Hi. My name is Rebecca Muller. During this session, we're going to look at straight lines. Now, straight lines are straight because they have a constant rate of change as you move from one point to another on that line. Even with vertical lines, we look at the same thing happening. That is, we're going up all the time from one point to another. Now let's look at some of the subtopics we'll consider. We're going to look at equations of lines in two variables. And we're going to consider something called the "general form" to start with. Then we'll look at slopes of lines. We'll consider parallel and perpendicular lines and then other forms of the equation of a line, slope-intercept form, point-slope form. And then we'll consider linear functions. How does this all play together? So let's begin with a definition of a linear equation. An equation in two variables is linear if it can be written in the following format-- Ax plus By equals C, where A, B, and C are real numbers. So let's look at our first example of that. So here's an equation that's going to be written in that standard general form. So we're going to have 2x plus 3y equals 12. Now, what I can do with this is think about what it would look like if I graphed it. And one way to graph this is to use a plotting of points method. I'm going to draw what can be called a "T chart" in that it looks like a T to start with. And I'm going to put x on the left and y on the right. And I'm going to substitute in different values for x or y to solve for the other variable. Now, when were plotting a straight line, if it's possible to find the x- and y-intercepts-- that would be the points where the line is going across the x- and y-axis-- those are going to be important to find in most instances. And also, they're the easiest ones to plot. If you're looking on an x-y plane, and you're interested in where it's going to hit the y-axis, it means that it hasn't moved to the right or the left. So therefore, my x value would be 0. Substitution of 0 into this equation will give me the y-intercept for this graph. So we're going to substitute in. We'll solve 2x, which is 2 times 0 in this case, plus 3y equals 12. 2 times 0 is 0. 3y equals 12 whenever we end up with y equal to 4. So our y-intercept will be the point 0 comma 4. Likewise, if we're looking for an x-intercept, which means we did not move up or down, we were staying along the x-axis, then our y value must be 0. And so substitution into the equation is going to give us 2 times x plus 3 times 0 equals 12. And so we end up with 2x equals 12 and x is equal to 6. So we'll have the point 6 comma 0. I'm going to do a really quick sketch of this graph. So here's an our axes, x and y. I'm going to need to plot 0 comma 4. So 1, 2, 3, 4 units up is going to be a point on that line. And then our x-intercept is at 6, 0-- so 1, 2, 3, 4, 5, 6. I have another point on the line. And then I am going to draw a straight line here. Now, I'm going to do it freehand. If you're at home, you might want to use a straightedge of some sort to make sure that you kind of get a straight line. Let's see how I'm able to do-- and not too bad. So in relation to this line, there's a concept that we call "slope." It has to do with the rate of change. How much is it sloping as I look at this line moving from left to right is really what I'm interested in. And so what I notice is that I can actually count that from this point that's on the y-axis to the point that's on the x-axis, I need to move down 4 units and then move across 6 units. So this is down 4 and then right 6. So now let's look at how we're going to designate what we've just found. We're going to designate the slope of a line using the letter m. And m is going to mean the change in y divided by the change in x. Well, since we're looking at a change in y, that's our down 4 here. Our change in x is our right 6. We can use some notational devices that you may see an instructor using. So I'm just going to show them to you now. I could write that same information by saying m equals delta y divided by delta x. The delta indicates "change in." Now, if I'm changing a y value, it means that I'm looking at a difference in y-coordinates. So my numerator becomes a difference in the y-coordinates, which can be stated as y sub 2 minus y sub 1. The change in your x-coordinates is going to be a difference in the x's. And I've written down x sub 2 minus x sub 1 to indicate that. One thing to note is that I always want to go from the same point to the same point so that a point would have coordinates x sub 1 comma y sub 1. And another point would have coordinates x sub 2 comma y sub 2. So that in essence, what you're looking at is taking the y-coordinate from point 2-- if you want to think of it like that-- minus the y-coordinate from point and then dividing that with the x-coordinate from point 2 minus the x-coordinate from point 1. We can also think of this as rise over run. And this is really the one that I prefer to use in general because it's going to encompass everything else that we see on that list. So back to the problem that we have at hand, we can see that we have this point, which we decided was 0 comma 4. We have a second point, which was 6 comma 0. And we know that we've gone down 4 and then to the right 6. So let's see how that plays out in the whole scheme of finding the slope. So using the slope, we can say that that's going to be the change in your y value-- and I can do it this way-- over the change in your x value. And that's also the same thing as your rise over your run. So I'm just going to