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Let me do that in blue color. Minus negative one squared plus y minus, y minus the y coordinate of the center, y minus one squared, is equal, is going to be equal to r squared, is going to be equal to the length of the radius squared. Well, r squared we already know is going to be 74. 74. And then if we want to simplify a little bit, you subtract a negative, this becomes a positive. So it simplifies to x plus one squared plus y minus one squared is equal to 74. Is equal to 74.
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Writing standard equation of a circle Mathematics II High School Math Khan Academy.mp3
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Now, we call the longest side of a right triangle, we call that side, and you can either view it as the longest side of the right triangle, or the side opposite the 90 degree angle, it is called the hypotenuse. It's a very fancy word for a fairly simple idea, just the longest side of a right triangle, or the side opposite the 90 degree angle. And it's just good to know that because someone might say hypotenuse, like, oh, they're just talking about this side right here, the side longest, the side opposite the 90 degree angle. Now, what I want to do in this video is prove a relationship, a very famous relationship, and you might see where this is going, a very famous relationship between the lengths of the sides of a right triangle. So let's say that the length of AC, so uppercase A, uppercase C, let's call that length lowercase a, let's call the length of BC, lowercase b right over here, I'll use uppercases for points, lowercases for lengths, and let's call the length of the hypotenuse, the length of AB, let's call that C. And let's see if we can come up with a relationship between A, B, and C. And to do that, I'm first going to construct another line, or another segment, I should say, between C and the hypotenuse, and I'm going to construct it so that they intersect at a right angle. And you can always do that, and we'll call this point right over here, we'll call this point capital D. And if you're wondering, how can you always do that, you can imagine rotating this entire triangle like this, and this isn't a rigorous proof, but it just kind of gives you the general idea of how you can always construct a point like this. So if I've rotated it around, so now our hypotenuse, we're now sitting on our hypotenuse, this is now point B, this is point A, and if I rotated the whole thing all the way around, this is point C, you can imagine just dropping a rock from point C, maybe with a string attached, and it would hit the hypotenuse at a right angle.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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Now, what I want to do in this video is prove a relationship, a very famous relationship, and you might see where this is going, a very famous relationship between the lengths of the sides of a right triangle. So let's say that the length of AC, so uppercase A, uppercase C, let's call that length lowercase a, let's call the length of BC, lowercase b right over here, I'll use uppercases for points, lowercases for lengths, and let's call the length of the hypotenuse, the length of AB, let's call that C. And let's see if we can come up with a relationship between A, B, and C. And to do that, I'm first going to construct another line, or another segment, I should say, between C and the hypotenuse, and I'm going to construct it so that they intersect at a right angle. And you can always do that, and we'll call this point right over here, we'll call this point capital D. And if you're wondering, how can you always do that, you can imagine rotating this entire triangle like this, and this isn't a rigorous proof, but it just kind of gives you the general idea of how you can always construct a point like this. So if I've rotated it around, so now our hypotenuse, we're now sitting on our hypotenuse, this is now point B, this is point A, and if I rotated the whole thing all the way around, this is point C, you can imagine just dropping a rock from point C, maybe with a string attached, and it would hit the hypotenuse at a right angle. So that's all we did here to establish segment CD, where we put our point D right over there. And the reason why I did that is now we can do all sorts of interesting relationships between similar triangles, because we have three triangles here. We have triangle ADC, we have triangle DBC, and then we have the larger original triangle, and we can hopefully establish similarity between those triangles.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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So if I've rotated it around, so now our hypotenuse, we're now sitting on our hypotenuse, this is now point B, this is point A, and if I rotated the whole thing all the way around, this is point C, you can imagine just dropping a rock from point C, maybe with a string attached, and it would hit the hypotenuse at a right angle. So that's all we did here to establish segment CD, where we put our point D right over there. And the reason why I did that is now we can do all sorts of interesting relationships between similar triangles, because we have three triangles here. We have triangle ADC, we have triangle DBC, and then we have the larger original triangle, and we can hopefully establish similarity between those triangles. And first I'll show you that ADC is similar to the larger one, because both of them have a right angle. ADC has a right angle right over here. Clearly if this angle is 90 degrees, then this angle is going to be 90 degrees as well.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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We have triangle ADC, we have triangle DBC, and then we have the larger original triangle, and we can hopefully establish similarity between those triangles. And first I'll show you that ADC is similar to the larger one, because both of them have a right angle. ADC has a right angle right over here. Clearly if this angle is 90 degrees, then this angle is going to be 90 degrees as well. They're supplementary. They have to add up to 180. And so they both have a right angle in them.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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Clearly if this angle is 90 degrees, then this angle is going to be 90 degrees as well. They're supplementary. They have to add up to 180. And so they both have a right angle in them. So the smaller one has a right angle, the larger one clearly has a right angle. That's where we started from. And they also both share this angle right over here.