is that to build a line using point-slope form we need to know a slope, and a point on the line.  In the problems we are usually given just enough information to find these things, but may have to use other formulas to do it.  Let me know if that helps out.  :^D

0 answers

Post by Araksya Fernandes on December 10, 2013

is there a little simple way to explain similar examples.

Linear Equations in Two Variables

  • To build the equation of a line you can use point-slope form ( y – y1 = m(x-x1) ). To use this you must know the slope of the line and one point on the line.
  • Parallel lines have the same slope. Perpendicular lines have slopes that are negative reciprocals of one another.
  • You can test if two lines are perpendicular by multiplying their slopes together. If you get -1 then you know they are perpendicular.
  • You may have to use other formulas such as the slope formula, before using the point-slope form.

Linear Equations in Two Variables

A line has a slope of - 4 and passes through the point ( - 4, - 3).
Find the equation of this line in point slope form.
  • y − y1 = m(x − x1)
  • y − ( − 3) = m[x − ( − 3)]
  • y + 3 = m(x + 3)
y + 3 = − 4(x + 3)
A line has a slope of - 7 and passes through the point (11, - 1).
Find the equation of this line in point slope form.
  • y − y1 = m(x − x1)
  • y − ( − 1) = − 7(x − 11)
y + 1 = − 7(x − 11)
A line has a slope of - 8 and passes through the point ( - 14,10).
Find the equation of this line in point slope form.
  • y − y1 = m(x − x1)
  • y − 10 = − 8[x − ( − 14)]
y − 10 = − 8(x + 14)
A horizontal line passes through the point (5,3). Find the equation of this line in point slope form.
  • slope = 0 for a horizontal line
    m = 0
  • y − y1 = m(x − x1)
y − 3 = 0(x − 5)
A horizontal line passes through the point ( - 12, - 13). Find the equation of this line in point slope form.
  • slope = 0 for a horizontal line
    m = 0
  • y − y1 = m(x − x1)
  • y − ( − 13) = 0[x − ( − 12)]
y + 13 = 0(x + 12)
The equation of a line is
y − 5 = [1/5](x + 3)
Find the equation of this line in slope intercept form.
  • y − 5 = [1/5](x + 3)
  • 5(y − 5) = [ [1/5](x + 3) ]5
  • 5y − 25 = 1(x + 3)
  • 5y − 25 = x + 3
  • 5y − 25 + 25 = x + 3 + 25
  • 5y = x + 28
  • y = [(x + 28)/5]
y = [1/5]x + [28/5]
The equation of a line is
y + 10 = [7/30](x − 2)
Find the equation of this line in slope intercept form.
  • y + 10 = [7/30](x − 2)
  • 30(y + 10) = [ [7/30](x − 2) ]30
  • 30y + 300 = 7(x − 2)
  • 30y + 300 = 7x − 14
  • 30y = 7x − 314
  • y = [(7x − 314)/30]
  • y = [7/30]x − [314/30]
y = [7/30]x − [157/15]
The equation of a line is y + 12 = [5/6](x + 9)
Find the equation of this line in slope intercept form.
  • y + 12 = [5/6](x + 9)
  • 6(y + 12) = [ [5/6](x + 9)]6
  • 6y + 72 = 5(x + 9)
  • 6y + 72 = 5x + 45
  • 6y = 5x − 27
y = [(5x − 27)/6]
A line passes through the points (7,9) and (2,5). Find the equation of this line in point slope form.
  • m = [(y2 − y1)/(x2 − x1)]
  • m = [(5 − 9)/(2 − 7)]
  • m = [( − 4)/( − 5)]
  • m = [4/5]
  • y − y1 = m(x − x1)
y − 9 = [4/5](x − 7)
A line passes through the points ( - 1,3) and ( - 7, - 11). Find the equation of this line in point slope form.
  • m = [(y2 − y1)/(x2 − x1)]
  • m = [( − 11 − 3)/( − 7 − ( − 1))]
  • m = [( − 14)/( − 6)]
  • m = [14/6] = [7/3]
  • y − y1 = m(x − x1)
  • y − 3 = [7/3][x − ( − 1)]
y − 3 = [7/3](x + 1)
A line is parallel to the line whose equation is
y = − [1/4]x − 3
This line also passes through the point ( - 5,7)