approach these in two different ways. The first part is really formula-driven. And what I mean by that is that I've just got to decide where I'm going to start with y sub 2. So let's just say I choose my y sub 2 to be the value 4. Then I'm going to have 4 minus my y sub 1, which would be the y-coordinate the other point, which is going to be 0, divided by-- and what's key here is that I go back to the same point to start with my x's. So I'll take the x-coordinate of the point 0, 4 to be and then minus the x-coordinate of my second point, which here is 6. And of course, I can come up with my value of 4 divided by negative 6, which is going to give me the negative 2/3. If I think of this as rise over run, I'm just going to really think of it in a visual manner. In other words, instead of substituting in my values for my coordinates, I'm just going to say, hey, from here, what did I do? I went down 4 units. So I'm going to think of this is a rise of negative 4. And then once I got down here, which happens to be at the origin, I moved over to the right 6 units. And that's in a positive direction. So that's going to be my denominator of 6. So I ended up, of course, at the same point where I say my slope is negative 2/3. But I'm thinking about it a little bit differently depending on which format I'm using. So now let's look at a typical problem you may be asked to solve. Find the slope of the line that passes through the points 5 comma negative 2 and 9 comma 18. So since I'm trying to find the slope of a line, I'm going to look at the slope formula. And again, my preferred version is to just think of it as rise over run because that tells me the same thing as the other ones do. The thing about the one with the subscripts is that you shouldn't get too caught up with what point you want to call x sub 1 y sub and what point you want to call x sub 2 y sub 2. Just realize those subscripts are there just to designate that it's two different points. It also indicates that we want to make sure we always come back to the same point. Now, a good rule of thumb is just to do it in the order in which you see the points given. So when I'm looking at my rise, which is the difference in my y-coordinates, I'm just going to start with the first point and write down negative 2 and subtract from that the y-coordinate of the second point, which is given as 18. In the denominator, the main thing I need to make sure I'm doing is coming back to the first point that I started with. So if negative 2 is the y-coordinate of the point I started with, I need to make sure I pick up the x-coordinate of that same point to begin. And here, I'm going to end up in my denominator with 5 minus 9. So now let's just simplify. We end up with m equals negative 20 divided by negative 4. And that gives us a value for m which is equal to 5. Now, what is that really telling us? Well, remember that 5 is equivalent to 5 divided by 1. So what it's saying is that for a change of 5 units in the vertical direction-- that's our rise-- that I'll end up with a change of 1 unit in the horizontal direction. And that will be true for all values across the board, which means that ratio is the same. So 5 to 1 is the same as 10 to 2 and so on and so forth. It's time for a quick quiz. In reference to the following graph, which one of the following inequalities would hold, where the notation m parentheses l sub 1 represents the slope of line 1. Well first of all we're given an x, y plane with 4 different lines graphed on it. l sub 1 looks like a horizontal line which is in quadrants 1 and 2. l sub 2, is a straight line that is increasing from quadrant 3 through the origin into quadrant one is doing it at a fairly rapid pace. l sub 3 is a straight line that starts in quadrant 2 and then decreases and then ends up through into quadrant 4. l sub 4 is a straight line that is increasing from quadrant 3 through quadrant 4, but at a fairly non rapid pace. Is the relationship between the slopes A. The slope of line 3 is less than the slope of line 1, is less than the slope of line 4, is less than the slope of line 2. Or B. The slope of line 1, is less than the slope of line 3, is less than the slope of line 2, is less than the slope of line 4. Or C. The slope of line 1, is less than the slope of line 4, which is less than the slope of line 3, which is less than the slope of line 2. So consider the 4 lines and then figure out which you should choose from A, B or C as to the relationship between the slopes. You're correct, the answer is A. The slope of line 3 is less than the slope of line 1, which is less than the slope of line 4, which is less than the slope of line 2. Sorry the correct answer is A. The slope of line 3 is less than the slope of line 1, which is less than the slope of line 4, which is less than the slope of line 2. Let's look at each line individually and think about the slopes. Line 1 is actually a horizontal a line, so the slope of this line is going to equals 0. Line 2 is the line which begins in quadric 3 goes to the origin and ends up in quadrant 1, but it's doing (inaudible) rapid pace. So this slope is definitely going to be positive. And I'll come back to that one in a minute. Line 3 is a line that is decreasing. So because it's a decreasing line we know that its slope, is going to be negative less than 0. The slope of line 4, well line 4 is a line that begins in quadrant 3 and enters into quadrant 4, but it is doing so increasing. As you move left or right but not at a very rapid pace. So it's slope is definitely going to be positive but when we compare this slope of line 