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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And so they both have a right angle in them. So the smaller one has a right angle, the larger one clearly has a right angle. That's where we started from. And they also both share this angle right over here. Angle DAC or BAC, however you want to refer to it. So we can actually write down that triangle. I'm going to start with the smaller one.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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And they also both share this angle right over here. Angle DAC or BAC, however you want to refer to it. So we can actually write down that triangle. I'm going to start with the smaller one. ADC, and maybe I'll shade it in right over here. So this is the triangle we're talking about. Triangle ADC, and I went from the blue angle to the right angle to the unlabeled angle from the point of view of triangle ADC.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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I'm going to start with the smaller one. ADC, and maybe I'll shade it in right over here. So this is the triangle we're talking about. Triangle ADC, and I went from the blue angle to the right angle to the unlabeled angle from the point of view of triangle ADC. This right angle isn't applying to that right over there. It's applying to the larger triangle. So we can say triangle ADC is similar to triangle.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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Triangle ADC, and I went from the blue angle to the right angle to the unlabeled angle from the point of view of triangle ADC. This right angle isn't applying to that right over there. It's applying to the larger triangle. So we can say triangle ADC is similar to triangle. Once again, you want to start at the blue angle, A. Then we went to the right angle, so we want to go to the right angle again. So it's ACB.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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So we can say triangle ADC is similar to triangle. Once again, you want to start at the blue angle, A. Then we went to the right angle, so we want to go to the right angle again. So it's ACB. And because they're similar, we can set up a relationship between the ratios of their sides. For example, we know the ratio of corresponding sides are going to, well, in general for a similar triangle, we know that the ratio of the corresponding sides are going to be a constant. So we could take the ratio of the hypotenuse of this side, of the smaller triangle.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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So it's ACB. And because they're similar, we can set up a relationship between the ratios of their sides. For example, we know the ratio of corresponding sides are going to, well, in general for a similar triangle, we know that the ratio of the corresponding sides are going to be a constant. So we could take the ratio of the hypotenuse of this side, of the smaller triangle. So the hypotenuse is AC. So AC over the hypotenuse over the larger one, which is AB. AC over AB is going to be the same thing as AD, as one of the legs.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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So we could take the ratio of the hypotenuse of this side, of the smaller triangle. So the hypotenuse is AC. So AC over the hypotenuse over the larger one, which is AB. AC over AB is going to be the same thing as AD, as one of the legs. AD, just to show that I'm just taking corresponding points on both similar triangles. This is AD over AC. You can look at these triangles yourself and show, look, AD, point AD is between the blue angle and the right angle.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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AC over AB is going to be the same thing as AD, as one of the legs. AD, just to show that I'm just taking corresponding points on both similar triangles. This is AD over AC. You can look at these triangles yourself and show, look, AD, point AD is between the blue angle and the right angle. Sorry, side AD is between the blue angle and the right angle. Side AC is between the blue angle and the right angle on the larger triangle. So both of these are from the larger triangle.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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You can look at these triangles yourself and show, look, AD, point AD is between the blue angle and the right angle. Sorry, side AD is between the blue angle and the right angle. Side AC is between the blue angle and the right angle on the larger triangle. So both of these are from the larger triangle. These are the corresponding sides on the smaller triangle. And if that is confusing looking at them visually, as long as we wrote our similarity statement correctly, you can just find the corresponding points. AC corresponds to AB on the larger triangle.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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So both of these are from the larger triangle. These are the corresponding sides on the smaller triangle. And if that is confusing looking at them visually, as long as we wrote our similarity statement correctly, you can just find the corresponding points. AC corresponds to AB on the larger triangle. AD on the smaller triangle corresponds to AC on the larger triangle. And we know that AC, we can rewrite that as lowercase a. AC is lowercase a. We don't have any label for AD or for AB.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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AC corresponds to AB on the larger triangle. AD on the smaller triangle corresponds to AC on the larger triangle. And we know that AC, we can rewrite that as lowercase a. AC is lowercase a. We don't have any label for AD or for AB. Sorry, we do have a label for AB. That is C right over here. We don't have a label for AD.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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We don't have any label for AD or for AB. Sorry, we do have a label for AB. That is C right over here. We don't have a label for AD. So let's just call that lowercase d. So lowercase d applies to that part right over there. C applies to that entire part right over there. And then we'll call DB, let's call that length E. That will just make things a little bit simpler for us.