Find the equation of this line in slop intercept form.
  • y = mx + b
    parallel lines have the same slope
    m = − [1/4]b = ?
  • 7 = − [1/4]( − 5) + b
  • 7 = [5/4] + b
  • 7 = 1[1/4] + b
  • 5[3/4] = b
y = − [1/4]x + 5[3/4]
A line is parallel to the line whose equation is
y = − [3/7]x + 6
This line also passes through the point ( - 4,2)
Find the equation of this line in slop intercept form.
  • y = mx + b
    parallel lines have the same slope
    m = − [3/7]b = ?
  • 2 = − [3/7]( − 4) + b
  • 2 = [12/7] + b
  • 2 = 1[5/7] + b
  • [2/7] = b
y = - [3/7]x + [2/7]
Are the lines determined by the equations
3x − 6y = 18
4x − 2y = 12
parallel, perpendicular, or neither?
  • 3x − 6y = 18
  • − 6y = 18 − 3x
  • y = [(18 − 3x)/( − 6)]
  • y = [18/( − 6)] − [3x/( − 6)]
  • y = − 3 + [x/2]m = [1/2]
  • 4x − 2y = 12
  • − 2y = − 4x + 12
  • y = [( − 4x + 12)/( − 2)]
  • y = 2x − 6m = 2
  • ([1/2]) ≠ − 1
  • ([1/2]) ≠ (2)
neither
Are the lines determined by the equations
5x − 2y = 10
2x + 12y = 24
parallel, perpendicular, or neither?
  • 5x − 2y = 10
  • − 2y = − 5x + 10
  • y = [( − 5x)/( − 2)] + [10/( − 2)]
  • y = [5/2]x − 5m = [5/2]
  • 2x + 12y = 24
  • 12y = − 2x + 24
  • y = [( − 2x)/12] + [24/12]
  • y = − [x/6] + 2m = − [1/6]
  • (−[1/6])([5/2]) ≠ − 1
  • (−[1/6]) ≠ ([5/2])
neither
A line passes through the point ( - 4, - 3) and is perpendicular to the line whose equation is y = 1/5 x − 6. Find the equation of this line in slope intercept form.
  • y = mx + b
    m = [1/5]
    negative reciprocal = - 5
  • − 3 = − 5( − 4) + b
  • − 3 = 20 + b
  • − 23 = b
y = − 5x − 23
A line passes through the point (5,4) and is perpendicular to the line whose equation is y = − 7/10 x + 3. Find the equation of this line in slope intercept form.
  • y = mx + b
    m = − [7/10]
    negative reciprocal = [10/7]
  • 4 = [10/7](5) + b
  • 4 = [50/7] + b
  • 4 = 7[1/7] + b
  • - 3[1/7] = b
y = [10/7]x − 3[1/7]
A line is perpendicular to the line whose equation is 2x − 4y = 8. This line also passes through the y - intercept of the graph of x − 5y = 20. Find the equation of this line in slope intercept form.
  • 2x − 4y = 8
  • − 4y = − 2x + 8
  • y = [( − 2x + 8)/( − 4)]
  • y = [1/2]x − 2
  • m = [1/2]
    slope of perpendicular line = - 2
  • x − 5y = 20
  • − 5y = − x + 20
  • y = [( − x + 20)/( − 5)]
  • y = [1/5]x − 4b = − 4
y = − 2x + 4
A line is perpendicular to the line whose equation is 3x − 7y = 42. This line also passes through the y - intercept of the graph of 6x + 3y = 30. Find the equation of this line in slope intercept form.
  • 3x − 7y = 42
  • − 7y = − 3x + 42
  • y = [( − 3x + 42)/( − 7)]
  • y = [3/7]x − 6
    negative reciprocal = − [7/3]
  • 6x + 3y = 30
  • 3y = − 6x + 30
  • y = [( − 6x + 30)/3]
  • y = − 2x + 10
    b = 10
y = − [7/3]x + 10
A line is parallel to the line whose equation is
y = − [2/5]x − 6
This line also passes through the point ( - 10,12)
Find the equation of this line in slop intercept form.
  • y = mx + b
    parallel lines have the same slope
    m = − [2/5]b = ?
  • 12 = − [2/5]( − 10) + b
  • 12 = [20/5] + b
  • 12 = 4 + b
  • 16 = b
y = − [2/5]x + 16
A line passes through the point ( - 6, - 4) and is perpendicular to the line whose equation is y = − 2/3 x + 8. Find the equation of this line in slope intercept form.
  • m = − [2/3]
    negative reciprocal = [3/2]
  • − 4 = [3/2]( − 6) + b
  • − 4 = [( − 18)/2] + b
  • − 4 = − 9 + b
  • 5 = b
y = [3/2]x + 5

