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And so you could see why that will be useful in proving that the slope is constant here. Because all we have to do is, look, if these two triangles are similar, then the ratio between corresponding sides is always going to be the same. We've picked two arbitrary sets of points. Then this would be true really for any two arbitrary sets of points across the line. It would be true for the entire line. So let's try to prove similarity. So the first thing we know is that both of these are right triangles.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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Then this would be true really for any two arbitrary sets of points across the line. It would be true for the entire line. So let's try to prove similarity. So the first thing we know is that both of these are right triangles. These green lines are perfectly horizontal. These purple lines are perfectly vertical. Because the green lines literally go along the x-axis or go in the horizontal direction.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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So the first thing we know is that both of these are right triangles. These green lines are perfectly horizontal. These purple lines are perfectly vertical. Because the green lines literally go along the x-axis or go in the horizontal direction. The purple lines go in the vertical direction. So let me make sure that we mark that. So we know that these are both right angles.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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Because the green lines literally go along the x-axis or go in the horizontal direction. The purple lines go in the vertical direction. So let me make sure that we mark that. So we know that these are both right angles. So we have one corresponding angle that is congruent. Now we have to show that the other ones are. And we can show that the other ones are using our knowledge of parallel lines and transversals.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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So we know that these are both right angles. So we have one corresponding angle that is congruent. Now we have to show that the other ones are. And we can show that the other ones are using our knowledge of parallel lines and transversals. Let's look at these two green lines. So now I'll continue them. These are line segments.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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And we can show that the other ones are using our knowledge of parallel lines and transversals. Let's look at these two green lines. So now I'll continue them. These are line segments. But if we view them as lines and we just continue them on and on and on. So let me do that just like here. So this line is clearly parallel to that one.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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These are line segments. But if we view them as lines and we just continue them on and on and on. So let me do that just like here. So this line is clearly parallel to that one. They essentially are perfectly horizontal. And now you can view our orange line as a transversal. And if you view it as a transversal, then you know that this angle corresponds to this angle.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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So this line is clearly parallel to that one. They essentially are perfectly horizontal. And now you can view our orange line as a transversal. And if you view it as a transversal, then you know that this angle corresponds to this angle. And we know from transversals of parallel lines that corresponding angles are congruent. So this angle is going to be congruent to that angle right over there. Now we make a very similar argument for this angle.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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And if you view it as a transversal, then you know that this angle corresponds to this angle. And we know from transversals of parallel lines that corresponding angles are congruent. So this angle is going to be congruent to that angle right over there. Now we make a very similar argument for this angle. But now we use the two vertical lines. We know that this segment, we could continue it as a line. So we could continue it if we wanted as a line.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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Now we make a very similar argument for this angle. But now we use the two vertical lines. We know that this segment, we could continue it as a line. So we could continue it if we wanted as a line. So just like that, a vertical line. And we could continue this one as a vertical line. We know that these are both vertical.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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So we could continue it if we wanted as a line. So just like that, a vertical line. And we could continue this one as a vertical line. We know that these are both vertical. They're just measuring. They're exactly in the y direction, the vertical direction. So this line is parallel to this line right over here.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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We know that these are both vertical. They're just measuring. They're exactly in the y direction, the vertical direction. So this line is parallel to this line right over here. Once again, our orange line is a transversal of it. And this angle corresponds to this angle right over here. And there we have it.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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So this line is parallel to this line right over here. Once again, our orange line is a transversal of it. And this angle corresponds to this angle right over here. And there we have it. They're congruent. Corresponding angles of transversal of two parallel lines are congruent. We learned that in geometry class.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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And there we have it. They're congruent. Corresponding angles of transversal of two parallel lines are congruent. We learned that in geometry class. And there you have it. All of the corresponding, this angle is congruent to this angle. This angle is congruent to that angle.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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We learned that in geometry class. And there you have it. All of the corresponding, this angle is congruent to this angle. This angle is congruent to that angle. And then both of these are 90 degrees. So both of these are similar triangles. So both of these are similar.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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This angle is congruent to that angle. And then both of these are 90 degrees. So both of these are similar triangles. So both of these are similar. So let me write that down. So we know that these are both similar triangles. And now we can use the common ratio of both sides.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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So both of these are similar. So let me write that down. So we know that these are both similar triangles. And now we can use the common ratio of both sides. So for example, if we called this side length a, and we said that this side has length b, and we said this side has length c, and this side has length d, we know for a fact that the ratio, because these are similar triangles, the ratio between corresponding sides, the ratio of a to b, is going to be equal to the ratio of c to d. And that ratio is literally the definition of slope, your change in y over your change in x. And this is constant, because any triangle that you generate, this right triangle that you generate between these two points, we've just shown that they are going to be similar. And if they are similar, then the ratio of this vertical line, the length of this vertical line segment to this horizontal line segment is constant.
