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Either way, this base angle is going to be x plus 10. Well, it's an isosceles triangle. So your two base angles are going to be congruent. So if this is x plus 10, then this is going to be x plus 10 as well. And now we have the three angles of a triangle expressed in terms of x. So when we take their sum, they need to be equal to 180. And then we can actually solve for x. | Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3 |
So if this is x plus 10, then this is going to be x plus 10 as well. And now we have the three angles of a triangle expressed in terms of x. So when we take their sum, they need to be equal to 180. And then we can actually solve for x. So we get 2x plus x plus 10 is going to be equal to 180 degrees. And then we can add up the x's. So we have a 2x there plus an x plus another x. | Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3 |
And then we can actually solve for x. So we get 2x plus x plus 10 is going to be equal to 180 degrees. And then we can add up the x's. So we have a 2x there plus an x plus another x. That gives us 4x. And then we have a plus 10 and another plus 10. So that gives us a plus 20 is equal to 180. | Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3 |
So we have a 2x there plus an x plus another x. That gives us 4x. And then we have a plus 10 and another plus 10. So that gives us a plus 20 is equal to 180. And we can subtract 20 from both sides of that. And we get 4x is equal to 160. Divide both sides by 4. | Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3 |
So that gives us a plus 20 is equal to 180. And we can subtract 20 from both sides of that. And we get 4x is equal to 160. Divide both sides by 4. And we get x is equal to 40. And we're done. We figured out what x is. | Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3 |
Divide both sides by 4. And we get x is equal to 40. And we're done. We figured out what x is. And then we can actually figure out what these angles are. If this is x plus 10, then you have 40 plus 10. This right over here is going to be a 50 degree angle. | Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3 |
We figured out what x is. And then we can actually figure out what these angles are. If this is x plus 10, then you have 40 plus 10. This right over here is going to be a 50 degree angle. This is 2x, so 2 times 40, this is an 80 degree angle. It doesn't look at it the way I've drawn it. And that's why you should never assume anything based on how a diagram is drawn. | Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3 |
This right over here is going to be a 50 degree angle. This is 2x, so 2 times 40, this is an 80 degree angle. It doesn't look at it the way I've drawn it. And that's why you should never assume anything based on how a diagram is drawn. So this right over here is going to be an 80 degree angle. And then these two base angles right over here are also going to be 50 degrees. So you have 50 degrees, 50 degrees, and 80. | Example involving an isosceles triangle and parallel lines Congruence Geometry Khan Academy.mp3 |
So that's negative four comma negative two and zero comma five. And zero comma five, so that's zero comma five right over there. The quadrilateral is left unchanged by a reflection over the line y is equal to x over two. So what does that line look like? Y is equal to x over two. I'll do that in blue. Y is equal to x over two. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
So what does that line look like? Y is equal to x over two. I'll do that in blue. Y is equal to x over two. So when x is equal to zero, y is zero. The y-intercept is zero here, and the slope is one half. Every time x increases by one, y will increase by one half, or when x increases by two, y will increase by one. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
Y is equal to x over two. So when x is equal to zero, y is zero. The y-intercept is zero here, and the slope is one half. Every time x increases by one, y will increase by one half, or when x increases by two, y will increase by one. So x increases by two, y increases by one. x increases by two, y increases by one. Another way to think about it, y is always one half of x. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
Every time x increases by one, y will increase by one half, or when x increases by two, y will increase by one. So x increases by two, y increases by one. x increases by two, y increases by one. Another way to think about it, y is always one half of x. When x is four, y is two. When x is six, y is three. When x is eight, y is four. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
Another way to think about it, y is always one half of x. When x is four, y is two. When x is six, y is three. When x is eight, y is four. So we could connect these. Let me try my best attempt to draw these in a relatively straight line, and then I can keep going. When x is negative two, y is negative one. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
When x is eight, y is four. So we could connect these. Let me try my best attempt to draw these in a relatively straight line, and then I can keep going. When x is negative two, y is negative one. When x is negative four, y is negative two. So it actually goes through that point right there, and it just keeps going with a slope of one half. So this line, and I can draw it a little bit thicker now, now that I've dotted it out. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
When x is negative two, y is negative one. When x is negative four, y is negative two. So it actually goes through that point right there, and it just keeps going with a slope of one half. So this line, and I can draw it a little bit thicker now, now that I've dotted it out. This is the line y is equal to x over two. And they also say that the quadrilateral is left unchanged by reflection over the line y is equal to negative two x plus five. So the y-intercept here is five. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
So this line, and I can draw it a little bit thicker now, now that I've dotted it out. This is the line y is equal to x over two. And they also say that the quadrilateral is left unchanged by reflection over the line y is equal to negative two x plus five. So the y-intercept here is five. When x is zero, y is five. So it actually goes through that point, and the slope is negative two. Every time we increase x by one, we decrease y by two. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
So the y-intercept here is five. When x is zero, y is five. So it actually goes through that point, and the slope is negative two. Every time we increase x by one, we decrease y by two. So we go there, we go there, and we keep going at a slope of negative two. So it's going to look something like this. It's going to look something like this. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
Every time we increase x by one, we decrease y by two. So we go there, we go there, and we keep going at a slope of negative two. So it's going to look something like this. It's going to look something like this. It actually goes through that point and just keeps going on and on. So this is my best attempt at drawing that line. So that is y is equal to negative two x plus five. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
