Hey, everyone. So in earlier videos, we saw the slope intercept form of a line, $y=mx+b$. And usually, we use this form whenever we were given information about the slope and intercept and asked for the graph of the equation or the other way around. So, for example, we were asked to graph something like $y=\frac{2}{3}x-1$. What I'm going to show you in this video is that sometimes you might be given an entirely different set of information. You might be told something like a slope, and then it passes through some point. So when you're asked to write an equation of a line that passes through a point that is not the y intercept, I'm going to show you that we actually use a different form of writing an equation called the point-slope form. Now I'm going to show you the differences and similarities between the two equations. So let's go ahead and get started here.

Again, I first want to talk about when you use these two different forms. We use $y=mx+b$ whenever we're given information about $m$ and $b$ and ask for the graph or the other way around. So, for example, $\frac{2}{3}x-1$, we're given the $m$ and the $b$, and we're asked to graph it. And we can see how to do that very quickly here. We just know that it's going to pass through the point (0, -1). That's the y intercept, and then we just sort of graph the next point by using rise over run. The slope is $\frac{2}{3}$, so you go up 2 over 3. And, basically, this line is going to look something like this. Alright. Now let's take a look at the equation the example that we're going to figure out solve here, which tells us that we're going to have to write the equation of a line that has a slope of $\frac{2}{3}$, and it passes through some random point (3, 1).

So, basically, whenever you're given information like this, whenever you're given the slope like we have here and some random points, which is going to be something like $(x_1,y_1)$, then you're going to use this new form called the point-slope form. You also could be given 2 points of information like $(x_1,y_1)$ and $(x_2,y_2)$. Alright? So that's the times where you're going to use this new point-slope form. And, basically, it's $y-y_1=m(x-x_1)$. So, really, with this equation, there are three numbers that you need to plug into this equation because these y's and x's actually don't get replaced with numbers. Alright? And so the reason we call it a point slope is because $x_1$ and $y_1$ really are just a point that it's giving you.

So let's get started here with part a. We're going to write the equation in this new form. So I'm going to take this equation here, $y-y_1=m(x-x_1)$. So I first need the slope of the line, and so I'm going to go ahead and figure that out first. And we are actually already told directly what the slope of this line is. It's just $\frac{2}{3}$. So we already have what that number is. So now all we need to do is just figure out the $x_1$ and the $y_1$. So what is that? Well, really, it's just the point that they're telling you that it passes through. When they say that it passes through the point (3, 1), what they're saying is that this line is going to pass through this point on the graph over here. That's just a coordinate. It's just they're really just giving you the $x_1$ and $y_1$ that you plug into the equation. So it's really straightforward. You just take these numbers and just pop them right into this equation over here, and you're basically done.

So if I want to just rewrite this equation, this is going to be $y-1=\frac{2}{3}(x-3)$. Alright? So this is the equation in point-slope form of the line that passes through this point over here and has a slope of $\frac{2}{3}$. That's it. That's the point-slope form. Alright? So we're basically done with part a. Now in part b, we're going to graph the line. So how do we do this? Well, again, we know this line is going to pass through this point (3, 1), but that's not enough for me to build the line because there's only one point. So the way we did this for slope intercept form was once we got the y intercept, we got the next point by using the slope. That's a very similar idea here. We know it passes through this point over here, so I can get the next point by using the rise over run in the slope. So I can go up 2 and over 3, and I'll end up over here, or I can go down 2 and to the left 3, I'm going to see that it passes through this point. So if I connect those points with a line, it's going to look something like this over here.

Now if you look at these two equations or these two graphs that we've ended up with, we've actually ended up with the exact same line. So, basically, what happens is that this equation over here, $\frac{2}{3}x-1$, and this equation that we wrote in point-slope form are actually the same exact equation written in 2 different ways. And the way I'm going to show that to you is we're going to use we're going to go through part c. We're going to rewrite my equation in slope intercept form. Alright? So how do I take this equation and now write this in $y=mx+b$? I basically just have to solve and isolate for this y term over here. So let's do that. I've got $y-1=\frac{2}{3}x-2$. So if I want $y$ by itself, first, I'm going to have to distribute the $\frac{2}{3}$ into everything that's inside the parentheses. So I get $y-1=\frac{2}{3}x-2$. So, now the last thing I have to do is just add 1 to both sides over here. Whatever you do to one side, you have to do to the other. And so I just get $y=\frac{2}{3}x-1$. So notice how now this is in slope-intercept form, and we've basically just gotten right back to the equation that we had on the left. So again, these two graphs over here and these two equations mean the exact same thing, but they're just written in 2 slightly different ways. Alright? So, hopefully, that made sense. Hopefully, you understand the difference between slope intercept and point-slope form. Let me know if you have any questions. Thanks for watching.