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Linear Equations in Two Variables

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Objectives 0:13
  • Linear Equations in Two Variables 0:36
    • Point-Slope Form
    • Substitute in the Point and the Slope
    • Parallel Lines: Two Lines with the Same Slope
    • Perpendicular Lines: Slopes are Negative Reciprocals of Each Other
    • Perpendicular Lines: Product of Slopes is -1
  • Example 1 6:02
  • Example 2 7:50
  • Example 3 10:49
  • Example 4 13:26
  • Example 5 15:30
  • Example 6 17:43

Transcription: Linear Equations in Two Variables

Welcome back to www.educator.com.0000

In this lesson we are going to continue on with our linear equations and look at how we can build these linear equations for ourselves.0002

Specifically, some of things that we will look at is we will look at the point slope form on a line.0014

It will help us when building those equations.0020

For some of our examples, we will also have to know a little bit more about parallel and perpendicular lines.0023

Watch out for my explanation on what these are and how you can tell if two lines are parallel or perpendicular.0029

Earlier, we looked at two forms of a line.0038

We looked at standard form and we looked at slope intercept form.0041

Remember both of these forms are important because they were good for graphing.0045

When we want to build our own line, sometimes we can do this as long as we have enough information.0057

To form that we use for building our line is known as point slope form.0064

The reason why it is called that is because those are the two bits of information that we need. 0068

We need to know at least one point on a line and we need to know what the slope of that line is.0072

Once we have both those bits of information and we can drop it into point slope form.0077

This form here requires a little bit of explanation.0086

You will see that there is an x and y that has some subscripts on it and that is where the point goes in terms of its x and y.0090

The slope is the n here so we can just put that in and then we also have a couple of other x’s and y’s which do not have subscripts.0098

With the ones that do not have any subscripts whatsoever, you will not be putting any values in for those.0108

Those simply stay in our equation. 0114

One last thing to note is even though we will use point slope to actually build the equation on a line.0116

Usually you take point slope and end up converting it into slope intercept form because you are looking to do some other things without a line,0122

other than just than build it or maybe you want to eventually graph it.0129

It is good to go ahead and move it in to one of those other forms.0132

To build lines using point slope form, we want to substitute in our point and we will go ahead and substitute in our slope.0143

I already have point slope form, I will put it over here and I have a point and I have a slope.0151

Let us go ahead and just substitute these in and see how it works.0157

I will have y - -3 then the m represent the slope, I will put in ¾, x – xy.0161

Notice how the x and y which did not have any subscripts are still in there.0176

Once we have everything substitute in here, we want to go ahead and start rewriting this0181

either into standard form or slope intercept form so we can do some other stuff with it.0186

I’m going to go ahead and put this into slope intercept form by just getting y all by itself and maybe cleaning up a few other terms.0190

I'm subtracting a negative on the left side and that will be the same as adding 3.0198

I will go ahead and distribute my ¾ here.0203

I have ¾ x and -3, looking much better.0206

I will go ahead and subtract 3 from both sides.0216

Y = 3/4x – 3 – 3 is 6.0219

Our point slope gets our foot in the door so we can actually build the equation and then put it into a different form, maybe a slope intercept down here.0227

We can go ahead and graph it and do some other things with it.0241

Sometimes we will be given information about another line in order to build the one we want.0247

And information about that other line, we may know that it is parallel or perpendicular to the one we want.0253

What exactly does that mean? 0258

We can say that two lines are parallel if they had exactly the same slope.0261

In my little picture here, you can see what that does.0266

You will end up with two lines and they usually have a little bit of a gap or space in between them, but they have exactly the same slope.0269

They are going in the same direction.0275

You might also have lines that are perpendicular.0279

With this one, they end up meeting at a right angle.0282

We do not talk much about angles in this course, another way that you can say two lines or perpendicular is 0286

if their slopes are negative reciprocals of one another.0291

Let me show you what that means.0295

I suppose this blue line had a slope of 4/5, if the red one was perpendicular then I wanted to be the negative reciprocal of the blue line.0296