It's going to look something like this. It actually goes through that point and just keeps going on and on. So this is my best attempt at drawing that line. So that is y is equal to negative two x plus five. Now let's think about it. Let's see if we can draw this quadrilateral. So let's first reflect the quadrilateral, or let's reflect the points we have over the line y is equal to x over two. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
So that is y is equal to negative two x plus five. Now let's think about it. Let's see if we can draw this quadrilateral. So let's first reflect the quadrilateral, or let's reflect the points we have over the line y is equal to x over two. So this is the line y is equal to x over two. This magenta point, the point negative four, two, is already on that line. So it's its own reflection, I guess you could say, or it's on the mirror, one way is to think about it. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
So let's first reflect the quadrilateral, or let's reflect the points we have over the line y is equal to x over two. So this is the line y is equal to x over two. This magenta point, the point negative four, two, is already on that line. So it's its own reflection, I guess you could say, or it's on the mirror, one way is to think about it. But we can easily reflect this line over here. This line, if we were to drop a perpendicular, and actually this line right over here, y is equal to negative two x plus five, is perpendicular to y is equal to x over two. How do we know? | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
So it's its own reflection, I guess you could say, or it's on the mirror, one way is to think about it. But we can easily reflect this line over here. This line, if we were to drop a perpendicular, and actually this line right over here, y is equal to negative two x plus five, is perpendicular to y is equal to x over two. How do we know? Well, if one line has a slope of m, then the line that's perpendicular would be the negative reciprocal of this. It would be negative one over m. So this first line has a slope of one half. What's the negative reciprocal of one half? | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
How do we know? Well, if one line has a slope of m, then the line that's perpendicular would be the negative reciprocal of this. It would be negative one over m. So this first line has a slope of one half. What's the negative reciprocal of one half? The reciprocal of one half is two over one, and you make that negative. So it is equal to negative two. So this slope is the negative reciprocal of this slope. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
What's the negative reciprocal of one half? The reciprocal of one half is two over one, and you make that negative. So it is equal to negative two. So this slope is the negative reciprocal of this slope. So these lines are indeed perpendicular. So we literally could drop a perpendicular, literally go along this line right over here in our attempt to reflect, and we see that we're going down two over one, down two over one twice. So let's go down two over one, down two over one twice again. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
So this slope is the negative reciprocal of this slope. So these lines are indeed perpendicular. So we literally could drop a perpendicular, literally go along this line right over here in our attempt to reflect, and we see that we're going down two over one, down two over one twice. So let's go down two over one, down two over one twice again. The reflection of this point across y is equal to x over two is this point right over there. So now we have three points of our quadrilateral. Let's see if we can get a fourth. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
So let's go down two over one, down two over one twice again. The reflection of this point across y is equal to x over two is this point right over there. So now we have three points of our quadrilateral. Let's see if we can get a fourth. So let's go to the magenta point. The magenta point we've already seen. It's sitting on top of y equals x over two. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
Let's see if we can get a fourth. So let's go to the magenta point. The magenta point we've already seen. It's sitting on top of y equals x over two. So trying to reflect it doesn't help us much, but we could try to reflect it across y is equal to negative 2x plus 5. So once again, these lines are perpendicular to each other. Actually, let me mark that off. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
It's sitting on top of y equals x over two. So trying to reflect it doesn't help us much, but we could try to reflect it across y is equal to negative 2x plus 5. So once again, these lines are perpendicular to each other. Actually, let me mark that off. These lines are perpendicular, so we can drop a perpendicular and try to find its reflection. So we're going to the right two and up one. We're doing that once, twice, three times on the left side. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
Actually, let me mark that off. These lines are perpendicular, so we can drop a perpendicular and try to find its reflection. So we're going to the right two and up one. We're doing that once, twice, three times on the left side. So let's do that once, twice, three times on the right side. So the reflection is right there. We essentially want to go to that line, and however far we were to the left of it, we want to go that bottom left direction. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
We're doing that once, twice, three times on the left side. So let's do that once, twice, three times on the right side. So the reflection is right there. We essentially want to go to that line, and however far we were to the left of it, we want to go that bottom left direction. We want to go in the same direction to the top right, the same distance to get the reflection. So there you have it. There is our other point. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
We essentially want to go to that line, and however far we were to the left of it, we want to go that bottom left direction. We want to go in the same direction to the top right, the same distance to get the reflection. So there you have it. There is our other point. So now we have the four points of this quadrilateral. Four points of this quadrilateral are the four sides. Let me actually just draw the quadrilateral. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
There is our other point. So now we have the four points of this quadrilateral. Four points of this quadrilateral are the four sides. Let me actually just draw the quadrilateral. We have our four points. This is one side right over here. This is one side right over here. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
Let me actually just draw the quadrilateral. We have our four points. This is one side right over here. This is one side right over here. This is another side right over here. And you can verify that these are parallel. How would you verify that they're parallel? | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