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Similar triangles to prove that the slope is constant for a line Algebra I Khan Academy.mp3
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We have three lines and we have to figure out which of the three are parallel. So line A, and it can't be parallel on its own, it has to be parallel to another of the three lines. So the equation for line A is y is equal to 3 4ths x minus 4. Line B is 4y minus 20 is equal to negative 3x. And then line C is negative 3x plus 4y is equal to 40. So to figure out if any of these lines are parallel to any of the other lines, we just have to compare their slopes. If any two of these lines have the same slope and they're different lines, they have different y-intercepts, then they're going to be parallel.
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Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3
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Line B is 4y minus 20 is equal to negative 3x. And then line C is negative 3x plus 4y is equal to 40. So to figure out if any of these lines are parallel to any of the other lines, we just have to compare their slopes. If any two of these lines have the same slope and they're different lines, they have different y-intercepts, then they're going to be parallel. Now line A, it's very easy to figure out its slope. It's already in slope-intercept form. This is mx plus b.
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Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3
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If any two of these lines have the same slope and they're different lines, they have different y-intercepts, then they're going to be parallel. Now line A, it's very easy to figure out its slope. It's already in slope-intercept form. This is mx plus b. The slope is 3 4ths and the y-intercept, which isn't as relevant when you're figuring out parallel lines, is negative 4. So let's see what the other character's slopes are. This isn't in any kind of standard form.
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Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3
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This is mx plus b. The slope is 3 4ths and the y-intercept, which isn't as relevant when you're figuring out parallel lines, is negative 4. So let's see what the other character's slopes are. This isn't in any kind of standard form. It's not in standard form, slope-intercept, or point-slope form. But let's see what the slope of this line is. So to get it into slope-intercept form, which is really the easiest one to pick out the slope from, let's add 20 to both sides of this equation.
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Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3
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This isn't in any kind of standard form. It's not in standard form, slope-intercept, or point-slope form. But let's see what the slope of this line is. So to get it into slope-intercept form, which is really the easiest one to pick out the slope from, let's add 20 to both sides of this equation. So let's add 20 to both sides. The left-hand side, those cancel out, that was the whole point. You get 4y is equal to negative 3x plus 20.
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Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3
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So to get it into slope-intercept form, which is really the easiest one to pick out the slope from, let's add 20 to both sides of this equation. So let's add 20 to both sides. The left-hand side, those cancel out, that was the whole point. You get 4y is equal to negative 3x plus 20. And now we can divide everything by 4. Just dividing both sides of this equation by 4. We are left with y is equal to negative 3 4ths x plus 5.
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Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3
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You get 4y is equal to negative 3x plus 20. And now we can divide everything by 4. Just dividing both sides of this equation by 4. We are left with y is equal to negative 3 4ths x plus 5. So in this case, y-intercept is 5, but most importantly, the slope is negative 3 4ths. So it's different than this guy. This is negative 3 4ths, this is positive 3 4ths.
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Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3
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We are left with y is equal to negative 3 4ths x plus 5. So in this case, y-intercept is 5, but most importantly, the slope is negative 3 4ths. So it's different than this guy. This is negative 3 4ths, this is positive 3 4ths. So these two guys definitely aren't parallel. Let's move on to this guy. This guy written in standard form.
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Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3
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This is negative 3 4ths, this is positive 3 4ths. So these two guys definitely aren't parallel. Let's move on to this guy. This guy written in standard form. Let's get the x term on the other side. So let's add 3x to both sides of this equation. Left-hand side, these cancel out.
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Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3
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This guy written in standard form. Let's get the x term on the other side. So let's add 3x to both sides of this equation. Left-hand side, these cancel out. We're just left with 4y is equal to 3x plus 40, or 40 plus 3x, either way. Now we can divide both sides by 4. We have to divide every term by 4.
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Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3
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Left-hand side, these cancel out. We're just left with 4y is equal to 3x plus 40, or 40 plus 3x, either way. Now we can divide both sides by 4. We have to divide every term by 4. The left-hand side, you're left with y. The right-hand side, you have 3 4ths x plus 10. So here, our slope is 3 4ths, and our y-intercept, if we care about it, is 10.
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Parallel lines from equation (example 2) Mathematics I High School Math Khan Academy.mp3
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What I want to do in this video is get some practice visualizing what happens if we were to try to rotate two-dimensional shapes in three dimensions. What do I mean by that? Let's say I started with a right triangle. So let's say my right triangle looks like this. So let's say it looks like that right over there. And so this is a right angle. And let's say that this width right over here is three units.