4, to the slope of line 2 which was increasing at a rapid pace. What we can say is that we know that the slope of line 2, is going to be greater than the slope, greater than the slope of line 4. So let's see how we can put all of this together. If we are going to rank them. The answer was A, the slope of line 3 was going to be the smallest. Now that makes sense, because we know the slope of line 3 was a negative number. So we have a negative number, represented by the slope of line 3. This is going to be smaller than all the other 3 slopes. The slope of line 1 is going to be equal to 0, so the negative number certainly smaller than 0. And 0 is certainly smaller than two positive slopes, which we concurred were going to be the slopes of lines of, line 2 and line 4. And we just determined that the slope of line 2 had to be greater than the slope of line and so that means that the slope of line 4 is less than the slope of line 2. So we have then even though they're both positive this is going to be a positive number which is smaller than another positive number and that relationship is underlined below. So A is going to be the correct solution. When we look at reading a graph, we want to do it from left to right just like we read words in a book. So in the picture that we have to my left, I have depicted a line that is increasing. Notice that as you move from left, right, the line goes up. Below that we have a line that is decreasing as you move from left to right, the line goes down. To the upper right we have depicted a horizontal line moving from left to right it stays horizontal. Think about the word horizon, like looking out over the horizon to keep track of which one's horizontal because the one below it is the vertical line. It's the one that goes straight up and down. Now we want to determine what this has to do with the slope of a line. So going back to the first graph that I have here, I have a line that is increasing. If I pick two points on this line and I think about rise over run because that's what slope is, we're going to, from the point on the left, we have to go up a certain man amount and then go to the right. So my rise is positive and my run is positive. When I have a line that is increasing, the slope is going to be greater than zero. What about align that is decreasing? I can again do the same procedure. If I pick a point on the line and then move to another point on the line and I move from left to right, then from this point I have to go down and then I have to go to the right. So my rise this time is going to be negative since I'm moving down, but my run is going to still be positive since I'm increasing the x value as I move in that direction. In this case, my slope is going to be a negative divided by a positive, which is going to give me a negative result. When we look at horizontal lines to move from one point to the other, we have to move, In this instance we don't move up at all. So the rise is going to be zero even though we do have a run that is going to be positive. So the horizontal lines, the slope is equal to zero again, because the numerator, the rise is going to be zero. For vertical lines notice that to move from this point that's, that I'm indicating, which is below another point, I would rise a certain amount that would be a positive value, but there's no run. I'm not moving to the right or left. So if I think about it, the slope would have to be a positive value divided by zero, and we know that division by zero is undefined. So we can't use slope in this instance. In fact, what we say is, let me just scratch that out completely because we need to understand that our slope is going to be something that is undefined. And so we now see in general, given us different indications of what lines look like, what the relationship is with the slope. Let's nail it back at the standard form of a line. We had Ax plus By equals C. And I want to investigate what happens if I make some substitutions. For instance, what happens if I allow A to equal 0? Then I'm going to end up with 0 times x, which is going to be 0, plus by equals C. And so this becomes B times y equals C. And y is equal to C divided by B. Now I'm just going to make a change of variable here. And I'm just going to say let's just call this y equals a b value. What kind of line would this be? Well, if my y value is the same for every point on the line, then what I'm ending up with is a horizontal line. So this is the basic format for the equation of a horizontal line. It is when we end up with y equal to some constant b. So what happens if, instead of letting a equal 0 in the standard form, we let capital B equal 0? So if capital B is equal to 0, then the equation is now going to read A times x plus B, which is now 0, times y-- so that's 0-- equals C. That can be simplified to A times x equals C. And we can divide by A on both sides to end up with C divided by A. And again, I'm just going to make a little change of variable. Remember, A and C are both just numbers. So x can be equal to some value that I'm going to call lowercase a. Well, in this instance, I'm going to have an equation of a line where all of the x-coordinates of the points on that line have to be the same, which means you had to move the same direction to the right or the left in order to get to a point on that line. This occurs when the line that we're working with is a vertical line. So what we can see is that the standard format for a horizontal line is going to be y equals a constant, which we can call B. And the standard form for a vertical line is going to