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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We don't have a label for AD. So let's just call that lowercase d. So lowercase d applies to that part right over there. C applies to that entire part right over there. And then we'll call DB, let's call that length E. That will just make things a little bit simpler for us. So AD we'll just call d. And so we have a over c is equal to d over a. If we cross multiply, you have a times a, which is a squared, is equal to c times d, which is cd. So that's a little bit of an interesting result.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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And then we'll call DB, let's call that length E. That will just make things a little bit simpler for us. So AD we'll just call d. And so we have a over c is equal to d over a. If we cross multiply, you have a times a, which is a squared, is equal to c times d, which is cd. So that's a little bit of an interesting result. Let's see what we can do with the other triangle right over here. So this triangle right over here. So once again, it has a right angle.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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So that's a little bit of an interesting result. Let's see what we can do with the other triangle right over here. So this triangle right over here. So once again, it has a right angle. The larger one has a right angle. And they both share this angle right over here. So by angle-angle similarity, the two triangles are going to be similar.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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So once again, it has a right angle. The larger one has a right angle. And they both share this angle right over here. So by angle-angle similarity, the two triangles are going to be similar. So we can say triangle BDC. We went from pink to right to not labeled. So triangle BDC is similar to triangle.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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So by angle-angle similarity, the two triangles are going to be similar. So we can say triangle BDC. We went from pink to right to not labeled. So triangle BDC is similar to triangle. Now we're going to look at the larger triangle. We're going to start at the pink angle, B. Now we go to the right angle, CA.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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So triangle BDC is similar to triangle. Now we're going to look at the larger triangle. We're going to start at the pink angle, B. Now we go to the right angle, CA. From pink angle to right angle to non-labeled angle, at least from the point of view here. We labeled it before with that blue. So now let's set up some type of a relationship here.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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Now we go to the right angle, CA. From pink angle to right angle to non-labeled angle, at least from the point of view here. We labeled it before with that blue. So now let's set up some type of a relationship here. We can say that the ratio on the smaller triangle, BC over BA. Once again, we're taking the hypotenuses of both of them. So BC over BA is going to be equal to BD.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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So now let's set up some type of a relationship here. We can say that the ratio on the smaller triangle, BC over BA. Once again, we're taking the hypotenuses of both of them. So BC over BA is going to be equal to BD. Let me do this in another color. So this is one of the legs. The way I drew it is the shorter legs.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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So BC over BA is going to be equal to BD. Let me do this in another color. So this is one of the legs. The way I drew it is the shorter legs. BD over BC. I'm just taking the corresponding vertices. Over BC.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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The way I drew it is the shorter legs. BD over BC. I'm just taking the corresponding vertices. Over BC. And once again, we know BC is the same thing as lowercase b. BA is lowercase c. And then BD we defined as lowercase e. This is lowercase e. We can cross multiply here and we get B times B. I've mentioned this in many videos. Cross multiplying is really the same thing as multiplying both sides by both denominators. B times B is B squared is equal to CE.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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Over BC. And once again, we know BC is the same thing as lowercase b. BA is lowercase c. And then BD we defined as lowercase e. This is lowercase e. We can cross multiply here and we get B times B. I've mentioned this in many videos. Cross multiplying is really the same thing as multiplying both sides by both denominators. B times B is B squared is equal to CE. And now we can do something kind of interesting. We can add these two statements. Let me rewrite the statement down here.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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B times B is B squared is equal to CE. And now we can do something kind of interesting. We can add these two statements. Let me rewrite the statement down here. So B squared is equal to CE. So if we add the left-hand sides, we get A squared plus B squared is equal to CD plus CE. And then we have a C in both of these terms so we can factor it out.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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Let me rewrite the statement down here. So B squared is equal to CE. So if we add the left-hand sides, we get A squared plus B squared is equal to CD plus CE. And then we have a C in both of these terms so we can factor it out. This is going to be equal to C times D plus E. Close the parentheses. Now what is D plus E? D is this length.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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And then we have a C in both of these terms so we can factor it out. This is going to be equal to C times D plus E. Close the parentheses. Now what is D plus E? D is this length. E is this length. So D plus E is actually going to be C as well. So this is going to be C. So you have C times C which is just the same thing as C squared.