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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So let's say my right triangle looks like this. So let's say it looks like that right over there. And so this is a right angle. And let's say that this width right over here is three units. And let's say that this length is five units. And I want to do something interesting. I'm going to take this two-dimensional right triangle and I'm going to try to rotate it in three dimensions around this line, around the line that I'm doing as a dotted magenta line.
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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And let's say that this width right over here is three units. And let's say that this length is five units. And I want to do something interesting. I'm going to take this two-dimensional right triangle and I'm going to try to rotate it in three dimensions around this line, around the line that I'm doing as a dotted magenta line. So I'm going to rotate it around this line right over there. So if I were to rotate it around this line, what type of a shape am I going to get? And I encourage you, once again, it's going to be a three-dimensional shape.
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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I'm going to take this two-dimensional right triangle and I'm going to try to rotate it in three dimensions around this line, around the line that I'm doing as a dotted magenta line. So I'm going to rotate it around this line right over there. So if I were to rotate it around this line, what type of a shape am I going to get? And I encourage you, once again, it's going to be a three-dimensional shape. I encourage you to think about it. Maybe take out a piece of paper, draw it, or just try to imagine it in your head. Well, to think about it in three dimensions, what I'm going to do is try to look at this thing in three dimensions.
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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And I encourage you, once again, it's going to be a three-dimensional shape. I encourage you to think about it. Maybe take out a piece of paper, draw it, or just try to imagine it in your head. Well, to think about it in three dimensions, what I'm going to do is try to look at this thing in three dimensions. So let me draw this same line, but I'm going to draw it in an angle so we can visualize the whole thing in three dimensions. So imagine if this was sitting on the ground. So that's our magenta line.
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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Well, to think about it in three dimensions, what I'm going to do is try to look at this thing in three dimensions. So let me draw this same line, but I'm going to draw it in an angle so we can visualize the whole thing in three dimensions. So imagine if this was sitting on the ground. So that's our magenta line. And then I can draw my triangle. So my triangle would look something like this. So it would look like this.
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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So that's our magenta line. And then I can draw my triangle. So my triangle would look something like this. So it would look like this. So once again, this is five units, this is three units, this is a right triangle. I'm going to rotate it around the line. So what's it going to look like?
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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So it would look like this. So once again, this is five units, this is three units, this is a right triangle. I'm going to rotate it around the line. So what's it going to look like? Well, this end right over here is going to rotate around, and it's going to form a circle with a radius of three. Right? So it's going to form, so when it intersects, if that was on the ground, it's going to be three again.
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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So what's it going to look like? Well, this end right over here is going to rotate around, and it's going to form a circle with a radius of three. Right? So it's going to form, so when it intersects, if that was on the ground, it's going to be three again. And let me draw it down. So it's going to keep going down. Whoops, don't want to press the wrong button.
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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So it's going to form, so when it intersects, if that was on the ground, it's going to be three again. And let me draw it down. So it's going to keep going down. Whoops, don't want to press the wrong button. So it's going to look something like this. That's what the base is going to look like. But then this end right over here is just going to stay at a point, because this is right on that magenta line.
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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Whoops, don't want to press the wrong button. So it's going to look something like this. That's what the base is going to look like. But then this end right over here is just going to stay at a point, because this is right on that magenta line. So it's just going to stay at a point. And so if you were to look at the intersect, so it would look something like this. So it would look like this, and then you'd have another thing that goes like this.
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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But then this end right over here is just going to stay at a point, because this is right on that magenta line. So it's just going to stay at a point. And so if you were to look at the intersect, so it would look something like this. So it would look like this, and then you'd have another thing that goes like this. And so if you were to take a section like this, you would have a little smaller circle here based on what this distance is. So what is the shape? What is the shape that I am drawing?
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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So it would look like this, and then you'd have another thing that goes like this. And so if you were to take a section like this, you would have a little smaller circle here based on what this distance is. So what is the shape? What is the shape that I am drawing? Well, what you see, what it is, it's a cone. It's a cone, and if I shade it in, you might see the cone a little bit better. So let me shade it in.
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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What is the shape that I am drawing? Well, what you see, what it is, it's a cone. It's a cone, and if I shade it in, you might see the cone a little bit better. So let me shade it in. So you see the cone. So what you end up getting is a cone where it's base, so I'm shading it in. So that hopefully helps a little bit.
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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So let me shade it in. So you see the cone. So what you end up getting is a cone where it's base, so I'm shading it in. So that hopefully helps a little bit. So what you end up getting is a cone where the base has a radius of three units. So let me draw this. So this right over here is the radius of the base, and it is three units.
