Okay, this is a lesson on distance, midpoint, and circles. The first two concepts, distance and midpoint, let's look at those. We're going to find the distance between two points and the midpoint of a segment. These two go very, very closely together.
And of course, this one fits in nicely with the distance formula. So, graphing an equation of a circle and writing the equation. Let's go ahead and begin with the distance formula.
So first thing I want to do is draw some axes, and then we need to label a point. We'll call this point X1, Y1. And then somewhere off over here, we'll put another point. Call it X2, Y2. And then I'd like to draw a line that...
connects those two and we're going to label this length d for distance. So make sure you're very clear with me. We've got a generic point, another generic point connected. I want to know how long this road is from here to here and I have to know these coordinates.
Okay where this is coming from now watch carefully this is coming from Draw a line, draw a line, make this a right angle, and it's coming off of the Pythagorean theorem. See, I can label this point. Now, help me with this point.
This point right here, it would come over in the x direction. It's going to come over the same amount as this one up here. So it's going to be an x2.
And then how high is it? What's the y value? It should be the same as this one.
So the point right here should be x2, y1. So the distance formula comes from, okay, let me just put up here, comes from the Pythagorean theorem. And the Pythagorean theorem says this distance across here, okay, so hold on. How would, what would it be like if I were just to draw a line? Just let me, let me just do some scratch paperwork down here.
Let's draw a line across there. And I say, oh, this is, just give it a number. This is 2 and this is 9. Then I would ask you.
How long is this piece? And hopefully you're going to say 7, because it's 9 minus 2. 7 units across here. So when I need to know this distance across here, it's what's the x value? See this x value is x2 lined up.
This x value is x1, so how long is this piece across here? And you say that's x2 minus x1. Now the same argument goes for this piece across here.
This would be the y values, y2 minus y1. So from the Pythagorean theorem, we get this length, which is x2 minus x1. Let's square it, because that's Pythagorean theorem.
This length, y2 minus y1, that quantity squared, equals d squared. So now if you'll take the square root of both sides, you'll get d is equal to, now we don't need a plus or minus because we're talking about a distance, so we just say, x2 minus x1 squared, i2 minus y1 squared. Add those up, take the square root, and you've got it.
That is the distance formula. So we want to take that formula and apply it. Let's do so.
Find the distance between the two points. Well, first thing I'm going to do... is the two points you gave me, here's an X and a Y, here's an X and a Y, I'll call this the first point.
and this the second point. Now recall with me, I'm just going to copy from the previous screen, that the distance formula is square root of x2 minus x1 squared plus y2 minus y1 squared. And so what I'm going to do is now just substitute these values in. So the distance is equal to have, let's just plug it in now. Next, I'm sorry, x2 is negative 2 minus a negative 4. So watch your signs.
That's going to be an easy little mistake if you're not careful. y2 is negative 5 minus, and then we have a 3 over here. Okay, so let's see what we get. It's going to be the square root of, this is a negative 2 plus 4. So negative 2 plus 4, that's 2 squared. Negative 5 minus 3 is negative 8 squared.
That's going to be a 4. That's a 64. And it looks like the square root of 68. Now you want to look to see if you can simplify that square root of 68. So what I'll do is I'll just start testing some things. Let me get a little scratch paper over here. Is a 4, does a 4 go into 68?
Let's check it. 4 goes into 6 one time. 4 goes into 28, 7. Yeah!
So 17 times. So we could come over here and rewrite this as... 4 times 17. So we're talking about that's the same thing as square root of 4 times square root of 17. So for algebra's sake, you want to simplify this to 2 square root of 17 units.
I mean, it is distance formula, so we need to be putting units on there of something. Meters, feet, miles, whatever. Okay, that's one great example. Let's look at another one.
Okay, so in this second example, I'm going to do the same thing. So I'm going to call this an x and a y, this an x and a y. This will be my first one, this will be my second one. And I'll start to put it all together. Let's see, so the formula says the distance is equal to the square root of...
And the only thing different in this example is we've got some square roots in there. We'll see what happens. So x2 is negative root 3 minus 2 root 3. We're going to square that plus y2 is 5 square roots of 6 minus square root of 6. All right, so begin simplifying. So this reads like negative 1 root 3. So we have negative 1 minus 2. So that's a negative 3 of the root 3s. And this one works the same way.
It's like 5 minus 1. So 5 minus 1 is 4. 4 of these root 6's. Okay. Continue simplifying.
Now, let me show you something. We're going to treat this really simple. You've got two things being multiplied in here. A negative 3 times the square root of 3. You want to square it.
So what you want is the negative 3 squared. So that's 9. Times. the square root of three squared, which is just three. So nine times three plus, do it again, 16 times six.
Let's see what that is. Nine times three is 27. 16 times six is 96. And 96 and 27 makes 13, 11, 123. So square root of 123. And I start to think, let me see, what's perfect inside of there? Is there a 4, 5, 6, nope, wait, nope, 7, nope. So that's as far as we can go.
It is already simplified. And so there's my answer. This is considered in math the exact value.
Now I can put that in my calculator. For those of you who are like, what is the square root of 123? Well, if I type it in, it's approximately 11.1 units.
I mean, that's a good practical answer. But anyway, in algebra class, the exact value is just fine. All right, one last example down here.
Let's take a look at that. Okay, I'm going to treat this one the same way. I encourage you to try to work ahead of me. Pause the video, maybe forward it up, see if you get the same answer. But I'm going to work it the same way.
I've got an X and a Y, so I'll call this my first one and my second one. And it doesn't matter which one. All right, so we come back over here and we say the distance is equal to...
square root of, all right, where's x2? 1 3rd minus 7 3rds, and we're going to square that. And, wow, this one's nice. There's no negatives in here.
So 6 5ths minus 1 5th, we're going to square that. So now let's start to simplify. So we even have common denominators. It just doesn't get any better.
1 minus 7 is negative 6 thirds. Negative 6 over 3, isn't that just negative 2 when you do the division? 6 fifths minus 1 fifth is 5 fifths, which is 1. So this is even better. So we end up with 4. And so we get square root of 5, and that won't simplify any further. So we're just going to call it square root of 5 units.
And there's your distance. So good. That takes care of the distance formula.