This is one side right over here. This is another side right over here. And you can verify that these are parallel. How would you verify that they're parallel? Well, they have the same slope. If you get from this point to that point, you have to go over, so your run has to be four, and you have to rise one, two, three, four, five, six, seven. So the slope here is 7 4ths. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
How would you verify that they're parallel? Well, they have the same slope. If you get from this point to that point, you have to go over, so your run has to be four, and you have to rise one, two, three, four, five, six, seven. So the slope here is 7 4ths. So slope here, rise over run, or change in y over change in x is 7 4ths. And over here, you go one, two, three, four. So you run four, and then you rise one, two, three, four, five, six, seven. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
So the slope here is 7 4ths. So slope here, rise over run, or change in y over change in x is 7 4ths. And over here, you go one, two, three, four. So you run four, and then you rise one, two, three, four, five, six, seven. So the slope here is also 7 over 4. So these two lines are going to be parallel. And then we could draw these lines over here. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
So you run four, and then you rise one, two, three, four, five, six, seven. So the slope here is also 7 over 4. So these two lines are going to be parallel. And then we could draw these lines over here. So this one at the top right over there. And what's its slope? Well, let's see. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
And then we could draw these lines over here. So this one at the top right over there. And what's its slope? Well, let's see. We go from x equals 0 to x equals 8. So we go down. Our change in y is negative 1 every time we increase x by 8. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
Well, let's see. We go from x equals 0 to x equals 8. So we go down. Our change in y is negative 1 every time we increase x by 8. So this is a slope of slope is equal to negative 1 over 8. And that's the exact same slope that we have right over here, negative 1 over 8. So these two lines are parallel as well. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
Our change in y is negative 1 every time we increase x by 8. So this is a slope of slope is equal to negative 1 over 8. And that's the exact same slope that we have right over here, negative 1 over 8. So these two lines are parallel as well. This line is parallel to this line as well. So at minimum, we're dealing with a parallelogram. Let's see if we can go even more specific because this kind of looks like a rhombus. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
So these two lines are parallel as well. This line is parallel to this line as well. So at minimum, we're dealing with a parallelogram. Let's see if we can go even more specific because this kind of looks like a rhombus. It looks like a parallelogram where all four sides have the same length. So there's a couple of ways that you could verify that this parallelogram is a rhombus. One way is you could actually find the distance between the points. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
Let's see if we can go even more specific because this kind of looks like a rhombus. It looks like a parallelogram where all four sides have the same length. So there's a couple of ways that you could verify that this parallelogram is a rhombus. One way is you could actually find the distance between the points. You could use that. We know the coordinates, so you could use the distance formula, which really comes straight out of the Pythagorean theorem. Or even better, you could look at the diagonals of this rhombus. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
One way is you could actually find the distance between the points. You could use that. We know the coordinates, so you could use the distance formula, which really comes straight out of the Pythagorean theorem. Or even better, you could look at the diagonals of this rhombus. You could look at the diagonals of this parallelogram. We're trying to figure out if it's a rhombus. And if the diagonals are perpendicular, then you're dealing with a rhombus. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
Or even better, you could look at the diagonals of this rhombus. You could look at the diagonals of this parallelogram. We're trying to figure out if it's a rhombus. And if the diagonals are perpendicular, then you're dealing with a rhombus. And we've already shown that these diagonals, that this diagonal, this diagonal, and this diagonal are perpendicular. They intersect at right angles. And so this must be a rhombus. | Constructing a shape by reflecting over 2 lines Transformations Geometry Khan Academy.mp3 |
It's a little bit of a hairy circle, but you get the idea. So this is a circle, this is the center of the circle. And let's say that I have an arc along this circle. So I'll do the arc in green. So I have an arc that is part of the circle, and it subtends an angle. So that's my arc right over there. And it subtends an angle, and the angle that it subtends, so what I mean by subtends, you take each of the end points of the arc, go to the center of the circle, go to the center of the circle, just like this. | Arc length as fraction of circumference Trigonometry Khan Academy.mp3 |
So I'll do the arc in green. So I have an arc that is part of the circle, and it subtends an angle. So that's my arc right over there. And it subtends an angle, and the angle that it subtends, so what I mean by subtends, you take each of the end points of the arc, go to the center of the circle, go to the center of the circle, just like this. And so it subtends angle theta right over here. So it subtends angle theta. And let's say that we know that angle theta is equal to two radians. | Arc length as fraction of circumference Trigonometry Khan Academy.mp3 |
And it subtends an angle, and the angle that it subtends, so what I mean by subtends, you take each of the end points of the arc, go to the center of the circle, go to the center of the circle, just like this. And so it subtends angle theta right over here. So it subtends angle theta. And let's say that we know that angle theta is equal to two radians. So my question to you is, what fraction of the entire circumference is this green arc? What fraction of the entire circumference is this green arc? And like always, pause the video and give it a go. | Arc length as fraction of circumference Trigonometry Khan Academy.mp3 |
And let's say that we know that angle theta is equal to two radians. So my question to you is, what fraction of the entire circumference is this green arc? What fraction of the entire circumference is this green arc? And like always, pause the video and give it a go. All right, so let's think through it a little bit. So you might say, well, how do I know that? I don't know what the radius of this thing is. | Arc length as fraction of circumference Trigonometry Khan Academy.mp3 |