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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So that hopefully helps a little bit. So what you end up getting is a cone where the base has a radius of three units. So let me draw this. So this right over here is the radius of the base, and it is three units. I could also draw it like this. So the cone is going to look like this, and this is the tip of the cone, and it's going to look just like this. And once again, let me shade it a little bit so that you can appreciate that this is a three-dimensional shape.
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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So this right over here is the radius of the base, and it is three units. I could also draw it like this. So the cone is going to look like this, and this is the tip of the cone, and it's going to look just like this. And once again, let me shade it a little bit so that you can appreciate that this is a three-dimensional shape. So draw the cone so you can shade it, and we can even construct the original so that, well, or we can construct this original shape so you see how it constructs, so it makes this, the line, that magenta line, is going to do this type of thing. It's going to go through the center of the base. It's going to go through the center of the base just like that, and our original shape, our original right triangle, if you just took a cross-section of it that included that line, you would have your original shape.
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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And once again, let me shade it a little bit so that you can appreciate that this is a three-dimensional shape. So draw the cone so you can shade it, and we can even construct the original so that, well, or we can construct this original shape so you see how it constructs, so it makes this, the line, that magenta line, is going to do this type of thing. It's going to go through the center of the base. It's going to go through the center of the base just like that, and our original shape, our original right triangle, if you just took a cross-section of it that included that line, you would have your original shape. Let me just say in orange. So the original shape is right over there. So what do you get?
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Rotating 2D shapes in 3D Perimeter, area, and volume Geometry Khan Academy.mp3
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The equation of a circle C is x plus 3 squared plus y minus 4 squared is equal to 49. What are its center hk and its radius r? So let's just remind ourselves what a circle is. You have some point. Let's call that hk. The circle is a set of all points that are equidistant from that point. So let's take the set of all points that are, say, r away from hk.
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Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3
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You have some point. Let's call that hk. The circle is a set of all points that are equidistant from that point. So let's take the set of all points that are, say, r away from hk. So let's say that this distance right over here, this distance right over here is r. And so we want all the set of points that are exactly r away. So all the points x comma y that are exactly r away. And so you could imagine you could rotate around and all of these points are going to be exactly r away.
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Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3
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So let's take the set of all points that are, say, r away from hk. So let's say that this distance right over here, this distance right over here is r. And so we want all the set of points that are exactly r away. So all the points x comma y that are exactly r away. And so you could imagine you could rotate around and all of these points are going to be exactly r away. And I'm going to try my best to draw at least a somewhat perfect looking circle. I won't be able to do a perfect job of it. But you get a sense.
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Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3
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And so you could imagine you could rotate around and all of these points are going to be exactly r away. And I'm going to try my best to draw at least a somewhat perfect looking circle. I won't be able to do a perfect job of it. But you get a sense. All of these are exactly r away, at least if I were to draw it properly. They are r away. So how do we find an equation in terms of r and hk and x and y that describes all these points?
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Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3
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But you get a sense. All of these are exactly r away, at least if I were to draw it properly. They are r away. So how do we find an equation in terms of r and hk and x and y that describes all these points? Well, we know how to find the distance between two points on a coordinate plane. And in fact, it comes straight out of the Pythagorean theorem. If we were to draw a vertical line right over here that essentially is the change in the vertical axis between these two points, up here we're at y, here we're k. So this distance is going to be y minus k. We can do the exact same thing on the horizontal axis.
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Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3
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So how do we find an equation in terms of r and hk and x and y that describes all these points? Well, we know how to find the distance between two points on a coordinate plane. And in fact, it comes straight out of the Pythagorean theorem. If we were to draw a vertical line right over here that essentially is the change in the vertical axis between these two points, up here we're at y, here we're k. So this distance is going to be y minus k. We can do the exact same thing on the horizontal axis. This x-coordinate is x, while this x-coordinate is h. So this is going to be x minus h is this distance. And this is a right triangle because by definition, we're saying, hey, we're measuring vertical distance here. We're measuring horizontal distance here.
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Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3
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If we were to draw a vertical line right over here that essentially is the change in the vertical axis between these two points, up here we're at y, here we're k. So this distance is going to be y minus k. We can do the exact same thing on the horizontal axis. This x-coordinate is x, while this x-coordinate is h. So this is going to be x minus h is this distance. And this is a right triangle because by definition, we're saying, hey, we're measuring vertical distance here. We're measuring horizontal distance here. So these two things are perpendicular. And so from the Pythagorean theorem, we know that this squared plus this squared must be equal to our distance squared. And this is where the distance formula comes from.
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Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3
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We're measuring horizontal distance here. So these two things are perpendicular. And so from the Pythagorean theorem, we know that this squared plus this squared must be equal to our distance squared. And this is where the distance formula comes from. So we know that x minus h squared plus y minus k squared must be equal to r squared. This is the equation for the set. This describes any x and y that satisfies this equation will sit on this circle.
