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Welcome to Algebra I: Writing the Equation of a Line When Given Two Points on the Line

25 minutes

Hey, guys, welcome to Algebra 1. Today's lesson focuses on writing the equation of a line when given two points on the line. You've written the equation of a line given its graph, given the slope and a point. This is a spin-off of what you already know. You ready to get going? Let's start. Okay, so just to jog your memory first, you know how to handle this kind of problem, right? Given the graph of a line, you can identify the Y intercept and count off that slope. You can figure out the equation of the line and write it in slope intercept form. And you've handled this type, right? You know given the slope of a line and given a point on the line, you can use point slope form and process that problem, and get to the slope intercept form of the line. For this kind of problem, you're going to be given two points on the line. There's two steps in the process. First, figure out what's the slope of the line that passes through those two points you were given. Then, after you have the slope, use that slope, and one of the points you were given, and the point slope form, and get to the slope intercept form. That sounds like a lot, but it'll make more sense when I show it to you. So here's our first example. We'll write the equation of the line

that passes through the points (8, 1) and (10, 2). First, figure out the slope of this line. There's a couple of ways you can find the slope of a line: use graph paper and plot these two points and count out that slope, or use the slope formula to get it. I don't have a graph paper, so I'll use the slope formula to find the slope of this line. I'll label one of these points X sub-1, Y sub-1. I'll label the other point X sub-2, Y sub-2, and I'm going to label-- this is my first step-- I'll determine the slope. Okay? Remember, the slope formula was M equals Y2 minus Y1 over X2. Oops--wrote another Y, let's get that cleaned up. There we go-- so used to writing Y's-- over X2 minus X1. There's our formula for slope. I'll fill in what needs to go where and get the slope of the line that passes through these two points. So Y2 minus Y1 would be 2 minus 1. Then, X sub-2 minus X sub-1 would be 10 minus 8. Okay. And let's clean this up. So 2 minus 1, that's 1. 10 minus 8, that's 2, right? So the slope of the line that passes through these two points is one-half. So I've got my slope. Now I'm going to use point slope form. I've got my slope. I can use either one of those points to get the equation of this line. Now, even though we've labeled this point X sub-1, Y sub-1, and this one X sub-2, Y sub-2, we could have done the opposite: labeled the (10, 2) with the X sub-1, Y sub-1, or labeled the (8, 1) with the X sub-2, Y sub-2. It doesn't matter, just stay consistent with what you use. When we do step two, it doesn't matter which point I use because they're both two points on my line. I'll use (8, 1) because I've already labeled it X sub-1, Y sub-1. But if I'd chosen to use (10, 2), I'd still have the same answer. Just know that while you're working. Let's keep going here. I'm going to label this step two and get my point slope form up here. Okay. I've got Y minus Y sub-1. Let's fill in our information here. Let's get some more workspace. We've got to do a little moving around first. Okay. That's good. All right, so Y minus-- our Y sub-1 is 1 here, but like I said, we could be using this point--equals M, which is a half, times X minus-- our X sub-1 is 8. Okay? We'll clean this up and get to that slope intercept form equation. I'll bring down the left side. I'll the distributive property on the right. So one-half times X, that's one-half X. One-half times negative 8, I'll come off to the side over here and do that. So one-half times negative 8 over 1. 1 times negative 8 is negative 8. 2 times 1 is 2. Negative 8 divided by 2, that's negative 4, okay? So back over here to my work. One-half times negative 8 is negative 4, right? I'm just going to clean this up. I need to get Y by itself, so I'll add 1 to both sides. So plus 1, plus 1. That's going to get wiped out. So Y equals one-half X. And negative 4 plus 1, that's negative 3. And you got it, you're all done. So it's two steps. It's lengthy, but when you get the swing of it, you can do these quickly. Remember, we were given two points. First, we calculated the slope, we used the slope formula, and after we had that slope, we used point slope form, and we got to that final equation of the line. Let's try another one. Write the equation of the line that passes