be the equation x equals a constant, which I've chosen to call a. Now let's look at a concrete example. For instance, what if I have the point that has coordinates 2 comma 3 and I tell you that I have a vertical line that is passing through that point? The equation of the vertical line that would pass through the point 2 comma would have to have the x-coordinate equal to the x-coordinate of that point. It would have to be the vertical line x equals 2. What if there's a horizontal line that passes through the point 2 comma 3? That horizontal line would have to have the equation where y is equal to the value that is the y-coordinate of that point. In other words, that horizontal line would have to have the equation y equals 3. Next we're going to look at two special cases for lines. The first one is going to be where I take a line l sub and I look at a parallel line. Now, to look at what's occurring with the slopes, I want to look at this triangle that's appearing. Look at the fact that the horizontal line, or the horizontal side of the triangle, is going to be the run. And the vertical side of that triangle is going to be the rise. So if I now look at that same triangle just shifting upward and I draw in a new line-- I'll call it l sub 2-- then see that I have the same rise and the same run. So for two lines that end up being parallel to each other, what we note is that the rise over run is the same, which means that the slope of line is equal to the slope of line 2. Now, we can say that what that means is that if two non-vertical lines are parallel, then their slopes are equal. Now, it doesn't take into account vertical lines because we've already determined that vertical lines are not going to have slopes that are undefined. Well, we've looked at the parallel lines. The other special case that we're interested in is going to be perpendicular lines. Perpendicular lines are lines that meet in a 90 degree angle. So again, let's look at a graphic. Here I have a line labeled l sub 1, which is line 1. And once again, I'm going to consider a triangle to represent the slope and see what's going on. In this case, the rise is going to be, again, the vertical side of that triangle. The run is going to be the horizontal side of that triangle. Now, to illustrate this, I am going to rotate this triangle 90 degrees. And it's going to end up giving you then when I draw in the line that is represented after that, l sub 2, a line that is perpendicular. Do you see that the horizontal part of that triangle became vertical and the vertical part of that triangle to begin with became horizontal? And so first of all, you can note that what was the rise became the run and what was the run became the rise. So we certainly have a reciprocal relationship. But also, l1 is going to be a line that is decreasing in our picture, which we know means that the slope is negative. l sub 2 is a line that is increasing in our picture, which we know means that the slope is positive. And so we have the following relationship. Whenever we have two lines that are perpendicular to each other, such as l sub 1 perpendicular to l sub 2, It turns out that the slopes are going to be negative reciprocals of each other. The slope of l sub 1 is going to equal the negative of 1 divided by the slope of l sub 2. And again, let's put that into words. If two non-vertical lines are perpendicular, then their slopes are negative reciprocals of each other. Next we're going to define what is meant by "linear functions." A linear function can be written in the format f of x equals mx plus b, where m is the slope of the line and b is the y-intercept. So if we go back to a line that we've already seen at the beginning of this session, you may recall that we had the line that we come up with. And we had a y-intercept at 4. And we had an x-intercept at 6. So again, just quickly sketching that line in. Now, this is going to be a linear function. It's certainly a function because it passes the vertical line test. And it is a line, hence a linear function. If I wanted to come up with an equation for this function, I could say it's going to have the format f of x equals mx plus b. The m in this format is going to be the slope. And the b is going to be the y-intercept. So I can look at the graph and see that in order to go from the y-intercept to the x-intercept, I have to go down 4. So therefore, my slope's going to move down 4 as my rise. And I move to the right 6. And we had done this a little bit earlier on. We see that our slope then is negative 2/3, so my f of x is going to equal negative 2/3 x. And then I need to put my y-intercept, which is going to be the value 4, plus 4. And I get the equation of the line. Alternatively, I'm given this functional format of the line, I could go back and start off-- and I'm just going to really quickly sketch this here. I could start at the y-intercept of 4. And then using the slope, I can go down 2 and over to get a second point on that line. And then I could go down 2 and over to get a third point on that line. And that's going to end up giving me, of course, the same line. But notice how I can use this format for the equation of the line in its functional form in order to come up with the slope. It's time for another quick quiz. Which of the following linear equations does not yield a linear function? Is it, A. One third x minus 2 fifths y equals B. 2x minus 4 fifths y equals one. C. 5 thirds x equals Or D. 7x equals 5y. Choose from A, B, C or D now. You're correct, the answer is C. 5 thirds x equals 20. Sorry the answer is C. 5 thirds x equals 