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Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3
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And this is where the distance formula comes from. So we know that x minus h squared plus y minus k squared must be equal to r squared. This is the equation for the set. This describes any x and y that satisfies this equation will sit on this circle. Now with that out of the way, let's go answer their question. The equation of the circle is this thing. And this looks awfully close to what we just wrote.
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Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3
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This describes any x and y that satisfies this equation will sit on this circle. Now with that out of the way, let's go answer their question. The equation of the circle is this thing. And this looks awfully close to what we just wrote. We just have to make sure that we don't get confused with the negatives. Remember, it has to be in the form x minus h, y minus k. So let's write it a little bit differently. Instead of x plus 3 squared, we can write that as x minus negative 3 squared.
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Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3
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And this looks awfully close to what we just wrote. We just have to make sure that we don't get confused with the negatives. Remember, it has to be in the form x minus h, y minus k. So let's write it a little bit differently. Instead of x plus 3 squared, we can write that as x minus negative 3 squared. And then plus, well, this is already in the form, plus y minus 4 squared is equal to, instead of 49, we can just call that 7 squared. And so now it becomes pretty clear that our h is negative 3. I wanted to do that in the red color.
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Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3
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Instead of x plus 3 squared, we can write that as x minus negative 3 squared. And then plus, well, this is already in the form, plus y minus 4 squared is equal to, instead of 49, we can just call that 7 squared. And so now it becomes pretty clear that our h is negative 3. I wanted to do that in the red color. That our h is negative 3, and that our k is positive 4, and that our r is 7. So we could say h comma k is equal to negative 3 comma positive 4. You might say, hey, there's a negative 4 here.
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Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3
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I wanted to do that in the red color. That our h is negative 3, and that our k is positive 4, and that our r is 7. So we could say h comma k is equal to negative 3 comma positive 4. You might say, hey, there's a negative 4 here. No, but look, it's minus k, minus 4. So k is 4. Likewise, it's minus h. You might say, hey, maybe h is a positive 3.
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Radius and center for a circle equation in standard form Algebra II Khan Academy.mp3
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What we're going to do in this video is think about a way to measure angles. And there's several ways to do this. You might have seen this leveraging things like degrees in other videos, but now we're going to introduce a new concept, or maybe you know this concept, but another way of looking at this concept. So we have this angle ABC, and we wanna think about what is a way to figure out a measure of angle ABC. Now, one way to think about it would be, well, this angle subtends some arc. In this case, it subtends arc AC. And we could see if this angle were smaller, if its measure were smaller, it would subtend a smaller arc.
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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So we have this angle ABC, and we wanna think about what is a way to figure out a measure of angle ABC. Now, one way to think about it would be, well, this angle subtends some arc. In this case, it subtends arc AC. And we could see if this angle were smaller, if its measure were smaller, it would subtend a smaller arc. The length of that arc would be smaller. And if the angle were wider or had a larger measure, if it looks something like that, then the arc length would be larger. So should we define the measure of an angle like this as being equal to the length of the arc that it subtends?
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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And we could see if this angle were smaller, if its measure were smaller, it would subtend a smaller arc. The length of that arc would be smaller. And if the angle were wider or had a larger measure, if it looks something like that, then the arc length would be larger. So should we define the measure of an angle like this as being equal to the length of the arc that it subtends? Is this a good measure? Well, some of you might immediately see a problem with that because this length, the length of the arc that is subtended is not just dependent on the angle, the measure of the angle, it also depends on how big of a circle you're dealing with. If the radius is larger, then you're gonna have a larger arc length.
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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So should we define the measure of an angle like this as being equal to the length of the arc that it subtends? Is this a good measure? Well, some of you might immediately see a problem with that because this length, the length of the arc that is subtended is not just dependent on the angle, the measure of the angle, it also depends on how big of a circle you're dealing with. If the radius is larger, then you're gonna have a larger arc length. For example, let me introduce another circle here. So we have the same angle measure, the central angle right over here. You could say angle ABC is still the same, but now it subtends a different arc in these two different circles.
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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If the radius is larger, then you're gonna have a larger arc length. For example, let me introduce another circle here. So we have the same angle measure, the central angle right over here. You could say angle ABC is still the same, but now it subtends a different arc in these two different circles. You have this arc right over here. Let's call this arc DE. And the length of arc DE is not equal to the length of AC.
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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You could say angle ABC is still the same, but now it subtends a different arc in these two different circles. You have this arc right over here. Let's call this arc DE. And the length of arc DE is not equal to the length of AC. And so we can't measure an angle just by the length of the arc that it subtends if that angle is a central angle in a circle. So we can get rid of that equality here. But what could we do?