through the points (negative 2, 7) and (0, 1). So what did we do first on that last one? Good-- we labeled these points so that we could get to that slope, right? So I'll label this one X sub-1, Y sub-1. I'll label this one X sub-2, Y sub-2. Then I'll go ahead and use the slope formula to calculate the slope. So M equals Y2 minus Y1 over X2 minus X1. So Y sub-2 minus Y sub-1, that would be 1 minus 7 over X sub-2 minus X sub-1, so 0 minus negative 2. I've got that double negative. I'll handle that. All right, so 1 minus 7, that's negative 6. 0 plus 2, that's 2, right? And negative 6 divided by 2 is negative 3. Now I know the slope is negative 3. Now, we can go ahead and... good, move on to step two. Use your point slope form, and then we can get at the final equation. Let's get point slope form down first. Y minus Y sub-1 equals M times X minus X sub-1. Scroll down a little bit, but still enough to see my point. It kind of disappeared, but I can see it. So Y minus--our Y sub-1 is 7--equals M, which is negative 3... X minus--and our X sub-1 is negative 2. Let's get some more room up here to work. We're always moving something around up here. So we can bring the left side down-- that Y minus 7 equals... Let's handle that double negative before anything else. All right, so negative 3 times X, that's negative 3X. Negative 3 times positive 2, that's negative 6. Then our final step on this one? Good--plus 7, plus 7. That's going to cancel out, and we have Y equals-- bring down that negative 3X, and then negative 6 plus 7 is a positive 1. You got it, guys, all done with that one. Bringing it all together one more time. Scroll back up. We were given two points. First, we labeled those points to use the slope formula and calculate the slope. Then, after we had the slope formula, we used the slope and one of our given points and put that in point slope form, worked that all out, and got to our final equation: Y equals negative 3X plus 1. Okay? All right, give this one a try. Press pause, take your time, work our process, and write the equation of the line that passes through these two points. To check your work, press play.

(female narrator) Write the equation of the line that passes through the points

(4, negative 3) and (12, negative 5). All right, let's see how you did with this one. I'll start out by labeling these points. First thing I'm going to do is get the slope. Let's get that formula written down. We've got that down. I've got the formula, let's plug in our values. So Y sub-2 minus Y sub-1 will be negative 5 minus negative 3. So negative 5 minus negative 3 over X sub-2 minus X sub-1, 12 minus 4. Let's clean up that double negative situation. All right, so negative 5 plus 3, that's negative 2. 12 minus 4, that's 8. I can reduce this fraction; it's not in simplest form yet. If I divided the numerator and the denominator by 2, I could reduce this to negative 1/4th, right? So the slope of this line is negative 1/4th. Now, let's jump into point slope form. So Y minus Y sub-1 equals M times X minus X sub-1. Okay? Scroll a little more. Hopefully, we can still see the point. So Y minus negative 3 equals-- our M is negative 1/4th, parentheses X minus-- and our X sub-1 is 4. Scroll down and get some more workspace. All right? Clean up that double negative. So on the left side, we have Y plus 3 equals-- and, on the right, we have negative 1/4th. Oop--I'm getting ahead of myself here. We need to apply the distributive property, get that right side under control. Okay, so negative 1/4th times X, that's negative 1/4th X, okay? Now, negative 1/4th times negative 4-- let's do some scratch work over here on the right. So negative 1/4th times negative 4. And I know I can put that negative 4 over 1. So negative 1 times negative 4 is positive 4. 4 times 1 is 4, and 4 divided by 4 is 1. Right? Back to our problem over here, negative 1/4th times negative 4 is a positive 1. Then, the last step, subtract 3 from both sides. Now, this is going to cancel out over here, and we have Y equals... I'll bring down that negative 1/4th X, and 1 minus 3, that's negative 2. And you got it, guys. All done with that one. Now, before we leave this, I'm always going through those special lines, and I'm going to show you how to interpret them so you can quickly get to the answer. "Write the equation of the line

that passes through the points (negative 3, 4) and (5, 4)." Okay. There's a couple ways to attack this. You might look at the points and initially notice that they have the same Y coordinate. That's a hint to you. If the Y's are constant, they didn't change, then that means I have a horizontal line. You could spin off from there and say, "Okay, if this line is horizontal, I'm only concerned with those Y values." They're both 4. So the equation of the line is Y equals 4. That's one way to attack this. And you should get in the habit of, every time you're given two points, check and see: Do they have constant Y's or constant X's? Is something not changing with those two points? If that doesn't immediately jump out at you, this is how you'll end up at the equation of that special line. You'll go through your normal process of first getting the slope. Let's erase here so we can get the workspace going. If that didn't initially jump out at you, then go through your process. Label X sub-1, Y sub-1. X sub-2, Y sub-2. Calculate your slope. This is step one. So M equals Y sub-2 minus Y sub-1 over X sub-2 minus X sub-1. Okay, so Y sub-2 minus Y sub-1, that'd be 4 minus 4... over X sub-2 minus X sub-1: 5 minus negative 3, right? You got that double negative situation in your denominator, so handle that. So 4 minus 4 is 0. 5 plus 3 is 8. 0 divided by 8 is 0. So that means this line that passes through these two points, has a 0 slope. And if it has a 0 slope, it has to be horizontal. So if this is a horizontal line, then you're only concerned with the Y values because it's the Y values that stay constant on a horizontal line. At that point, you could jump to your answer. The equation of this line has to be, because those Y's are constant, that Y equals 4. And you can get it that way, okay? If you don't initially see-- when the points are given to you, if it doesn't jump out that the Y values are constant, go through your process. You'll see, at the end, the slope will be 0. Then think, "Well, if this slope is 0, "then this line is horizontal, and I'm only concerned with the Y values here." Then get your answer, which in this case would be Y equals 4. Now give this one a try. Actually, let me work this one out with you. I see here that we're given the points

