Common Core Algebra II.Unit 3.Lesson 3.Forms of a Line

Hello and welcome to another common core algebra two lesson by E math instruction. My name is Kirk weiler and today we're going to be doing unit number three lesson number three on the forms of a line. You see, linear equations and linear functions come in many, many different forms. There are two primary ones and we're going to review those right away and then get lots of practice today and lots of review on a variety of things that you probably already feel pretty comfortable with in terms of equations of lines. But it pays to review them early on in this course. So let's jump right into it. Two common forms of a line. You've got what's known as your slope intercept form. And your point slope form. The one that, of course, you're more kind of comfortable with and you've been seeing for more years is Y equals MX plus B where M is the slope of the line, also the average rate of change. And B is the Y intercept where it crosses the Y axis. Something that you were introduced to last year, and you may have even been introduced in common core algebra one, is what's known as the point slope form of a line. I like this form a lot. I use it a lot when I teach calculus. The M is still that slope of the line. But then Y one and X one are a specific point on the line. Which point it doesn't matter. That's one of the beautiful things that can be any point whatsoever. So we're going to get we're going to get some use with both the slope intercept and the point slope form. Generally speaking, if a problem asks you for the equation of a line. But they don't specify which form to use. You can use either one of them. All right. But sometimes one of them is more just easy than the other one, you know? Like let's say you know the Y intercept, then slope intercept forms great. Let's say you don't. Let's say you just know a point on the line and the slope of the line, then the point slope form is really nice. And we're going to use both throughout this course. So let me clear that out. And let's jump into some problems. All right, exercise one. Consider the linear function F of X equals three X plus 5. So here we have a linear function given in the form Y equals MX plus B letter ace is determined the Y intercept of this function by evaluating F of zero. So any Y intercept we can find for any function linear or not. By substituting in zero for X and we get 5. Now, the purpose of this exercise is really to illustrate why that number is the Y intercept, right? If I put X equals zero in, M times X just goes away. Now, let her be says find the average rate of change over the interval negative two to three. Well, let's just be very mechanical about this. We know that the average rate of change is delta F divided by delta X so let's calculate F of negative two. That's three times negative two plus 5. Negative 6 plus 5 gives me negative one. So there's my output when my input is negative two. Let's try this one with my input is three. It'll be 9 plus 5. And that's 14. So now my change in F divided by my change in X watch out 14 minus negative one. Divided by three minus negative two gives me 15 divided by 5 and that gives me three. We also knew that. We knew that it should come out to be the slope of the line, but we have to check anyway. So the main purpose of this exercise was to give you a line in slope intercept form. And then verify that, in fact, B is the Y intercept. And M is the slope. All right. So just kind of verifying something that we hopefully were already pretty comfortable with. Pause the video down, write down anything you need to. Okay, here we go. Exercise two. Now let's work with a little point slope form. Exercise two says consider a line whose slope is 5. And which passes through the point negative two 8. Write the equation of this line in point slope form. So in this case, what we have is we have M is equal to 5. X one is equal to negative two, and Y one is equal to 8. So when we literally just plug all of this information into this form, we would get Y -8 equals 5 times X minus negative two. Now this is an okay way to leave it. But if this was, let's say, a multiple choice question or something like that, they wouldn't leave it that way. Any time that we have a double negative, that's going to get turned into a positive. So X minus negative two becomes X plus two. And that's really the way we should leave it. Now what's interesting is that the point slope form of a line is not unique because of course we could use any point that's on the line. In this case, we were only given the point negative two 8. What is unique about every line, though, is its slope intercept form. So it says write this equation in slope intercept form. Now you might say, well, I don't know what the Y intercept is. That's okay. All we have to do is use properties. To rearrange this equation, like we'll use the distributive property. Then we'll use the additive property of equality. And we'll rearrange it into MX plus B form. And what's cool about that then is you can look at it and say, yeah, sure, yeah, the Y intercept is 18. Right. We don't have to do anything fancy to figure it out. We just rearrange it. And there it is. Every point slope form of a line can be rearranged into the Y into the slope intercept form. I'm going to clear this out. So pause the video if you need to. Okay. Let's go on to the next problem. Exercise three. Which of the following represents an equation for the line that is parallel to Y equals three halves X -7. And passes through the .6 comma negative 8. So something about parallel lines. Why don't you pause the video now and think about this for a moment? All right. Well, each one of these is written obviously in point slope form. Y minus Y one equals M times X minus X one. What do we know about parallel lines? This is one of the things we want to review today. Parallel lines have equal slopes. And that should make complete and utter sense to you, right? The whole point of slope is to measure how steep align is. And when two lines run parallel to each other, they have to have the same steepness. In which case they have to have the same slope. So one thing I know for certain is that my slope must be three halves. Now that immediately eliminates this where the slope is negative two thirds and this where the slope is negative two thirds. The question is then which one of these tours the correct answer? Well, given that X one is equal to 6, and Y one is equal to negative 8. We can now say Y minus Y one equals the slope three halves times X minus X one. Again, we'll change that double negative into a positive Y plus 8. Equals three halves times X -6. And that's choice three. All right. So what we wanted to do in that problem was to review what's called the parallel line condition. So parallel lines are parallel, if and only if they have equal slopes. This is assuming we don't have vertical lines where the slopes are nonexistent. Okay? So I'm going to clear this out, write down anything you need to. Okay, here we