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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And the length of arc DE is not equal to the length of AC. And so we can't measure an angle just by the length of the arc that it subtends if that angle is a central angle in a circle. So we can get rid of that equality here. But what could we do? Well, you might realize that these two pi's that I just created, you could kind of say pi ABC and pi DBE, these are similar pi's. Now we're not used to talking in terms of similar pi's, but what does it mean to be similar? Well, you have similarity if you can map one thing onto another, one shape onto another through not just rigid transformations, but also through dilations.
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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But what could we do? Well, you might realize that these two pi's that I just created, you could kind of say pi ABC and pi DBE, these are similar pi's. Now we're not used to talking in terms of similar pi's, but what does it mean to be similar? Well, you have similarity if you can map one thing onto another, one shape onto another through not just rigid transformations, but also through dilations. And in this situation, if you were to take pi ABC and just dilate it by a scale factor larger than one, there's some scale factor where you would dilate it out to pi DBE. And what's interesting about that is if two things are similar, that means the ratio between corresponding parts are going to be the same. So for example, the ratio of the length of arc AC to the length of segment BC is going to be equal to the ratio of the length of arc DE, DE, to the length of segment BE.
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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Well, you have similarity if you can map one thing onto another, one shape onto another through not just rigid transformations, but also through dilations. And in this situation, if you were to take pi ABC and just dilate it by a scale factor larger than one, there's some scale factor where you would dilate it out to pi DBE. And what's interesting about that is if two things are similar, that means the ratio between corresponding parts are going to be the same. So for example, the ratio of the length of arc AC to the length of segment BC is going to be equal to the ratio of the length of arc DE, DE, to the length of segment BE. So maybe this is a good measure, maybe this is a good measure for an angle. And it is indeed a measure that we use in geometry and trigonometry and throughout mathematics. And we call it the radian measure of an angle, of an angle.
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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So for example, the ratio of the length of arc AC to the length of segment BC is going to be equal to the ratio of the length of arc DE, DE, to the length of segment BE. So maybe this is a good measure, maybe this is a good measure for an angle. And it is indeed a measure that we use in geometry and trigonometry and throughout mathematics. And we call it the radian measure of an angle, of an angle. And it equals, equals the ratio of the arc length subtended by that angle subtended by that angle to the radius. We just saw that in both of these situations. So let's see if we can make this a little bit more tangible.
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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And we call it the radian measure of an angle, of an angle. And it equals, equals the ratio of the arc length subtended by that angle subtended by that angle to the radius. We just saw that in both of these situations. So let's see if we can make this a little bit more tangible. Let's say we had a circle here and it has a central point, let's just call that point F. And then let me create an angle here. And actually I can make a right angle. So let's call it F, let's call this point G and let's call this H. And let's say that the radius over here is two meters.
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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So let's see if we can make this a little bit more tangible. Let's say we had a circle here and it has a central point, let's just call that point F. And then let me create an angle here. And actually I can make a right angle. So let's call it F, let's call this point G and let's call this H. And let's say that the radius over here is two meters. And now what would be the length of the arc subtended by angle GFH? Well, it's going to be 1 4th the circumference of this entire circle, the way that I've drawn it. So the entire circumference, I could write it here, the circumference is going to be equal to two pi times the radius, which is going to be two pi times two meters, is going to be four pi meters.
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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So let's call it F, let's call this point G and let's call this H. And let's say that the radius over here is two meters. And now what would be the length of the arc subtended by angle GFH? Well, it's going to be 1 4th the circumference of this entire circle, the way that I've drawn it. So the entire circumference, I could write it here, the circumference is going to be equal to two pi times the radius, which is going to be two pi times two meters, is going to be four pi meters. And so if this arc length is 1 4th of that, this is going to be pi meters. And so based on this arc length and this radius, what is going to be the measure of angle GFH in radians? Well, we could say the measure of angle GFH in radians is going to be the ratio between the length of the arc subtended and the radius.
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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So the entire circumference, I could write it here, the circumference is going to be equal to two pi times the radius, which is going to be two pi times two meters, is going to be four pi meters. And so if this arc length is 1 4th of that, this is going to be pi meters. And so based on this arc length and this radius, what is going to be the measure of angle GFH in radians? Well, we could say the measure of angle GFH in radians is going to be the ratio between the length of the arc subtended and the radius. And so it's going to be pi meters over two meters. The meters, you could view those as canceling out, which equals pi over two. And pi over two, what?
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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Well, we could say the measure of angle GFH in radians is going to be the ratio between the length of the arc subtended and the radius. And so it's going to be pi meters over two meters. The meters, you could view those as canceling out, which equals pi over two. And pi over two, what? Well, we would say this is equal to pi over two radians. Now, one thing to think about is why do we call it radians? It seems close to the word radius.