And like always, pause the video and give it a go. All right, so let's think through it a little bit. So you might say, well, how do I know that? I don't know what the radius of this thing is. I don't, I, you know, how do I think through this? And we just have to remind ourselves what radians mean. What radians mean. | Arc length as fraction of circumference Trigonometry Khan Academy.mp3 |
I don't know what the radius of this thing is. I don't, I, you know, how do I think through this? And we just have to remind ourselves what radians mean. What radians mean. If an arc subtends an angle of two radians, that means that the arc itself is two radiuses long. So this right over here, so let me make this a little clearer. If the radius is r, if this radius, I already used that color. | Arc length as fraction of circumference Trigonometry Khan Academy.mp3 |
What radians mean. If an arc subtends an angle of two radians, that means that the arc itself is two radiuses long. So this right over here, so let me make this a little clearer. If the radius is r, if this radius, I already used that color. If this radius, I have trouble switching colors. All right. If this radius is length r, then the length, if this angle is two radians, then the arc that subtends it is going to be two radiuses long. | Arc length as fraction of circumference Trigonometry Khan Academy.mp3 |
If the radius is r, if this radius, I already used that color. If this radius, I have trouble switching colors. All right. If this radius is length r, then the length, if this angle is two radians, then the arc that subtends it is going to be two radiuses long. So this length right over here is two radiuses. Now, what fraction of the entire circumference is that? Well, the entire circumference, we know this from basic geometry, the entire circumference is two pi times the radius, or you could say it's two pi radii, two pi radiuses. | Arc length as fraction of circumference Trigonometry Khan Academy.mp3 |
If this radius is length r, then the length, if this angle is two radians, then the arc that subtends it is going to be two radiuses long. So this length right over here is two radiuses. Now, what fraction of the entire circumference is that? Well, the entire circumference, we know this from basic geometry, the entire circumference is two pi times the radius, or you could say it's two pi radii, two pi radiuses. Two pi radii is the correct way to say it. So what fraction is it? It's two radii, it's two radii over two pi radii, over two pi radii. | Arc length as fraction of circumference Trigonometry Khan Academy.mp3 |
The measure of angle P is 0.4 radians, and the length of the radius is five units. Length of the radius is five units, that's this length right over here. Find the length of the green arc. So just to kind of conceptualize this a little bit, P, you can imagine, is the center of this larger circle, is the larger circle, and this angle right over here that has a measure of 0.4 radians, it intercepts this green arc right over here. In order to figure this out, and actually, I encourage you to pause this video now and try to think about this question on your own. How long is this arc, given the information that they've given us? Well, all we have to do is remind ourselves what a radian is. | Finding arc length from radian angle measure Trigonometry Khan Academy.mp3 |
So just to kind of conceptualize this a little bit, P, you can imagine, is the center of this larger circle, is the larger circle, and this angle right over here that has a measure of 0.4 radians, it intercepts this green arc right over here. In order to figure this out, and actually, I encourage you to pause this video now and try to think about this question on your own. How long is this arc, given the information that they've given us? Well, all we have to do is remind ourselves what a radian is. One way to think about a radian is, if you look at the arc that the angle intercepts, which is this green arc, and you think about its length, the length of this green arc is going to be 0.4 radii. One way to think about radians is, if this angle is 0.4 radians, that means that the arc that it intercepts is going to be 0.4 radii long. So this length, we could write as 0.4 radii, or radiuses, but radii is the proper term. | Finding arc length from radian angle measure Trigonometry Khan Academy.mp3 |
Well, all we have to do is remind ourselves what a radian is. One way to think about a radian is, if you look at the arc that the angle intercepts, which is this green arc, and you think about its length, the length of this green arc is going to be 0.4 radii. One way to think about radians is, if this angle is 0.4 radians, that means that the arc that it intercepts is going to be 0.4 radii long. So this length, we could write as 0.4 radii, or radiuses, but radii is the proper term. Now, we don't want our length in terms of radii, we want our length in terms of whatever units the radius is, these kind of five units. Well, we know that each radius has a length of five, that our radius of the circle has a length of five. So this is going to be 0.4 radii times five, and you know, they just call it units right over here, times five, I'll put it in quotes because it's kind of a generic term, five units per radii. | Finding arc length from radian angle measure Trigonometry Khan Academy.mp3 |
So this length, we could write as 0.4 radii, or radiuses, but radii is the proper term. Now, we don't want our length in terms of radii, we want our length in terms of whatever units the radius is, these kind of five units. Well, we know that each radius has a length of five, that our radius of the circle has a length of five. So this is going to be 0.4 radii times five, and you know, they just call it units right over here, times five, I'll put it in quotes because it's kind of a generic term, five units per radii. So the radii cancel out, we're left with just the units, which is what we want, so 0.4 times five is two. So this is going to give us, this is going to be equal to two. So just as a refresher again, when the angle measure in radians, one way to think about it is the arc that it intercepts, that's going to be this many radii long. | Finding arc length from radian angle measure Trigonometry Khan Academy.mp3 |
And this is true for all triangles. And so to set up this proof, I've put an arbitrary triangle here, but I've put one vertex at the origin. That'll simplify the math. And then I put another vertex on the x-axis. And I've given them coordinates. So this one right over here is at zero, zero. This one over here, we're just saying that the x-coordinate is a, and so it's a comma zero. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
And then I put another vertex on the x-axis. And I've given them coordinates. So this one right over here is at zero, zero. This one over here, we're just saying that the x-coordinate is a, and so it's a comma zero. And then this one up here has some x and some y-coordinate. We're just calling them b and c. This is some arbitrary triangle. And if you had some other triangle that had the same dimensions as this one, and this one can have any dimensions because we haven't defined a, b, and c, you could go from this triangle to any of those other ones using rigid transformations. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
