Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Pythagorean theorem. In a right triangle, the sum of the squares of the sides of a right triangle equal the square of the hypotenuse. So a ^2 + b ^2 = C ^2, where C is opposite the right angle and A&B make up the right angle. Hence A&B are the sides, C is the hypotenuse. If we look at a equal 3C equal 5. If we draw our triangle, A&B make up the right angle, so we'll put A at the bottom, the hypotenuse, where C is always opposite the right angle. So in this case we have 3 ^2 plus our unknown, which is going to be b ^2 is going to equal 5 ^2. 3 ^2 is 9 B squared. We don't know. We're trying to solve 5 ^2 is 25. If we subtract 925 -, 9 is 16, so b ^2 = 16, or B is 4 sqrt 16, which is 4. If we look at the next example, we have our A and our B, so we know the two sides that are making up our right angle. We're trying to find our C, so a ^2, 3 ^2 + b ^2 has got to equal C ^2, 3 ^2 is 93 squared again is 9/9 plus 9 is 18, and that's C ^2. So we're going to square root it. And if we think about 18, can we simplify that? Well, 18 is really 3 * 6 or 3 * 2 * 3, so we would get sqrt 2 * 3 ^2. We don't know sqrt 2, but we do know sqrt 3 ^2 because the index of two goes into the exponent of two one time. So our C is going to be 3 square roots of two. Looking at a few more examples, we have our A and our B again. So our A is root 3, our B is 2. We're trying to find our C, so root 3 ^2 plus b ^2 it's got equals C ^2. Root 3 ^2 is just going to be 3. Two squared is four, 3 + 4 is 7. To get the C, we're going to square root. So C is the distance of sqrt 7. The next one, we know A and we know C, so we know one of the sides that makes up the right angle, and we know the hypotenuse or the side opposite the right angle. So we want to find our B. This time 1 ^2 + b ^2 = sqrt 5 ^2. Well, the square root and the square are going to cancel, so we're going to get 1 + b ^2 equal 5. If we subtract the 1B squared equal, 4B is sqrt 4 which is 2. The distance formula is actually a special formula made-up from the idea of the Pythagorean theorem. If we give ourselves any two points X1Y1 and X2Y2, if I thought about forming a right triangle, we actually would know what this right angle vertice would be. We've gone over the same amount as this point up here, the X2, and we've gone up the same amount as this other point X1 Y 1. So if I wanted to think about just the horizontal here, how do I think about a distance on a number line? We take the farthest to the right and we subtract the farthest to the left, so X 2 -, X one. If I wanted to think about the height here, how do we figure out a height? We take the highest minus the lowest, so the Y 2 -, y one. If I'm trying to find D the distance between the original 2 points, I would know that d ^2 would equal X2 minus X 1 ^2 + y two minus Y 1 ^2. Well, how do I get rid of the square? If I want to get D by itself, I want to know the actual distance we're going to square root it. So D equal the square root of X2 minus X 1 ^2 + y two minus Y 1 ^2. That is the distance formula. That's a very important formula. Let's look at a few examples. If we're given two points 47 and 8:10, we could think of this as X1Y1 and X2Y2. So the distance formula says take and subtract the X's 8 -, 4 ^2 and subtract the YS and square it. So 8 - 4 ^2, 4 ^2. So 1610 - 7 ^2 is 3/3 squared is 9. So 4 ^2 + 3 ^2 16 + 9 sqrt 25. So we know the distance here is going to be 5. Now, doesn't matter which one we call X1 and X2. And the answer is no. If we had said 4 - 8 ^2 + 7 - 10 ^2, well, 4 - 8 is -4 ^2, 7 - 10 is -3 ^2 -4 ^2 is 16 negative 3 ^2 is 9 sqrt 16 + 9 is 25 or five. So once again, it doesn't matter which one the X1 and X2 represent when we square it. If it's a negative, it's going to turn positive anyway. So on this next one, 6 - 4 ^2 + -4 - -2 ^2 6 - 4 is 2 and we're going to square it -4 - -2 is going to be -2 ^2. So 2 ^2 4 -2 ^2 is another four. So the distance is sqrt 8 but sqrt 8 we can simplify 8 is 4 * 2 or 2 * 2 * 2. So we have sqrt 2 ^3. That understood index of 2 is going to go into three one time with one leftover. So we're going to get 2 square roots of two as our final distance. Thank you and have a wonderful day.