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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D is this length. E is this length. So D plus E is actually going to be C as well. So this is going to be C. So you have C times C which is just the same thing as C squared. So now we have an interesting relationship. We have that A squared plus B squared is equal to C squared. Let me rewrite that.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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So this is going to be C. So you have C times C which is just the same thing as C squared. So now we have an interesting relationship. We have that A squared plus B squared is equal to C squared. Let me rewrite that. A squared. I'll do that in an arbitrary new color. I deleted that by accident.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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Let me rewrite that. A squared. I'll do that in an arbitrary new color. I deleted that by accident. So let me rewrite it. So we've just established that A squared plus B squared is equal to C squared. And this is just an arbitrary right triangle.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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I deleted that by accident. So let me rewrite it. So we've just established that A squared plus B squared is equal to C squared. And this is just an arbitrary right triangle. This is true for any two right triangles. We've just established that the sum of the squares of each of the legs is equal to the square of the hypotenuse. And this is probably what's easily one of the most famous theorems in mathematics named for Pythagoras.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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And this is just an arbitrary right triangle. This is true for any two right triangles. We've just established that the sum of the squares of each of the legs is equal to the square of the hypotenuse. And this is probably what's easily one of the most famous theorems in mathematics named for Pythagoras. Not clear if he's the first person to establish this, but it's called the Pythagorean theorem. And it's really the basis of, well, not all of geometry, but a lot of the geometry that we're going to do. And it forms the basis of a lot of the trigonometry we're going to do.
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Pythagorean theorem proof using similarity Geometry Khan Academy (2).mp3
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Her character is on a quest to vanquish an evil sorcerer and his minions from the land. Her character is a wizard whose spells have a range of six meters. The locations of objects in the game are stored by the computer in terms of x and y coordinates. So 5, 4 is the location of Alyssa's wizard. 8, 7 is the location of minion A. 2, negative 1 is the location of minion B. 9, 0 is the location of minion C. So what I want to do, and I want you to pause this video, and I want you to think about, given that her wizard has a range of six meters, which of these minions can the wizard actually reach?
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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So 5, 4 is the location of Alyssa's wizard. 8, 7 is the location of minion A. 2, negative 1 is the location of minion B. 9, 0 is the location of minion C. So what I want to do, and I want you to pause this video, and I want you to think about, given that her wizard has a range of six meters, which of these minions can the wizard actually reach? I'm assuming you've given a go at it. And we just have to remind ourselves, to figure out which of these minions are in reach, we have to say, well, which of these points are within six units? We're assuming that these units are in meters right over here.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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9, 0 is the location of minion C. So what I want to do, and I want you to pause this video, and I want you to think about, given that her wizard has a range of six meters, which of these minions can the wizard actually reach? I'm assuming you've given a go at it. And we just have to remind ourselves, to figure out which of these minions are in reach, we have to say, well, which of these points are within six units? We're assuming that these units are in meters right over here. Which of these points are within six units of 5, 4? And to think about that, we just have to calculate the distance between this point and this point, that point and that point, that point and that point, and see if they are greater than or less than six meters. And how do we calculate a distance between two points?
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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We're assuming that these units are in meters right over here. Which of these points are within six units of 5, 4? And to think about that, we just have to calculate the distance between this point and this point, that point and that point, that point and that point, and see if they are greater than or less than six meters. And how do we calculate a distance between two points? Well, if this is some point right over here, that's x1 comma y1. And then this is another point right over here, x2 comma y2. And we want to calculate this distance right over here.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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And how do we calculate a distance between two points? Well, if this is some point right over here, that's x1 comma y1. And then this is another point right over here, x2 comma y2. And we want to calculate this distance right over here. The distance formula comes straight out of the Pythagorean theorem. The Pythagorean theorem tells us if this side right over here is our change in y, and let's actually just write that as the absolute value of our change in y. And let's say that this side right over here is the absolute value of our change in x.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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And we want to calculate this distance right over here. The distance formula comes straight out of the Pythagorean theorem. The Pythagorean theorem tells us if this side right over here is our change in y, and let's actually just write that as the absolute value of our change in y. And let's say that this side right over here is the absolute value of our change in x. The Pythagorean theorem tells us that this one, the hypotenuse, is going to be the square root of the sum of the squares of the two sides. So change in x squared plus change in y squared. You might say, hey, what happened to the absolute value?