(1, negative 6) and (1, 2). We'll write the equation of the line that passes through these two points. If you look at the points, notice that they both have an X coordinate of 1. So the X coordinates stayed constant. And if the X coordinates stayed constant, then that means this line has to be vertical. And if this is a vertical line, then I'm only concerned with the X values. And if they're both 1, then the equation of this line is X equals 1. You can be done that fast. That's only if you notice, when you start, that these equations have the same value for X. If that doesn't immediately jump out at you, that's okay. We'll go through our process, and it'll jump out at you. Let me show you what I mean. Okay. Let's get my pen back. Here we go. I'll go through my process. I'll label the points X sub-1, Y sub-1, X sub-2, Y sub-2. Okay, so step one, let's get that slope. So M equals Y sub-2 minus Y sub-1 over X sub-2 minus X sub-1, all right? So Y sub-2 minus Y sub-1 would be 2 minus negative 6 over X sub-2 minus X sub-1, would be 1 minus 1. I'll handle this double negative first. Okay, 2 plus 6, that's 8. 1 minus 1, that's 0. Here you'll see it. You cannot divide a number by 0-- that's undefined. You can't have a numerical value and then divide it into zero parts, it's not possible. In a situation where you have 0 in the denominator, you are in a situation where your slope, or your value, is undefined, right? In this case, because we are calculating slope, we know that the slope of the line that passes through these two points is undefined. So then, again, we get to thinking, right, if the slope of this line is undefined, then this line is vertical. It has to run just straight up and down. If this line is vertical, then I'm only concerned with the X's. The X's are both 1. So the equation of my line is X equals 1. And you're done, okay? Again, that's another way to get at the equations of these special lines. If it doesn't jump out when you see the points, go through your process. If at the end of step one, you see your slope is undefined, you have a vertical line and the equation of that line is only concerned with those X values. So in this case, it was X equals 1. Okay? Let's keep going. It's time for you to try a couple. You knew it was coming, so press pause. Take a few minutes and work through these problems. See if it jumps out or if you go through the process. Either one is fine. To check your answers, press play.

(female narrator) Write the equation of the line that passes through the given points. Letter A: (8, 1) and (8, negative 5).

Letter B: (2, negative 3) and (4, negative 3). All right, let's see how you did. So for A, the equation of the line that passes through (8, 1) and (8, negative 5) is X equals 8. For B, the equation of the line that passes through

(2, negative 3) and (4, negative 3) is Y equals negative 3. Now, to see how I got these, I'll show you. Here we go. Like I said, it may have jumped out at you here that your X coordinates were constant, so that this had to be a vertical line. If it didn't, then you went through the process. So label the points-- let's get that pen back-- X sub-1, Y sub-1; X sub-2, Y sub-2. Get your slope formula written down. Now, fill in your values. So Y sub-2 minus Y sub-1 would be negative 5 minus 1 over X sub-2 minus X sub-1, would be 8 minus 8. Okay? So negative 5 minus 1, that's negative 6. 8 minus 8, that's 0. Again, I see 0 in the denominator. It's not possible to divide a number by 0, so this slope is undefined. And if I have a line with an undefined slope, then that means I have a vertical line. And if I have a vertical line, then I'm only concerned with these X's. These X's are both 8, so the equation of my line: X equals 8, and you got it. You're all done with that one. Let me show you how I got the next one. Either it jumped out that the Y values were both negative 3, so this was a horizontal line, making the equation Y equals negative 3, or you went through the process. So label your points. Okay? Now, go ahead and get your slope. Let me get that slope formula up there. So Y sub-2 minus Y sub-1 would be negative 3 minus negative 3 over X sub-2 minus X sub-1: 4 minus 2. I'm going to handle that double negative situation up there. There we go. So negative 3 plus 3, that's 0. 4 minus 2, that's 2. 0 divided by 2, that is possible to calculate. It's 0, okay? So if I have a line that has a slope of 0, then I have a horizontal line. And if I have a horizontal line, then I am concerned with the Y's. So if my Y's are both negative 3, the equation of this line is Y equals negative 3. And you're done. All right, excellent job, guys, writing equations of lines given two points and still persisting and working through those special lines, and writing the equations of those lines when given two points. Hope to see you back here soon for more Algebra 1. Bye!

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The equation of a line is typically written as y=mx+b. If students know two points that a line passes through, then they can write the equation for the line in standard form (y=mx+b). Part of the "Welcome to Algebra I" series.

Media Details

Runtime: 25 minutes

Welcome to Algebra I
Episode 1
31 minutes
Grade Level: 7 - 12
Welcome to Algebra I
Episode 2
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Episode 3
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Episode 4
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Episode 5
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Episode 6
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Episode 7
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Episode 8
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Episode 9
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Episode 10
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