This one over here, we're just saying that the x-coordinate is a, and so it's a comma zero. And then this one up here has some x and some y-coordinate. We're just calling them b and c. This is some arbitrary triangle. And if you had some other triangle that had the same dimensions as this one, and this one can have any dimensions because we haven't defined a, b, and c, you could go from this triangle to any of those other ones using rigid transformations. So if we can prove that the medians of this triangle, this general triangle, always intersect at one point, this will be true for all triangles. So let's do a little bit more. Let's draw the medians. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
And if you had some other triangle that had the same dimensions as this one, and this one can have any dimensions because we haven't defined a, b, and c, you could go from this triangle to any of those other ones using rigid transformations. So if we can prove that the medians of this triangle, this general triangle, always intersect at one point, this will be true for all triangles. So let's do a little bit more. Let's draw the medians. So what we're going to do is draw lines from each of the vertex to the midpoint of the opposite side. So if we do that, we've drawn all the medians, and it for sure looks like they intersect at one point, but to prove that, let's think about what the coordinates are of the midpoints of each of these sides. So what is the coordinate right over here? | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
Let's draw the medians. So what we're going to do is draw lines from each of the vertex to the midpoint of the opposite side. So if we do that, we've drawn all the medians, and it for sure looks like they intersect at one point, but to prove that, let's think about what the coordinates are of the midpoints of each of these sides. So what is the coordinate right over here? Pause this video and think about that. Well, this is going to be the midpoint of this top point and this bottom right point. So this length is equal to that length. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
So what is the coordinate right over here? Pause this video and think about that. Well, this is going to be the midpoint of this top point and this bottom right point. So this length is equal to that length. And for the midpoint, you can really just think about it as you're taking the average of each of the coordinates. So the x-coordinate here is going to be the average of b and a. So you could just write that as a plus b over two. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
So this length is equal to that length. And for the midpoint, you can really just think about it as you're taking the average of each of the coordinates. So the x-coordinate here is going to be the average of b and a. So you could just write that as a plus b over two. And then the y-coordinate is going to be the average of c and zero. That would be c plus zero over two, or just c over two. And we could do that for each of these points. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
So you could just write that as a plus b over two. And then the y-coordinate is going to be the average of c and zero. That would be c plus zero over two, or just c over two. And we could do that for each of these points. So this point right over here, its x-coordinate is going to be the average of zero and a. So that's just a over two. And its y-coordinate is going to be the average of zero and zero. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
And we could do that for each of these points. So this point right over here, its x-coordinate is going to be the average of zero and a. So that's just a over two. And its y-coordinate is going to be the average of zero and zero. You can see that it sits on the x-axis, so its y-coordinate is zero. And then last but not least, what's the coordinate of this point? Pause the video and try to figure that out. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
And its y-coordinate is going to be the average of zero and zero. You can see that it sits on the x-axis, so its y-coordinate is zero. And then last but not least, what's the coordinate of this point? Pause the video and try to figure that out. All right, well, the x-coordinate is going to be the average of b and zero, which is just going to be b over two. And then the y-coordinate is going to be the average of c and zero, which is just going to be c over two. So the way that I'm gonna prove that all three of these medians intersect at a unique point is by showing you a coordinate that sits on all three lines. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
Pause the video and try to figure that out. All right, well, the x-coordinate is going to be the average of b and zero, which is just going to be b over two. And then the y-coordinate is going to be the average of c and zero, which is just going to be c over two. So the way that I'm gonna prove that all three of these medians intersect at a unique point is by showing you a coordinate that sits on all three lines. If it sits on all three lines, that must be the point of intersection. And that interesting point is 2 3rds along the way of any one of the medians. So one way to think about it is the distance between the vertex and that point is 2 3rds of the length of the median. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
So the way that I'm gonna prove that all three of these medians intersect at a unique point is by showing you a coordinate that sits on all three lines. If it sits on all three lines, that must be the point of intersection. And that interesting point is 2 3rds along the way of any one of the medians. So one way to think about it is the distance between the vertex and that point is 2 3rds of the length of the median. So if we just look at this blue median, the coordinate of this point that is twice as far away from the vertex as it is from the opposite side, it will be based on a weighted average of the x and y-coordinates. When we did a midpoint and things were equally far away, you equally weighted the coordinates. So you just took their average. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
So one way to think about it is the distance between the vertex and that point is 2 3rds of the length of the median. So if we just look at this blue median, the coordinate of this point that is twice as far away from the vertex as it is from the opposite side, it will be based on a weighted average of the x and y-coordinates. When we did a midpoint and things were equally far away, you equally weighted the coordinates. So you just took their average. But if you are closer to this side, you will take a weighted average accordingly. So it's going to be 2 3rds times a plus b over two plus 1 3rd times zero. And then the y-coordinate is going to be 2 3rds times c over two plus 1 3rd times the y-coordinate here, which is just going to be zero. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