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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And let's say that this side right over here is the absolute value of our change in x. The Pythagorean theorem tells us that this one, the hypotenuse, is going to be the square root of the sum of the squares of the two sides. So change in x squared plus change in y squared. You might say, hey, what happened to the absolute value? Well, when I square it, it's going to be positive anyway, so I don't have to write down the absolute value. So really, I just need to figure out between each of these two points, what is the change in x, what's the change in y, square them, add them together, and then take the square root. So for example, let's find, if I were to call this P1, if I were to call this P2, let me call this P3.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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You might say, hey, what happened to the absolute value? Well, when I square it, it's going to be positive anyway, so I don't have to write down the absolute value. So really, I just need to figure out between each of these two points, what is the change in x, what's the change in y, square them, add them together, and then take the square root. So for example, let's find, if I were to call this P1, if I were to call this P2, let me call this P3. I'm going to do them in different colors so you can keep track of what I'm doing. This is P3, and let's say this is P4. So let's first think about the distance.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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So for example, let's find, if I were to call this P1, if I were to call this P2, let me call this P3. I'm going to do them in different colors so you can keep track of what I'm doing. This is P3, and let's say this is P4. So let's first think about the distance. The distance between P1 and P2, well, that's going to be equal to the square root of our change in x squared. So our change in x is 3. That squared is 9.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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So let's first think about the distance. The distance between P1 and P2, well, that's going to be equal to the square root of our change in x squared. So our change in x is 3. That squared is 9. Plus our change in y squared. Our change in y is also 3. That squared is 9.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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That squared is 9. Plus our change in y squared. Our change in y is also 3. That squared is 9. So this is going to be square root of 18, which is the same thing as 3 square roots of 2. Now, is this more or less than 6? Well, 3 times 2 is equal to 6.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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That squared is 9. So this is going to be square root of 18, which is the same thing as 3 square roots of 2. Now, is this more or less than 6? Well, 3 times 2 is equal to 6. Square root of 2 is less than 2. It's 1 point something. So this right over here is going to be less than 6.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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Well, 3 times 2 is equal to 6. Square root of 2 is less than 2. It's 1 point something. So this right over here is going to be less than 6. So P2 is in range. Alyssa's wizard can get minion A. Minion A she can attack. Now let's think about minion B.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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So this right over here is going to be less than 6. So P2 is in range. Alyssa's wizard can get minion A. Minion A she can attack. Now let's think about minion B. So the distance between P1 and P3 is going to be equal to the square root of. So your change in x, it's negative 3. Negative 3 squared is positive 9.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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Now let's think about minion B. So the distance between P1 and P3 is going to be equal to the square root of. So your change in x, it's negative 3. Negative 3 squared is positive 9. Our change in y, to go from 4 to negative 1, it's negative 5. That squared is 25. So 9 plus 25, which is equal to the square root of 34.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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Negative 3 squared is positive 9. Our change in y, to go from 4 to negative 1, it's negative 5. That squared is 25. So 9 plus 25, which is equal to the square root of 34. Now, is this greater than or equal to 6? Well, the square root of 36 is 6. So this is a square root of a lower number.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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So 9 plus 25, which is equal to the square root of 34. Now, is this greater than or equal to 6? Well, the square root of 36 is 6. So this is a square root of a lower number. So this is going to be less than 6 as well. So P minion B is also in reach. Now let's think about this last point, the distance.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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So this is a square root of a lower number. So this is going to be less than 6 as well. So P minion B is also in reach. Now let's think about this last point, the distance. The distance between P1 and P4 is going to be equal to the square root of our change in x squared. Change in x is 4, squared is 16. Plus our change in y squared.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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Now let's think about this last point, the distance. The distance between P1 and P4 is going to be equal to the square root of our change in x squared. Change in x is 4, squared is 16. Plus our change in y squared. Our change in y is negative 4, but you square that, you get another 16. So this is going to be square root of 32, which could be written as, actually, we could just leave it as square root of 32. Square root of 32 is clearly less than the square root of 36, which is 6.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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Plus our change in y squared. Our change in y is negative 4, but you square that, you get another 16. So this is going to be square root of 32, which could be written as, actually, we could just leave it as square root of 32. Square root of 32 is clearly less than the square root of 36, which is 6. So this is also going to be less than 6. So she can get at all of the minions. They're all within 6 units of her.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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Square root of 32 is clearly less than the square root of 36, which is 6. So this is also going to be less than 6. So she can get at all of the minions. They're all within 6 units of her. Now, which of these is the furthest away? Well, actually, if we were to write this, we simplified this, but we could write this as the square root of 18. Square root of 18 is clearly the smallest out of square root of 18, square root of 32, and square root of 34.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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They're all within 6 units of her. Now, which of these is the furthest away? Well, actually, if we were to write this, we simplified this, but we could write this as the square root of 18. Square root of 18 is clearly the smallest out of square root of 18, square root of 32, and square root of 34. So minion A is the closest, and minion B, square root of 34, is the furthest. Or I should say the farthest. is the farthest.