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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And pi over two, what? Well, we would say this is equal to pi over two radians. Now, one thing to think about is why do we call it radians? It seems close to the word radius. And one way to think about it is when you divide this length by the length of the radius, you figure out how many of the radii is equivalent to the arc length in question. So for example, in this situation, one radii would look something like this. If you took the same length and you just went around like this, so you can see it's going to be one point something radii.
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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It seems close to the word radius. And one way to think about it is when you divide this length by the length of the radius, you figure out how many of the radii is equivalent to the arc length in question. So for example, in this situation, one radii would look something like this. If you took the same length and you just went around like this, so you can see it's going to be one point something radii. And that's why you could also say it's one point something radians. If you took pi divided by two, you're going to get a little bit over one. You're going to get 1.07 something.
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Radians as ratio of arc length to radius Circles High school geometry Khan Academy.mp3
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So what are these things? Well, a parabola can be defined as the set of all points. Let me draw an arbitrary axis right over here. So that's my y-axis. This is my x-axis. This is my x-axis. And so a parabola can be defined as the set of all points that are equidistant to a point and a line.
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Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3
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So that's my y-axis. This is my x-axis. This is my x-axis. And so a parabola can be defined as the set of all points that are equidistant to a point and a line. And that point is the focus of that parabola, and that line is the directrix of the parabola. So what am I talking about? So let's give ourselves a point.
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Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3
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And so a parabola can be defined as the set of all points that are equidistant to a point and a line. And that point is the focus of that parabola, and that line is the directrix of the parabola. So what am I talking about? So let's give ourselves a point. So let's say this point right over here. And we could even say that that is the point, let's say that's the point, let's say the x-coordinate is a, and the y-coordinate is b right over here. So that is the point a, comma, b.
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Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3
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So let's give ourselves a point. So let's say this point right over here. And we could even say that that is the point, let's say that's the point, let's say the x-coordinate is a, and the y-coordinate is b right over here. So that is the point a, comma, b. And then let's give ourselves a line for the directrix. And actually, let me do this in a different color instead of just white, because I already did the coordinates in white. So I will do it in this magenta color.
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Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3
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So that is the point a, comma, b. And then let's give ourselves a line for the directrix. And actually, let me do this in a different color instead of just white, because I already did the coordinates in white. So I will do it in this magenta color. So that's a, comma, b is the focus. And let's say y equals c is the directrix. So this right over here is the line.
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Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3
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So I will do it in this magenta color. So that's a, comma, b is the focus. And let's say y equals c is the directrix. So this right over here is the line. This right over here is the line y is equal to c. So this on the y-axis right over there, that is c. This is the line y is equal to c. So a parabola, what does it mean to be the set of all points that are equidistant between a point and this line? Let's think about what those points might be. Well, this point right over here would be halfway between this point, between the focus and the directrix.
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Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3
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So this right over here is the line. This right over here is the line y is equal to c. So this on the y-axis right over there, that is c. This is the line y is equal to c. So a parabola, what does it mean to be the set of all points that are equidistant between a point and this line? Let's think about what those points might be. Well, this point right over here would be halfway between this point, between the focus and the directrix. And then as we move away from x equals a, you're going to get points anywhere along this curve, which is a parabola. And you might be saying, wait, I don't get this. I don't get why points along this curve are going to be equidistant.
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Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3
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Well, this point right over here would be halfway between this point, between the focus and the directrix. And then as we move away from x equals a, you're going to get points anywhere along this curve, which is a parabola. And you might be saying, wait, I don't get this. I don't get why points along this curve are going to be equidistant. Well, let's just eyeball the distances. So this one, this distance, and obviously I'm drawing it by hand, so it's not going to be completely precise. That distance needs to be equal to that distance.
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Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3
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I don't get why points along this curve are going to be equidistant. Well, let's just eyeball the distances. So this one, this distance, and obviously I'm drawing it by hand, so it's not going to be completely precise. That distance needs to be equal to that distance. Well, that seems believable. And now if we take this point right over here on the parabola, this distance needs to be the same as that distance. Well, that seems believable.
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Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3
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That distance needs to be equal to that distance. Well, that seems believable. And now if we take this point right over here on the parabola, this distance needs to be the same as that distance. Well, that seems believable. If you take this point on the parabola, this distance needs to be the same as this distance. So hopefully you get what I'm talking about when I say that the parabola is a set of all points that are equidistant to this focus and this directrix. So any point along this parabola, this point right over here, the distance to the focus should be the same as the distance to the directrix.