20. When we look at the question. Which of the following linear equations does not yield a linear function? We need to recall that all straight lines are going to be functions except those that fail the vertical line test. The only straight line is going to fail a vertical line test is going to be an actual vertical line. So that means we're going to be able to draw a vertical line right on top of that vertical line that's drawn and end up hitting the line in more than one point, which means for every x, there's going to be more than one y. The vertical line equation has the format x equals a constant number which I'm going to call A. So when I look at A, B and D, as my selections I noticed there's a variable y in all of those equations. In the equation 3 or the third equation which is under C, I can actually solve for x. And I can do that by multiplying both sides equation by the reciprocal of the coefficient. So I'm going to multiply both sides the equation by 3 fifths, that will give us 3 fifths times 5 thirds x equals 3 fifths times 20. On the left hand side multiplying a number by its reciprocal gives us the value one. So one times x is x. On the right hand side 3 fifths times 20 give us result of 12. So I can see explicitly that I have a vertical line given to us in part C and that's going to be the only one that's going to end up not being a linear function. Now, there's another form for an equation of a line that you'd like to be familiar with. And it's a direct derivative of what's going on when we talk about the slope formula. So again, let's recall the slope formula. Our slope is equal to our rise over run. Remember, if we go ahead and put in y values for the rise, I can think of our slope as equal to a difference in our y values. And I'll use the subscripts here. And our run is a difference in our x values. And I'll use the x values with the subscripts there. Now, I'm going to do something here that's a little bit different in that I'm going to think about this without the subscript of 2. And my reasoning is I'd like to leave it as a point x comm y without designating a particular point. So I'm going to rewrite this slope formula. And all I'm going to do is just drop the subscript. So it's going to be y minus y sub divided by x minus x sub 1. Next what I'm going to do is multiply both sides of this equation by the denominator. So I end up with m times x minus x sub 1 equals y minus y sub 1. This is called the "point-slope form" of the equation of a line. In this form, what would happen is I would go ahead and substitute for the slope of the line as m. And I would substitute for a point that I was given in the place of x sub 1, y sub 1. The x and the y in the equation I want to leave in variable format so that it would take the place of any point that would end up being on the line. So we're going to use that in an example next. Here's the problem we're going to consider. Find the linear function that passes through the points 7 comma negative 1 and 9 comma negative 5. Well, we want a linear function. And recall that the format for a linear function is f of x equals mx plus b. So this is where we're heading. Well, what can we do? We know for sure we need to have the slope. So we know that slope is equal to rise over run. And I'm going to choose to pick up my points in the order in which I see them. That is, my rise is going to be the difference in my y values. I'm going to pick up the y-coordinate from the first point I see and subtract from it the y-coordinate from the second point. So that numerator becomes negative 1 minus negative 5. And as long as I get the same structure going all the time, I won't make a mistake here. I know I'm going to come back to the first point that I saw here and pick up my x-coordinate. In my denominator is 7. And then subtract from that the x-coordinate of the second point that I see, which is 9. So let's find our value for m. It's going to equal a numerator, which is negative plus 5, which is going to give us 4. And the denominator 7 minus 9 gives us negative 2. So we come up with a slope of negative 2. Now I notice that I have two points that are given to me for this line. And so I can use the point-slope form of the equation of a line in order to go a little bit further. Recall we just saw that's going to be our m times x minus x sub 1 equals y minus y sub 1. In this formula, it really doesn't matter what you called x sub 1 and y sub from your previous work. Remember, that's just saying you use a point that's on the line, the point as the coordinates x sub 1, y sub 1. So I can choose to use either of the points that are given to me in the problem and substitute for the x-coordinate here, x sub 1, and the y-coordinate there, y sub 1, in order to work through it. Our m is going to be negative 2 no matter what. So for instance, I can choose to use the 9, negative in this equation. So I can write down negative 2 times x. And it's going to be minus 9 equals y minus the negative 5. We can simplify this. We're going to have negative 2x plus 18 equals y plus 5. And then I'm going to do what next? Well, remember we're looking for the linear function of the format f of x equals mx plus b. That means I want to have this equation solved for y so I can substitute that y equals f of x. I'll subtract 5 from both sides of the equation now to give me negative 2x plus 13 equals y. And now I have the correct format to designate the linear function f of x equals negative 2x plus 13 as my result. Next we're going to look at a good skill you want to develop when it comes to finding slopes of