To write the equation of a line with a positive slope, calculate the slope as the change in Y divided by the change in X (rise over run), then use the point-slope form (y - y₁)/(x - x₁) = slope, where (x₁, y₁) is any point on the line; since similar triangles are formed by any two points on the line, the slope remains constant regardless of which two points are chosen, and algebraic manipulation of different point-slope forms yields the same final equation.
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Writing the equation of a line with a positive slope
Added:So, we're asked which equation represents this line, and they give us a line here, and they give us two points that sit on it. And like always, pause the video and see if you can figure it out on your own before we do it together.
Okay. So, before I even look at these choices, I'm actually going to try to figure out the equation of this line based on the slope. So, we have two points. We can calculate a slope between those two points. So, the definition of slope is our change in Y divided by change in X. Sometimes people will say rise over run, which is the same thing as change in Y. Greek letter delta represents change over change in X. So, what is our change in Y when we have a Y that is at two, and then we go to a Y that it is at eight, or sorry, a Y that is at 10. Our Y value is going from two to 10. Well, we can see that we went 1 2 3 4 5 6 7 8 up. You could also see it because the difference between 10 and two is eight. So, this is change in Y is equal to eight. And then when we had that change in Y, what was our change in X? Well, we went from X equals two to X equals eight. Whoops.
We went from X equals two to X equals eight. So, this is going to be equal to positive six. You can even count the squares. 1 2 3 4 5 6. So, our slope, our slope, which is defined as change in Y over change in X, or rise over run, is equal to 8 over 6, which we can rewrite. Both the numerator and the denominator is divisible by two. That's the same thing as 4 over 3.
Now, what we know about lines are, if I take any two points on this line, and I were to calculate the slope, I need to get the same slope. The slope is constant throughout this line. So, let's say I have some arbitrary point right over here, x, y.
x, y right over here. And x, y I could have drawn it anywhere cuz I'm just saying some arbitrary x and y. If I were to take the slope of between that and 8, so if I were to do something similar over here where I tried to do a I tried to do a rise over run, I should get the same I should get the same thing. So, what would be my change in y here?
And I'm not going to literally try to calculate this point. I mean, you can try to look at it and say, "That looks like 14." But I want you to just say, "Let's assume this is some arbitrary x and y." Well, if you had to calculate this change in y, you would say, "Change in y is equal to y - 10."
y - 10. And then what's this change in x right over here? Our change in x is going to be equal to this x, whatever this value is, - 8. x - 8. So here, the slope is going to be y - 10 y - 10 over x - 8.
The slope over here and we know that that has to equal 4/3 cuz we just calculated the slope of this line. So that equals 4 thirds. Now, do we see that as a choice anywhere over here? y - 10 over x - 8 is equal to 4/3.
y - 10 over x - 8 is equal to 4/3. I see 10s and 8s, but they don't seem to be in exactly the right place.
Now, maybe instead of using this point, they're using this point to calculate the equation. And to be clear, no matter how you calculate it, if you algebraically manipulate it, you should get the same thing. In fact, I'm going to show you that in a second. So, let's say we wanted to calculate the slope between x {comma} y and 2 {comma} 2.
Well, then our slope is going to be our change in y is going to be y minus this two and our change in x is going to be x minus this two and that also has to be equal to 4/3.
Well, luckily we do see that choice over here. So, for the sake of solving this problem, we know it is going to be B.
But, I want you to feel good about that this is actually the same equation as what I wrote over here. I'll do that algebraically in a second. And I also want you to appreciate that you pick any two points on this line. In fact, we could even pick these two points all the way over here, which is I just which is what I actually did just do. All of these triangles I drew are similar triangles. I can dilate from one of these triangles to the other and we could talk more about that in other videos. But, let me show you what I promised that I can manipulate both of these equations to get to the same place. So, if I multiply both sides of this equation by x minus 8, I get y minus 10 is equal to 4/3 * x 8 and then I could add 10 to both sides.
I get y is equal to 4/3 x - 8 + 10 and then this is getting a little bit messy here, but let me actually multiply out the the this part.
Actually, let me skip this so I don't get too messy. I get y is equal to 4/3 x minus what is this? 4 * 8 is 32 minus 32 over 3 + 10 and I can calculate this in a second. Let's see, this is 30 -52 over 3 is -10 and 2/3.
-10 and 2/3. So, if I add 10 to that, I I'm just left with -2/3.
So, all of this stuff before that simplifies to Y is equal to 4/3 X - 2/3.
Now, let me do it with this part here.
Maybe I'll do it on this part of my screen, so it's less messy.
If I multiply both sides of this equation by X - 2, I get Y - 2 is equal to 4/3 * X - 2. Add 2 to both sides, I get Y is equal to 4/3 * X - 2 + 2.
Distribute this 4/3. That's equal to 4/3 X and then 4/3 * -2 is -8/3 + 2. -8/3 is the same thing as 3 and 2 Sorry, 2 and 2/3.
So, this is -2 and 2/3. Well, if I have -2 and 2/3, and then I add 2, that adds up to -2/3. So, I have This is all equal to 4/3 X 2/3, which is exactly what I got from that other equation. So, even if they looked pretty different to begin with, they both are actually the same fundamental equation.
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