I made it negative and I flipped it over.0312

With perpendicular lines, they will always be different in sign, one will be positive and one will be negative and they will be reciprocal when flipped over.0314

Another way that you can test if two lines are perpendicular or not is actually you take both of their slopes and just multiply them together.0325

Two things could happen if you multiply them together and their slope is negative or their combined value is -1 0333

then you will know for sure that they are perpendicular.0342

If you multiply them together and you get anything else other than -1 then you will know that they are not perpendicular.0346

It is a nice and easy task you can use to know how the two lines are related.0352

We will definitely know this as we get into more of those examples.0358

Let us start off with example 1, in this one we want to write the equation on a line using the given point and the slope.0363

Since the two bits of information that I need for point slope, I will just go ahead and nearly drop it into the formula.0372

y -y1 equals slope x - x1.0378

Let us put in the y value first, y - 3 equals my slope is -2/3x - -6.0386

You will know the formula has a minus sign in there since the x value is a negative, go ahead and put that negative sign in there as well.0399

Now all we got to do is clean it up a little bit and maybe turned it into a different form.0408

I will see what we can do.0412

y – 3 = -2/3, when I subtract the negative that is the same as addition.0413

x + 6 and I think I will go ahead and distribute this -2/3 in there.0422

-2/3x – 12 ÷ 3 and a -12 ÷ 3 = -4, it looks pretty good. 0430

Let us add 3 to both sides and now we have that same line written into slope intercept form.0449

Again we use point slope form to go ahead and create the equation of the line and then probably put it into some other form.0462

This one is a little different, in this one we simply want to determine whether the two lines are parallel0472

or maybe the perpendicular or maybe they do not fall into either of those two cases.0477

With these ones, notice how many of them are written not in one of the forms that we have covered. 0482

In other words they are not in slope intercept form but the second one is in standard form but we have to be able to figure out what their slope is.0489

A way to determine what the slope of the line is to go ahead and rewrite it into our slope intercept form and we can just read it out of the equation.0501

Let us do that first before we actually compare what these slopes are.0509

Starting with this first one here, I can move the 2 to the other side by subtracting 2 so y =5x – 2.0512

Let us see what the second one, let us go ahead and move the x to the other side, -x – 15 and then we will divide it by 5.0521

Here is our first line, here is our second one.0539

Now they are both in slope intercept form we can say that the first one has a slope of 5 0545

since it is right next to x and the other one has a slope of -1/5.0550

Looking at the two slopes, they are definitely not the same, they are not parallel.0555

It looks like one is positive and one is negative and they are reciprocals.0562

I think these two are perpendicular.0567

If you want to test out feel free to just take the two slopes and multiply them together and see that when you do this you get a -1.0570

You will know that these two lines are perpendicular.0579

Let us try this process with the next pair of lines.0590

I will begin by just putting this into slope intercept form.0595

This one is almost in slope intercept form.0599

I just have to reverse the y and putting it first.0602

This one I will go ahead and move the 3x to the other side, so now I have my two lines right here.0606

y = 2x + 1 and y = -3x + 4, in this form, I can read off what both of their slopes are.0616

Looking at their slopes I can see that they are not the same, they are definitely not parallel.0627

One is positive and one is negative, but they are not reciprocal so they are not perpendicular either.0633

They are not parallel or not perpendicular, I will put this in the neither categories.0638

They are just two lines hanging around on the graph.0643

Let us try that same process with another pair of lines.0650

Let us see if we can find a couple of that which actually are parallel.0652

We will begin by putting these into a better form so we can read off that slope.0656

I'm moving the 4x to the other side, this one the y is still not completely all by itself.0662

Let us go ahead and divide everything through by 2, now we have one of our lines.0668

The second one let us go ahead and rearrange things.0678

The y is on the left and let us get it completely isolated by multiplying everything through by -1.0683

Now we have both of our lines.0693

Now that they are in this form, the slope of the first one is -2 and the slope of the second one is -2.0702

I can see that they are exactly the same and we will call these pair of lines parallel.0709

Onto one more pair of lines, let us see what they are, parallel, perpendicular, or neither, we will find out.0719

I need to get my y isolated I will start off by moving the 4x to the other side.0726