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Which minions can the wizard reach Analytic geometry Geometry Khan Academy.mp3
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1, 2, 3, 4. So that's 3, negative 4. And I also have the point 6, 1. So 1, 2, 3, 4, 5, 6, 1. So just like that. 6, 1. In the last video, we figured out that we could just use the Pythagorean theorem if we wanted to figure out the distance between these two points.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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So 1, 2, 3, 4, 5, 6, 1. So just like that. 6, 1. In the last video, we figured out that we could just use the Pythagorean theorem if we wanted to figure out the distance between these two points. We just drew a triangle there and realized that this was the hypotenuse. In this video, we're going to try to figure out what is the coordinate of the point that is exactly halfway between this point and that point. So this right here is kind of the distance, the line that connects them.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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In the last video, we figured out that we could just use the Pythagorean theorem if we wanted to figure out the distance between these two points. We just drew a triangle there and realized that this was the hypotenuse. In this video, we're going to try to figure out what is the coordinate of the point that is exactly halfway between this point and that point. So this right here is kind of the distance, the line that connects them. Now what is the coordinate of the point that is exactly halfway in between the two? What is this coordinate right here? It's something comma something.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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So this right here is kind of the distance, the line that connects them. Now what is the coordinate of the point that is exactly halfway in between the two? What is this coordinate right here? It's something comma something. And to do that, let me draw it really big here. Because I think you're going to find out that it's actually pretty straightforward. At first it seems like a really tough problem.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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It's something comma something. And to do that, let me draw it really big here. Because I think you're going to find out that it's actually pretty straightforward. At first it seems like a really tough problem. Gee, let me use the distance formula with some variables. But you're going to see it's actually one of the simplest things you'll learn in algebra and geometry. So let's say that this is my triangle right there.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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At first it seems like a really tough problem. Gee, let me use the distance formula with some variables. But you're going to see it's actually one of the simplest things you'll learn in algebra and geometry. So let's say that this is my triangle right there. That is my triangle right there. This right here is the point 6, 1. This down here is the point 3 comma negative 4.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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So let's say that this is my triangle right there. That is my triangle right there. This right here is the point 6, 1. This down here is the point 3 comma negative 4. And we're looking for the point that is smack dab in between those two points. What are its coordinates? It seems very hard at first.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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This down here is the point 3 comma negative 4. And we're looking for the point that is smack dab in between those two points. What are its coordinates? It seems very hard at first. But it's easy when you think about it in terms of just the x and the y coordinates. What's this guy's x coordinate going to be? This line out here represents x is equal to 6.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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It seems very hard at first. But it's easy when you think about it in terms of just the x and the y coordinates. What's this guy's x coordinate going to be? This line out here represents x is equal to 6. This over here, let me do it in a little darker color, this over here represents x is equal to 6. This over here represents x is equal to 3. What will this guy's x coordinate be?
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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This line out here represents x is equal to 6. This over here, let me do it in a little darker color, this over here represents x is equal to 6. This over here represents x is equal to 3. What will this guy's x coordinate be? Well, his x coordinate is going to be smack dab in between the two x coordinates. This is x is equal to 3. This is x equal to 6.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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What will this guy's x coordinate be? Well, his x coordinate is going to be smack dab in between the two x coordinates. This is x is equal to 3. This is x equal to 6. He's going to be right in between. This distance is going to be equal to that distance. His x coordinate is going to be right in between the 3 and the 6.