So you just took their average. But if you are closer to this side, you will take a weighted average accordingly. So it's going to be 2 3rds times a plus b over two plus 1 3rd times zero. And then the y-coordinate is going to be 2 3rds times c over two plus 1 3rd times the y-coordinate here, which is just going to be zero. Now once again, why do we have this 2 3rd and this 1 3rd weighting? Because we are twice as close to this point as we are to that point. Now if we wanted to simplify it, what would we get? | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
And then the y-coordinate is going to be 2 3rds times c over two plus 1 3rd times the y-coordinate here, which is just going to be zero. Now once again, why do we have this 2 3rd and this 1 3rd weighting? Because we are twice as close to this point as we are to that point. Now if we wanted to simplify it, what would we get? So this two would cancel with that two, so and this is zero, so we would get a plus b over three for the x-coordinate. And for the y-coordinate, this is zero. That two cancels with that two, and we get c over three. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
Now if we wanted to simplify it, what would we get? So this two would cancel with that two, so and this is zero, so we would get a plus b over three for the x-coordinate. And for the y-coordinate, this is zero. That two cancels with that two, and we get c over three. So we just found a point that for sure sits on this blue median. Now let's do a similar exercise with this pink-colored median. So that pink-colored median, what is the coordinate of the point that sits on that median that is twice as far from the vertex as it is from the opposite side? | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
That two cancels with that two, and we get c over three. So we just found a point that for sure sits on this blue median. Now let's do a similar exercise with this pink-colored median. So that pink-colored median, what is the coordinate of the point that sits on that median that is twice as far from the vertex as it is from the opposite side? Well, it would be the exact same exercise. We would doubly weight these coordinates, so the x-coordinate would be 2 3rds times b over two plus 1 3rd times a, 1 3rd times a, and then the y-coordinate would be 2 3rds times c over two, 2 3rds times c over two plus 1 3rd times zero, 1 3rd times zero, and what does that get us? Well, let's see, this two cancels out with that two, and we are left with b over three plus a over three, so that's the same thing as a plus b over three, and over here, that's zero, that cancels out. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
So that pink-colored median, what is the coordinate of the point that sits on that median that is twice as far from the vertex as it is from the opposite side? Well, it would be the exact same exercise. We would doubly weight these coordinates, so the x-coordinate would be 2 3rds times b over two plus 1 3rd times a, 1 3rd times a, and then the y-coordinate would be 2 3rds times c over two, 2 3rds times c over two plus 1 3rd times zero, 1 3rd times zero, and what does that get us? Well, let's see, this two cancels out with that two, and we are left with b over three plus a over three, so that's the same thing as a plus b over three, and over here, that's zero, that cancels out. That is c over three. So notice, this exact coordinate sits on both the blue median and this pink median, so that must be the place that they intersect. Let's see if that's also true for this orange median. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
Well, let's see, this two cancels out with that two, and we are left with b over three plus a over three, so that's the same thing as a plus b over three, and over here, that's zero, that cancels out. That is c over three. So notice, this exact coordinate sits on both the blue median and this pink median, so that must be the place that they intersect. Let's see if that's also true for this orange median. So in the orange median, same exact exercise, and I encourage you to pause this video and try to calculate the value of this point on your own. What's the coordinate of that point on the orange line where this distance is twice as large as this distance? Well, same idea, we will doubly weight these points right over here, so the x coordinate would be 2 3rds times a over two plus 1 3rd times b. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
Let's see if that's also true for this orange median. So in the orange median, same exact exercise, and I encourage you to pause this video and try to calculate the value of this point on your own. What's the coordinate of that point on the orange line where this distance is twice as large as this distance? Well, same idea, we will doubly weight these points right over here, so the x coordinate would be 2 3rds times a over two plus 1 3rd times b. The y coordinate would be 2 3rds times zero, 2 3rds times zero plus 1 3rd times c, 1 3rd times c, and what does that get us? Let's see, that cancels with that, so we have a over three plus b over three. Well, that's the same thing as a plus b over three, and then over here, that's just zero, and we're left with c over three. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
Well, same idea, we will doubly weight these points right over here, so the x coordinate would be 2 3rds times a over two plus 1 3rd times b. The y coordinate would be 2 3rds times zero, 2 3rds times zero plus 1 3rd times c, 1 3rd times c, and what does that get us? Let's see, that cancels with that, so we have a over three plus b over three. Well, that's the same thing as a plus b over three, and then over here, that's just zero, and we're left with c over three. So notice, we have just shown that this exact coordinate sits on all three medians, and so therefore, all three medians must intersect at that point because that point exists on all the lines. We've just shown that, and that's true for this arbitrary triangle. You can make this triangle have arbitrary dimensions by changing your value for a, b, or c, and if you see a triangle that has the same dimensions but it's shifted or it's in a different orientation, you can do a rigid transformation which doesn't change any of the dimensions, and you can show that that would be true for that triangle as well. | Proving triangle medians intersect at a point Analytic geometry Geometry Khan Academy.mp3 |
And the point is called the focus of the parabola, and the line is called the directrix of the parabola. What I want to do in this video, it's gonna get a little bit of hairy algebra, but given that definition, I want to see, and given that definition and given a focus at the point x equals a, y equals b, and align a directrix at y equals k to figure out what is the equation of that parabola actually going to be. And it's going to be based on a's, b's, and k's. So let's do that. So let's take a arbitrary point on the parabola. Let's say we take this point right over here, and its x-coordinate is x and its y-coordinate is y. And by definition, in order for this to be a parabola, it has to be equidistant to its focus and its directrix. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