go. Exercise four. A line passes through the points, 5 comma negative two, and 20 comma four. Letter a determine the slope of this line in simplest rational form. Why don't you pause the video and go ahead and do that? All right, let's go through it. We want to be nice and careful. The slope is the change in Y divided by the change in X Y two minus Y one divided by X two minus X one. If you took common core geometry, then obviously you did a lot of coordinate geometry proofs. So four minus negative two divided by 20 -5. Let's see, four minus negative two is 6, 20 -5 is 15. Simplest rational form, we can divide both of those by three and our slope is two fifths. Great. Write an equation of this line in point slope form. All right, oh, well point slope form, Y minus Y one equals M times X minus X one, which point do I choose for X one, Y one. And the answer is, it doesn't matter. Now most people will very understandably pick this first point because that's the first point. So why wouldn't you? So I'll get Y minus negative two equals two fifths times X -5. And I'll rewrite this as usual. Y plus two equals two fifths. Times X -5. There's a nice point slope form. Now again, an alternative would be this one. Y minus four equals two fifths times X -20. That would also be a completely acceptable answer based on this point. Either one of those were great. Let her see write an equation from this line in slope, intercept form. So in that case, what I want is Y equals MX plus B and you can take either one of these two and algebraically rearrange it like we did in the last problem to get it in Y equals MX plus B four. So I'd like you to try that. Pause the video now. I know there's a fraction involved, but you can do it. Trust me. All right, let's go through the manipulation. I'm going to go with the first one, but the great thing is, it doesn't matter which one you use. You'll come out with exactly the same answer. All right, well, if I want to rearrange this, what I'm going to do is distribute the two fifths. Now again, I know fractions are fractions. Nobody likes them. The old joke is that four out of three people don't understand them. Now, I have to do two fifths times 5. That's actually pretty easy. That just gives me two. Now I'll subtract two from both sides, be careful. Negative two minus two isn't zero. It's negative four. And there's my Y equals MX plus B four. All right, last but not least, letter D, for one X value. Well, this line passed through a Y value of 12. Wow. That's easy enough. But what I'd like you to do is pause the video and try to figure that out. All right, let's go through it. We're going to put 12 in for Y and we're just going to simply solve for X we could do that with any one of the three equations we have now. This is probably going to be the easiest one, a little bit of an additive property. Whoops additive property of equality. I don't know where that four went. Gives me 16 equals two fifths X and remember always easier to solve this by multiplying by the reciprocal rather than dividing. Although with the modern calculator, it doesn't make too much of a difference, and we find X is four. All right. So that was nice because it really walked us through how we can take two points on a line. And come up with its equation in a variety of different ways. Pause the video now right now. Anything you need to, and then we'll move on. Okay, one last problem. The graph of a linear function is shown below. Letter a asks us to write the equation of this line and Y equals MX plus B form. So take a moment to do that. All right. Well, we should easily be able to look at this line and say, ah, okay, B is one. And then we should be able to go, all right, well, the slope, if I go to the right one and go up to, well, then M, the change in Y divided by the change in X is two to one, or two. So Y is going to be two X plus one. There's my equation. Easy. Let her be. What must be the slope of a line perpendicular to the one shown? Wow, so if I, if I drew a line in, and it was perpendicular, let me do that. Let me draw it in red. Am I right? Let's see. What would that look like a little bit challenging? But if I kind of did that, right? So that I had a nice perpendicular line. What would it slope be? Can you remember this from last year? Or can you just calculate it from the line I just drew? Well, the slope of the line will be negative one half. This is what's called the negative reciprocal. Whoops. Now, letter C, I already did. It's a draw the line perpendicular to the one shown that passes through the .1 comma three. So just as remember, a negative reciprocal, if I've got a slope to one, I'll flip it, and I'll negate it. So if I had a slope of 5 thirds, it's negative reciprocal would be negative three fifths. Something like that. So we already did draw a line that was through the .1 comma three perpendicular. And now letter D says, well, write the equation of the line you just drew in point slope form. No problem. We've got everything we need, right? M is negative one half. And then my X one Y one is one comma three. So it'll be Y minus three. Equals negative one half times X minus one. Now, letter E asks, does the line that you drew contain the .3 comma negative 15, justify? Well, pause the video now and think about how to answer that question clearly that point is not on the graph grid that we have up on the screen. But that doesn't mean it's not on the line somewhere way out there or there. Whatever. So play around with this for a moment and see if you can come up with the right answer. All right. Well, at the end of the day, it's this simple. A point lies on a line if it makes the equation true. So I am going to see right now if I plug this point into this equation doesn't make it true. Now I'm going to be careful. I'm putting negative 15 in for Y and I'm going to put in 30 for X. All right, that's simple, right? That's going to give me negative 18. This is going to be a mi 29. And the question is, does negative 18 equal negative 29 halves? Well, it's not too far off, but nope. Negative 29 halves is negative 14 and a half. And that's not equal to 8 negative 18. And so the answer. Is no. All right. Okay, well, there's a lot on this screen. So pause the video now if you need to to write something down. And then we'll finish up. Okay. All clear. So in today's lesson, what we did is we looked at two primary ways that we write equations of lines. The slope intercept or Y equals MX plus B form. And the point slope form. Y minus Y one equals M times X minus X one. Both of these forms of align are important, and they both get used quite a bit. How often though it's up to the context of the problem. What's the easiest one to use? And sometimes we like to originally write an equation in point slope form and then rearrange it into slope intercept form. More on that later. For now, I'd like to thank you for joining me for another common core algebra two lesson by E math instruction. My name is Kirk weiler. And until next time, keep thinking. And keep solving problems.