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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This is x equal to 6. He's going to be right in between. This distance is going to be equal to that distance. His x coordinate is going to be right in between the 3 and the 6. So what do we call the number that's right in between the 3 and the 6? Well, we could even call that the midpoint. Or we could call it the mean or the average, or however you want to talk about it.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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His x coordinate is going to be right in between the 3 and the 6. So what do we call the number that's right in between the 3 and the 6? Well, we could even call that the midpoint. Or we could call it the mean or the average, or however you want to talk about it. We just want to know what's the average of 3 and 6. So to figure out this point, the point halfway between 3 and 6, you literally just figure out 3 plus 6 over 2, which is equal to 4.5. So this x coordinate is going to be 4.5.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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Or we could call it the mean or the average, or however you want to talk about it. We just want to know what's the average of 3 and 6. So to figure out this point, the point halfway between 3 and 6, you literally just figure out 3 plus 6 over 2, which is equal to 4.5. So this x coordinate is going to be 4.5. Let me draw that on this graph. 1, 2, 3, 4.5. And you see it's smack dab in between.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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So this x coordinate is going to be 4.5. Let me draw that on this graph. 1, 2, 3, 4.5. And you see it's smack dab in between. That's its x coordinate. Now, by the exact same logic, this guy's y coordinate is going to be smack dab between y is equal to negative 4 and y is equal to 1. So it's going to be right in between those.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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And you see it's smack dab in between. That's its x coordinate. Now, by the exact same logic, this guy's y coordinate is going to be smack dab between y is equal to negative 4 and y is equal to 1. So it's going to be right in between those. So this is the x right there. The y coordinate is just going to be right in between y is equal to negative 4 and y is equal to 1. So you just take the average.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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So it's going to be right in between those. So this is the x right there. The y coordinate is just going to be right in between y is equal to negative 4 and y is equal to 1. So you just take the average. 1 plus negative 4 over 2, that's equal to negative 3 over 2, or you could say negative 1.5. So you go down 1.5. It is literally right there.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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So you just take the average. 1 plus negative 4 over 2, that's equal to negative 3 over 2, or you could say negative 1.5. So you go down 1.5. It is literally right there. So just like that, you literally take the average of the x's, take the average of the y's, or maybe I should say the mean to be a little bit more specific, a mean of only two points, and you will get the midpoint of those two points, the point that's equidistant from both of them. It's the midpoint of the line that connects them. So the coordinates are 4.5 comma negative 1.5.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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It is literally right there. So just like that, you literally take the average of the x's, take the average of the y's, or maybe I should say the mean to be a little bit more specific, a mean of only two points, and you will get the midpoint of those two points, the point that's equidistant from both of them. It's the midpoint of the line that connects them. So the coordinates are 4.5 comma negative 1.5. Let's do a couple more of these. These actually, you're going to find are very, very straightforward. But just to visualize it, let me graph it.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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So the coordinates are 4.5 comma negative 1.5. Let's do a couple more of these. These actually, you're going to find are very, very straightforward. But just to visualize it, let me graph it. Let's say I have the point 4, negative 5. So 1, 2, 3, 4, and then go down 5. 1, 2, 3, 4, 5.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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But just to visualize it, let me graph it. Let's say I have the point 4, negative 5. So 1, 2, 3, 4, and then go down 5. 1, 2, 3, 4, 5. So that's 4, negative 5. And I have the point 8 comma 2. So 1, 2, 3, 4, 5, 6, 7, 8 comma 2.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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1, 2, 3, 4, 5. So that's 4, negative 5. And I have the point 8 comma 2. So 1, 2, 3, 4, 5, 6, 7, 8 comma 2. So what is the coordinate of the midpoint of these two points, the point that is smack dab in between them? Well, we just average the x's, average the y's. So the midpoint is going to be the x values are 8 and 4.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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So 1, 2, 3, 4, 5, 6, 7, 8 comma 2. So what is the coordinate of the midpoint of these two points, the point that is smack dab in between them? Well, we just average the x's, average the y's. So the midpoint is going to be the x values are 8 and 4. So it's going to be 8 plus 4 over 2. And the y value is going to be, well, we have a 2 and a negative 5, so you get 2 plus negative 5 over 2. And what is this equal to?