So let's do that. So let's take a arbitrary point on the parabola. Let's say we take this point right over here, and its x-coordinate is x and its y-coordinate is y. And by definition, in order for this to be a parabola, it has to be equidistant to its focus and its directrix. So what does that mean? That means that the distance to the directrix, which I'm drawing here in blue, has to be the same as the distance to the focus, which I am drawing in magenta. And when we take the distance to the directrix, we literally just drop a perpendicular, I guess you could say. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
And by definition, in order for this to be a parabola, it has to be equidistant to its focus and its directrix. So what does that mean? That means that the distance to the directrix, which I'm drawing here in blue, has to be the same as the distance to the focus, which I am drawing in magenta. And when we take the distance to the directrix, we literally just drop a perpendicular, I guess you could say. That is the, that's going to be the shortest distance to that line. And then when we, but the distance to the focus, well that's, we see that's at a bit of an angle, and we might have to use the distance formula, which is really just the Pythagorean theorem. So let's do that. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
And when we take the distance to the directrix, we literally just drop a perpendicular, I guess you could say. That is the, that's going to be the shortest distance to that line. And then when we, but the distance to the focus, well that's, we see that's at a bit of an angle, and we might have to use the distance formula, which is really just the Pythagorean theorem. So let's do that. This distance has to be the same as that distance. So what's this blue distance? Well that's just going to be our change in y. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
So let's do that. This distance has to be the same as that distance. So what's this blue distance? Well that's just going to be our change in y. It's going to be this y, it's going to be this y minus k. It's just this distance. So it's going to be y minus k. Now we have to be careful. The way I've just drawn it, yes, y is greater, y is greater than k, so this is going to give us a positive value, and you need a non-negative value if you're talking about distances. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
Well that's just going to be our change in y. It's going to be this y, it's going to be this y minus k. It's just this distance. So it's going to be y minus k. Now we have to be careful. The way I've just drawn it, yes, y is greater, y is greater than k, so this is going to give us a positive value, and you need a non-negative value if you're talking about distances. But there are, you can definitely have a parabola where the y coordinate of the focus is lower than the y coordinate of the directrix, in which case this would be negative. So what we really want is the absolute value of this, or you could say we could square it, and then we could take the square root, the principal root, which would be equivalent to taking the absolute value of y minus k. So that's this distance right over here. And by the definition of a parabola, in order for x comma y to be sitting on the parabola, that distance needs to be the same as the distance from x comma y to a comma b, to the focus. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
The way I've just drawn it, yes, y is greater, y is greater than k, so this is going to give us a positive value, and you need a non-negative value if you're talking about distances. But there are, you can definitely have a parabola where the y coordinate of the focus is lower than the y coordinate of the directrix, in which case this would be negative. So what we really want is the absolute value of this, or you could say we could square it, and then we could take the square root, the principal root, which would be equivalent to taking the absolute value of y minus k. So that's this distance right over here. And by the definition of a parabola, in order for x comma y to be sitting on the parabola, that distance needs to be the same as the distance from x comma y to a comma b, to the focus. So what's that going to be? Well, we just apply the distance formula, or really just the Pythagorean theorem. It's going to be our change in x, so x minus a squared plus the change in y, y minus b squared, and the square root of that whole thing. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
And by the definition of a parabola, in order for x comma y to be sitting on the parabola, that distance needs to be the same as the distance from x comma y to a comma b, to the focus. So what's that going to be? Well, we just apply the distance formula, or really just the Pythagorean theorem. It's going to be our change in x, so x minus a squared plus the change in y, y minus b squared, and the square root of that whole thing. The square root of all of that business. Now, this right over here is an equation of a parabola. It doesn't look like it. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
It's going to be our change in x, so x minus a squared plus the change in y, y minus b squared, and the square root of that whole thing. The square root of all of that business. Now, this right over here is an equation of a parabola. It doesn't look like it. It looks really hairy, but it is the equation of a parabola. And to show you that, we just have to simplify this. And if you get inspired, I encourage you to try to simplify this on your own. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
It doesn't look like it. It looks really hairy, but it is the equation of a parabola. And to show you that, we just have to simplify this. And if you get inspired, I encourage you to try to simplify this on your own. It's just going to be a little bit of hairy algebra, but it really is not too bad. You're going to get an equation for a parabola that you might recognize, and it's going to be in terms of kind of a general focus, a comma b, and a general directrix, y equals k. So let's do that. So the simplest thing to start here is let's just square both sides so we get rid of the radicals. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
And if you get inspired, I encourage you to try to simplify this on your own. It's just going to be a little bit of hairy algebra, but it really is not too bad. You're going to get an equation for a parabola that you might recognize, and it's going to be in terms of kind of a general focus, a comma b, and a general directrix, y equals k. So let's do that. So the simplest thing to start here is let's just square both sides so we get rid of the radicals. So if you square both sides, on the left-hand side, you're going to get y minus k squared. You're going to get y minus k squared is equal to, is equal to x minus a squared plus y minus b, plus y minus b squared. Fair enough. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
So the simplest thing to start here is let's just square both sides so we get rid of the radicals. So if you square both sides, on the left-hand side, you're going to get y minus k squared. You're going to get y minus k squared is equal to, is equal to x minus a squared plus y minus b, plus y minus b squared. Fair enough. Now, what I want to do is I just want to end up with just a y on the left-hand side and just x's, a, b's, and k's on the right-hand side. So the first thing I might want to do is let's expand each of these expressions that involve with y. So this blue one on the left-hand side, that is going to be, that's going to be y squared minus two yk plus k squared, and that is going to be equal to, I'm going to keep this first one the same, so it's going to be x minus a squared. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