I will divide both sides by 3, y =4/3x +2.0734

The second one, let us go ahead and move the 2 over.0745

I have 3x + 2 and I will divide everything by the 4, ¾ x + ½.0749

Here is equation 1 and here is equation 2.0760

If you look at them in this form, we can pick out what each of their slopes are, I have 4/3 and 3/4.0767

These ones are pretty close, they are definitely reciprocals of one another since you would take 4/3 and flip it over and get ¾.0775

When those have been both positive so we can not say that these are perpendicular.0783

Even though they are reciprocals they are not negative reciprocals of one another.0788

They are definitely so they are not parallel.0792

Even though they are close, they do end up in the neither category.0795

They are not parallel or not perpendicular, they are simply neither.0799

They are just two lines.0802

With this one we want to write the equation of a line that happens to be parallel to the given line and also goes through the given point.0809

This one is a little different, notice how we need to know what the slope of the line is0818

and we need to know what the point is but they have not quite given us what the slope is.0822

We are going to have to extract that out of this other line that they have given us by knowing that it is parallel to the one we want to build.0827

Let us hunt down what the slope of this other line is.0834

We will do that by dividing everything through by 4.0837

It looks good so y = ¼ x + 5.0845

Let us see, I know that the slope is ¼.0850

Now that I have a point, I have a slope, I can use point slope form and build our line.0855

Y - -3 = ¼ x – xy and now let us clean it up and see what line we have.0861

y + 3 = ¼ x, ¼ of 2 would be -1/2.0872

Now we will subtract 3 from both sides.0890

y = ¼ x – ½ - 3, get a common denominator over there.0893

Our line is y = ¼ x – 7/2.0908

The way we built this line, I know for sure that it goes through the given point 2, -3.0917

I already explored what the slope of the other line was so I know it has a slope of 1/4.0923

Let us try this one, write the equation of a line which is perpendicular to the given line and it goes through the given point.0932

Similar to the other one, but it have not given us the slope directly.0939

It just told us it is perpendicular to this other line over here.0943

Let us figure out what slope is, y equals, I will move the 8 to the other side, -8x + 3.0948

The slope of this line is -8 which is good but that is not exactly the slope we want to use.0958

Our line is perpendicular to this line, so we need to use the slope -1, 1/8. 0966

That way it is the negative reciprocal of the other.0979

Now we will drop it into our formula.0984

Let us go ahead and put in our y value. 0994

We have our slope and our x value.0997

We will clean it up and see what line we have.1002

y - 8 = 1/8 - 4/8, my x in there.1005

y – 8, 1/8x – ½ and we will go ahead and 8 to both sides, find a common denominator and combine the last of our like terms.1017

1/8x + 15/2.1044

We can see that its slope is the negative reciprocal of the other line.1053

I know it is perpendicular and the way we build that it, for sure it goes through the point 4/8.1057

This last one is a little bit more of the word problem but you can see how we can still use these techniques in order to build the equation of the line.1066

We want to build a line that represents the following problem.1074

It cost a $20 flat fee to rent a drill + $2 every day starting with the first day.1077

Let x represent the number of days that we rent this drill and y represent the charge to the users.1085

How much will we end up willing to get from them?1091

When we are all done, let us see if we can write this line in a slope intercept form.1095

Using the techniques that we know about so far, we need to figure out what our slope is and what point this will go through.1100

Let us see, one thing that we can think of is the slope being what changes or your variable costs.1108

The $20 that you have to pay up front is not a variable.1120

It is going to be there no matter what, what does change is a $2 every single day.1123

The variable cost, we can think of that as our slope.1134

Let us see what we can do with that.1144

What happens with that flat cost?1146

We will be charged of that no matter what, that is like our y intercept.1151

If this one, we can just drop in our variable cost and we can drop in our flat fee.1158

What we are left here is the equation of the line but it represents how much we end up charging the person1177

and we can see this by substituting some values.1183

Suppose they rent this drill for one day, it cost $2 for that day + $20 flat fee, it cost them $22.1186

If they rent it for two days, we can put that in for x, $4 for the variable + $20 flat fee, $24.1196

You can go on and on figuring out how much it cost for each of the different days.1206

But in the end, this formula right here would represent how much we need to charge them.1211

Just simply plug in the number of days for x and y would give you the cost.1215

You can see that building a line is not so bad and using point slope form is handy in that entire process.1221

If you do use point slope form, you need to know the slope of the line and you need to know a point on that line.1228

Thanks for watching www.educator.com.1234