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Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3
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Well, that seems believable. If you take this point on the parabola, this distance needs to be the same as this distance. So hopefully you get what I'm talking about when I say that the parabola is a set of all points that are equidistant to this focus and this directrix. So any point along this parabola, this point right over here, the distance to the focus should be the same as the distance to the directrix. Now what you might realize is, when you're taking the distance between a point and a point, the distance, it could be at an angle. This one's straight up and down. This one is going from the top left to the bottom right.
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Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3
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So any point along this parabola, this point right over here, the distance to the focus should be the same as the distance to the directrix. Now what you might realize is, when you're taking the distance between a point and a point, the distance, it could be at an angle. This one's straight up and down. This one is going from the top left to the bottom right. But when you take the distance from a point to a line, you essentially drop a perpendicular. You essentially go straight down, or if the parabola was down here, you would go straight up to find that distance. These are all right angles right over here.
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Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3
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This one is going from the top left to the bottom right. But when you take the distance from a point to a line, you essentially drop a perpendicular. You essentially go straight down, or if the parabola was down here, you would go straight up to find that distance. These are all right angles right over here. So that's all a focus and a directrix is. And every parabola is going to have a focus and directrix because every parabola is the set of all points that are equidistant to some focus and some directrix. So that's what they are.
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Focus and directrix introduction Conic sections Algebra II Khan Academy.mp3
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Let me draw my best diameter. That's pretty good. This right here is the diameter of the circle, or it's a diameter of the circle. That's a diameter. And let's say I have a triangle where the diameter is one side of the triangle and the angle opposite that side, its vertex, sits someplace on the circumference. So let's say the angle opposite of this diameter sits on that circumference, so the triangle looks like this. What I'm going to show you in this video is that this triangle is going to be a right triangle.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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That's a diameter. And let's say I have a triangle where the diameter is one side of the triangle and the angle opposite that side, its vertex, sits someplace on the circumference. So let's say the angle opposite of this diameter sits on that circumference, so the triangle looks like this. What I'm going to show you in this video is that this triangle is going to be a right triangle. And the 90 degree side is going to be the side that is opposite this diameter. I don't want to label it just yet because that would ruin the fun of the proof. Now let's see what we can do to show this.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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What I'm going to show you in this video is that this triangle is going to be a right triangle. And the 90 degree side is going to be the side that is opposite this diameter. I don't want to label it just yet because that would ruin the fun of the proof. Now let's see what we can do to show this. We have in our toolkit the notion of an inscribed angle. It's a relation to a central angle that subtends the same arc. So let's look at that.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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Now let's see what we can do to show this. We have in our toolkit the notion of an inscribed angle. It's a relation to a central angle that subtends the same arc. So let's look at that. Let's say that this is an inscribed angle right here. Let's call this theta. Let's say that's the center of my circle right there.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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So let's look at that. Let's say that this is an inscribed angle right here. Let's call this theta. Let's say that's the center of my circle right there. Then this angle right here would be a central angle. Let me draw another triangle right here, another line right there. This is a central angle right here.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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Let's say that's the center of my circle right there. Then this angle right here would be a central angle. Let me draw another triangle right here, another line right there. This is a central angle right here. This is a radius. This is the same radius. Actually, this distance is the same.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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This is a central angle right here. This is a radius. This is the same radius. Actually, this distance is the same. But we've learned several videos ago that this angle, this inscribed angle, it subtends this arc up here. The central angle that subtends that same arc is going to be twice this angle. We proved that several videos ago.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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Actually, this distance is the same. But we've learned several videos ago that this angle, this inscribed angle, it subtends this arc up here. The central angle that subtends that same arc is going to be twice this angle. We proved that several videos ago. So this is going to be 2 theta. It's the central angle subtending the same arc. Now this triangle right here, this one right here, this is an isosceles triangle.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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We proved that several videos ago. So this is going to be 2 theta. It's the central angle subtending the same arc. Now this triangle right here, this one right here, this is an isosceles triangle. I could rotate it and draw it like this. If I rotated it, I could draw it like this. If I flipped it over, it would look like that, that, and then the green side would be down like that.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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Now this triangle right here, this one right here, this is an isosceles triangle. I could rotate it and draw it like this. If I rotated it, I could draw it like this. If I flipped it over, it would look like that, that, and then the green side would be down like that. Both of these sides are of length r. This top angle is 2 theta. All I did is I took it and I rotated it around to draw it for you this way. This side is that side right there.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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If I flipped it over, it would look like that, that, and then the green side would be down like that. Both of these sides are of length r. This top angle is 2 theta. All I did is I took it and I rotated it around to draw it for you this way. This side is that side right there. Since its two sides are equal, this is isosceles, so these two base angles must be the same. These two base angles must be the same. That and that must be the same, or if I were to draw it up here, that and that must be the exact same base angle.
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Proof Right triangles inscribed in circles High School Math Khan Academy.mp3
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