lines. We want to find the slope of the line 6x minus 5y equals 24. Now, one way we could do this is to find two points on that line and then use the slope formula. But here's another method that's going to serve you well. If we look at this equation and we solve for y, we'd be putting it into the format of the slope-intercept form, which is what we saw when we looked at the linear function. So given the equation 6x minus 5y equals 24, I will simply solve for y. I can do that by subtracting 6x from both sides of the equation. So negative 5y equals negative 6x plus 24. And then I'll divide both sides of the equation by negative 5. So y is going to equal-- and I'm going to go ahead and write it as a division by negative 5 of each term so that it has the format that I want. And that's going to be plus 24 divided by negative is my second term. Now we're going to see that y equals-- and if I rewrite negative 6x divided by negative 5 as 6/5 x, and then we'll have 24 divided by negative is going to be a subtraction of 24/5, it's not the negative 24/5 that I'm interested in. Since I'm only looking to find the slope, I'll want to see that this has the format where I have y equals mx plus b. So I can read off the slope of this line as the coefficient of x. And so the answer to the problem is the slope of the line m is equal to 6/5. Now we're going to look at a problem that you can try on your own. Here's what it reads as. "Find the slope-intercept form for the equation of a line that passes through the point negative 9 comma and is perpendicular to the line 3x plus 8y equals 2." Now, I'll tell you at the outset, this combines a lot of the things that we've discussed on this video so far. So just a couple of hints-- you're given a point on the line. And you're also given some information about the slope of the line. The thing is you're given the information about the slope of the line in that it's perpendicular to this other given line. So try to think about the things that we've discussed so for and put those together in order to come up with a equation for this particular line. So pause the tape, give it some thought process, and then work through the problem. Restart the tape when you're ready to look at it together. OK. Let's see how you did. Again, we want to think about the fact that we want to have some information about the slope of the line. We can find that information from the fact that we're told that the line that we're looking at is perpendicular to a line whose equation is 3x plus 8y equals 2. Now, a few moments ago, we talked about a quick way to find a slope of a line. And that is to take an equation and solve for y. So that's the process I'm going to use. Once I find the slope of this line, it's not going to be the slope of the line I'm looking for. But I do know that the slopes of perpendicular lines are negative reciprocals of each other. And I'll use that fact. Again, I'll start by solving for y. So 8y equals 2 minus 3x. And I'll divide by 8. So I'll divide each term by 8. y is going to equal 2 divided by 8, which is 1/4, minus 3x divided by 8. And I'm going to write it in the format minus 3/8 x. Now, I wrote it in that order because I wanted to point out to you your slope is not going to be the first number that you see. It's going to be the coefficient of your x term. So notice that my slope for the perpendicular line is going to be negative 3/8, which means that the line I'm looking for is going to have a slope that is the negative reciprocal of negative 3/8. That means the slope is going to be 8/3. The point on the line is given to us. And that point has coordinates negative 9 comma 1. Now I want to put that information together. And I can use the point-slope form of an equation of a line. And again, I'll just rewrite it for you. It's m times x minus x sub 1 equals y minus y sub 1. Let's substitute in the values that we found. We have 8/3 times x minus a negative 9, which I can go ahead and write in as plus 9, equals y minus the y-coordinate of the point, which is 1. We want to have this final answer in slope-intercept form, which means I need to solve for y. And I don't want to have this set of parentheses here. So let's multiply out. We're going to end up with 8/3 x plus-- I'll go ahead and write it in. It's 8/3 times 9 is going to equal y minus 1. Here we can think of this as being 9 over 1. There's going to be a common factor of that we'll divide out. So dividing the denominator by 3 and the numerator by 3, I end up with 8/3 x plus-- now I end up with 8 times 3, which is 24 divided by 1. And that's equal to y minus 1. And finally, to solve for y, I'm going to add 1 to both sides of the equation to give me 8/3 x plus 25. And so this is going to be the slope-intercept form of the equation of this line. So I hope you were able to put all of those facts together in order to come up with this solution yourself. Right now it's time for you to try more problems dealing with straight lines, reviewing the ideas of slope, reviewing the ideas of parallel and perpendicular lines, and also bringing into the fold the idea of a linear function. Good luck.
Table of contents
- 0. Review of Algebra4h 16m
- 1. Equations & Inequalities3h 18m
- 2. Graphs of Equations43m
- 3. Functions2h 17m
- 4. Polynomial Functions1h 44m
- 5. Rational Functions1h 23m
- 6. Exponential & Logarithmic Functions2h 28m
- 7. Systems of Equations & Matrices4h 6m
- 8. Conic Sections2h 23m
- 9. Sequences, Series, & Induction1h 19m
- 10. Combinatorics & Probability1h 45m
2. Graphs of Equations
Lines
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