 So, again, one important thing we want to be able to do is to work back and forth between algebra and geometry. So, distance and midpoint are geometric concepts, and so the question we want to ask is how do we find these algebraically? So, the distance problem is the following. Given two points, we can try to find the distance between them. So, a useful strategy in pre-calculus, and calculus, and calculus two, and, well, actually every math course after this one, is to look for the right triangle. Now, there will be several different reasons for one to look for the right triangle, but for now, the important one is that given a right triangle, we have a relationship between the lengths of the sides. And that relationship is either known as the Pythagorean theorem, or the right triangle theorem. Let a and b be the lengths of the sides of a right triangle, and c the length of the hypotenuse. Then, a squared plus b squared equals c squared. So, let's see if we can find the distance between the points three, negative one, and negative one, five. So, the great secret about how to solve problems involving graphing easily is this. If a problem involves graphing, graph. We want to find the distance between these two points. Well, let's graph those two points and draw what that distance is. And now, let's look for the right triangle. The key here is to not make things too difficult. The lengths we're interested in runs between these two points, so we can make the distance between the points the hypotenuse of a right triangle. And notice if I drop horizontal and vertical lines from these points, they'll form the other two sides of the right triangle. So now I have to figure out my lengths. So, let's consider our horizontal side. This horizontal line begins here at a point with x-coordinate, negative one, and ends here above a point with x-coordinate, three. And that means the length of this horizontal segment is going to be four units. On the other hand, this vertical side starts at the same level as a point with y-coordinate, five, and drops down to a point with y-coordinate, negative one. So the length of this vertical side will be six. So I have two sides of the right triangle, and I can find the third side, the hypotenuse c will satisfy c squared equals a squared plus b squared. So we're placing our values for a and b, then solving for c. And since c is a distance, we only need the principal square root of 52. And so we find the distance between the two points, square root of 52. We can convert this into a formula as follows. If our points are located at x1, y1, and x2, y2, then we see that one leg of our right triangle will have length absolute difference between x2 and x1. The other will have length absolute difference between y2 and y1. So the hypotenuse c will satisfy c squared equals... Now, since we're squaring the absolute difference, we don't actually need to keep the absolute value symbols, so we'll drop them out. And since we actually want the distance, we'll take the square root. And this will give us our distance formula. And again, an important reminder, remember concepts, not formulas. While it's useful to know the distance formula, it's important to keep in mind that it's just based on the Pythagorean theorem, and it's more important to understand this idea of looking for the right triangle. So again, we actually solved this problem without having to use a formula. If we were to use the formula, we can find the distance between the two points. So our x1, y1 are the coordinates of one set of points, and x2, y2 are the coordinates of the other set of points, substituting those in and get square root of 52 once again. So once I can talk about the distance between two points, I can also talk about the midpoint, the point that is halfway between the two. So again, the secret to solving problems involving graphing is to graph. So remember, the secret to graphing is to graph first, then label. So we know this point 1, 2 is over some, up some, and we know it's in the right place because we say it's in the right place. Likewise, the point 3, 8 is over more, up more. And again, we label it. Now, fair warning, this works when we're drawing a graph to organize our thoughts. But if we actually need to draw an accurate graph, we'll have to be a little bit more careful. So let's find that midpoint. Well, it's someplace in between the two, so let's draw the line between them. And our midpoint is here. So now let's look for the right triangle. And again, it doesn't pay to be too fancy. We'll draw our sides horizontally and vertically. And we'll use the points that we have as the vertices of a right triangle. So we might begin by going horizontally and vertically from our two given points. But then if we include the midpoint, there's two others. Now, because this is supposed to be the midpoint, then the x-coordinate will be midway between this x-coordinate, 1, and this x-coordinate, 3. So the x-coordinate of this point will be 2. Likewise, our y-coordinate, well, that must be midway between the y-coordinate of 2 and the y-coordinate of 8. So it must be 5. And so now we have the coordinates of the midpoint. And again, we can encapsulate this in another formula. If our points are x1, y1, and x2, y2, then the x-coordinate of the midpoint will be halfway between the two of them. And similarly for the y-coordinate. And so the coordinates of the midpoint will be... So again, we already found this midpoint by using the concept of the midpoint, but we can also use the formula. And again, remember it's more important to understand the concept than it is to memorize the formula. Well, here's the formula anyway. So x1, y1 are the coordinates of one point, and x2, y2 are the coordinates of the other point, so we'll substitute those in. And now let's verify that this is the midpoint. So the key idea is that if 2, 5 is really the midpoint, then it will be the same distance from both points and halfway between them. So we need to calculate what those distances are. So first, let's find the distance between 1, 2, and 3, 8. So using our distance formula, we find the distance between 1, 2, and 3, 8 is... square root of 40. The distance between 1, 2, and 2, 5 is... square root of 10. And the distance between 2, 5, and 3, 8 is... also square root of 10. And so we see that 2, 5 is the same distance between 1, 2, and 3, 8. And if we reduce it, we find that square root of 40 is, in fact, 2 times the square root of 10. So our point 2, 5 is, in fact, halfway between 1, 2, and 3, 8, which confirms that 2, 5 is, in fact, the midpoint.