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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So the midpoint is going to be the x values are 8 and 4. So it's going to be 8 plus 4 over 2. And the y value is going to be, well, we have a 2 and a negative 5, so you get 2 plus negative 5 over 2. And what is this equal to? This is 12 over 2, which is 6 comma 2 minus 5 is negative 3. Negative 3 over 2 is negative 1.5. So that right there is the midpoint.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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And what is this equal to? This is 12 over 2, which is 6 comma 2 minus 5 is negative 3. Negative 3 over 2 is negative 1.5. So that right there is the midpoint. You literally just average the x's and average the y's or find their mean. So let's graph it just to make sure it looks like the midpoint. 6, negative 5.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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So that right there is the midpoint. You literally just average the x's and average the y's or find their mean. So let's graph it just to make sure it looks like the midpoint. 6, negative 5. 1, 2, 3, 4, 5, 6. Negative 1.5. Negative 1, negative 1.5.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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6, negative 5. 1, 2, 3, 4, 5, 6. Negative 1.5. Negative 1, negative 1.5. Yep, looks pretty good. It looks like it's equidistant from this point and that point up there. Now, that's all you have to remember.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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Negative 1, negative 1.5. Yep, looks pretty good. It looks like it's equidistant from this point and that point up there. Now, that's all you have to remember. Average the x or take the mean of the x or find the x that's right in between the two. Average the y's. You've got the midpoint.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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Now, that's all you have to remember. Average the x or take the mean of the x or find the x that's right in between the two. Average the y's. You've got the midpoint. What I'm going to show you now is what's in many textbooks. They'll write, oh, if I have the point x1, y1, and then I have the point, actually I'm just sticking yellow. It's kind of painful to switch colors all the time.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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You've got the midpoint. What I'm going to show you now is what's in many textbooks. They'll write, oh, if I have the point x1, y1, and then I have the point, actually I'm just sticking yellow. It's kind of painful to switch colors all the time. And then I have the point x2, y2. Many books will give you something called the midpoint formula, which once again, I think is kind of silly to memorize. Just remember, you just average.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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It's kind of painful to switch colors all the time. And then I have the point x2, y2. Many books will give you something called the midpoint formula, which once again, I think is kind of silly to memorize. Just remember, you just average. Find the x in between, find the y in between. So midpoint formula. What they'll really say is the midpoint.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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Just remember, you just average. Find the x in between, find the y in between. So midpoint formula. What they'll really say is the midpoint. So maybe we'll say the midpoint x, or maybe I'll call it this way. I'm just making up notation. The x midpoint and the y midpoint is going to be equal to, and they'll give you this formula, x1 plus x2 over 2.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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What they'll really say is the midpoint. So maybe we'll say the midpoint x, or maybe I'll call it this way. I'm just making up notation. The x midpoint and the y midpoint is going to be equal to, and they'll give you this formula, x1 plus x2 over 2. And then y1 plus y2 over 2. And it looks like something you have to memorize. But all you have to say is, look, that's just the average or the mean of these two numbers.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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The x midpoint and the y midpoint is going to be equal to, and they'll give you this formula, x1 plus x2 over 2. And then y1 plus y2 over 2. And it looks like something you have to memorize. But all you have to say is, look, that's just the average or the mean of these two numbers. This is just the average or the mean of these two numbers. I'm just saying I'm adding the two together, dividing by 2. Adding these two together, dividing by 2.
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Midpoint formula Analytic geometry Geometry Khan Academy.mp3
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Several videos ago, I very quickly went through why side-side angle is not a valid postulate. What I want to do in this video is explore it a little bit more. It's not called angle-side-side for obvious reasons, because then the acronym would make people giggle in geometry class. I guess we don't want people giggling while they're doing mathematics. Let's just think about a triangle here. Let's say I have a triangle that looks something like this. If I have a triangle that looks something like that.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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I guess we don't want people giggling while they're doing mathematics. Let's just think about a triangle here. Let's say I have a triangle that looks something like this. If I have a triangle that looks something like that. Let's say that we've found another triangle that has a congruent side. A side that is congruent to this side right over here. That is next to any side on the triangle.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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If I have a triangle that looks something like that. Let's say that we've found another triangle that has a congruent side. A side that is congruent to this side right over here. That is next to any side on the triangle. Next to that is a side that is congruent to this side right over here. That side is one of the sides of an angle. It forms one of the parts of an angle right over here.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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That is next to any side on the triangle. Next to that is a side that is congruent to this side right over here. That side is one of the sides of an angle. It forms one of the parts of an angle right over here. That other triangle has a congruent angle right over here. This is the angle that that first side is not a part of. Only that second side is part of this angle.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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It forms one of the parts of an angle right over here. That other triangle has a congruent angle right over here. This is the angle that that first side is not a part of. Only that second side is part of this angle. This is side-side-angle, or you could call it angle-side-side. And giggle a little bit about it. How do we know that this doesn't by itself show that this is congruent?
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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Only that second side is part of this angle. This is side-side-angle, or you could call it angle-side-side. And giggle a little bit about it. How do we know that this doesn't by itself show that this is congruent? We'd have to show that this could actually imply two different triangles. To think about that, let's say we know that the angle, we know that this other triangle has that same yellow angle there. Which means that the blue side is going to have to look something like that.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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How do we know that this doesn't by itself show that this is congruent? We'd have to show that this could actually imply two different triangles. To think about that, let's say we know that the angle, we know that this other triangle has that same yellow angle there. Which means that the blue side is going to have to look something like that. Just the way we drew it over here. This side down here, I'll make it a green side. This green side down here, we know nothing about.
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More on why SSA is not a postulate Congruence Geometry Khan Academy.mp3
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