Fair enough. Now, what I want to do is I just want to end up with just a y on the left-hand side and just x's, a, b's, and k's on the right-hand side. So the first thing I might want to do is let's expand each of these expressions that involve with y. So this blue one on the left-hand side, that is going to be, that's going to be y squared minus two yk plus k squared, and that is going to be equal to, I'm going to keep this first one the same, so it's going to be x minus a squared. And now let me expand, let me expand, I'm going to find a color, expand this in green, so plus y squared minus two yb plus b squared. All I did is I multiplied y minus b times y minus b. Now let's see if we can simplify things. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
So this blue one on the left-hand side, that is going to be, that's going to be y squared minus two yk plus k squared, and that is going to be equal to, I'm going to keep this first one the same, so it's going to be x minus a squared. And now let me expand, let me expand, I'm going to find a color, expand this in green, so plus y squared minus two yb plus b squared. All I did is I multiplied y minus b times y minus b. Now let's see if we can simplify things. So I have a y squared on the left, I have a y squared on the right. Well, if I subtract y squared from both sides, so I can do that. Well, that simplified things a little bit. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
Now let's see if we can simplify things. So I have a y squared on the left, I have a y squared on the right. Well, if I subtract y squared from both sides, so I can do that. Well, that simplified things a little bit. And now I can, I can, let's see what I can do. So let's get the k squared on this side, so let's subtract k squared from both sides. So subtract k squared, subtract k squared from both sides, so that's going to get rid of it on the left-hand side. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
Well, that simplified things a little bit. And now I can, I can, let's see what I can do. So let's get the k squared on this side, so let's subtract k squared from both sides. So subtract k squared, subtract k squared from both sides, so that's going to get rid of it on the left-hand side. And now let's add two yb to both sides, so we have all the y's on the left-hand side. So plus two yb, that's going to move, that's going to give us a two yb on the left-hand side, plus two yb. So what is this going to be equal to? | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
So subtract k squared, subtract k squared from both sides, so that's going to get rid of it on the left-hand side. And now let's add two yb to both sides, so we have all the y's on the left-hand side. So plus two yb, that's going to move, that's going to give us a two yb on the left-hand side, plus two yb. So what is this going to be equal to? I'm starting to run into my graph, so let me give myself a little bit more real estate over here. So on the left-hand side, what am I going to have? This is the same thing as two yb minus two yk, which is the same thing, actually let me just write that down. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
So what is this going to be equal to? I'm starting to run into my graph, so let me give myself a little bit more real estate over here. So on the left-hand side, what am I going to have? This is the same thing as two yb minus two yk, which is the same thing, actually let me just write that down. That's going to be two y, do it in green. Actually, well, yeah, why not green? That's going to be, actually let me start a new color. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
This is the same thing as two yb minus two yk, which is the same thing, actually let me just write that down. That's going to be two y, do it in green. Actually, well, yeah, why not green? That's going to be, actually let me start a new color. That's going to be two yb minus two yk. You can factor out a two y, and it's going to be two y times b minus k. So let's do that. So we could write this as two, or we could write it two times b minus ky, if you factor out a two and a y. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
That's going to be, actually let me start a new color. That's going to be two yb minus two yk. You can factor out a two y, and it's going to be two y times b minus k. So let's do that. So we could write this as two, or we could write it two times b minus ky, if you factor out a two and a y. So that's the left-hand side, so that's that piece right over there. These things cancel out. Now on our right-hand side, I promised you a little bit of hairy algebra, so hopefully you see that I'm delivering on that promise. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
So we could write this as two, or we could write it two times b minus ky, if you factor out a two and a y. So that's the left-hand side, so that's that piece right over there. These things cancel out. Now on our right-hand side, I promised you a little bit of hairy algebra, so hopefully you see that I'm delivering on that promise. On the right-hand side, you have x minus a squared, and then, let's see, these characters cancel out, and you're left with b squared minus k squared. So these two are going to be b squared minus k squared, plus b squared minus k squared. Now, I said all I want is a y on the left-hand side, so let's divide everything by two times b minus k. So let's divide everything times two times b minus k. So two times b minus k, and I'm actually going to divide this whole thing by two times b minus k. Now, obviously, on the left-hand side, this all cancels out. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |
Now on our right-hand side, I promised you a little bit of hairy algebra, so hopefully you see that I'm delivering on that promise. On the right-hand side, you have x minus a squared, and then, let's see, these characters cancel out, and you're left with b squared minus k squared. So these two are going to be b squared minus k squared, plus b squared minus k squared. Now, I said all I want is a y on the left-hand side, so let's divide everything by two times b minus k. So let's divide everything times two times b minus k. So two times b minus k, and I'm actually going to divide this whole thing by two times b minus k. Now, obviously, on the left-hand side, this all cancels out. You're left with just a y, and then it's going to be y equals, y is equal to one over two times b minus k, and notice b minus k is the difference between the y-coordinate of the focus and the y-coordinate, I guess you could say, of the line, y equals k. So it's one over two times that, times x minus a squared. x minus a squared. So if you knew what b minus k was, this would just simplify to some number, some number that's being multiplied times x minus a squared. | Equation for parabola from focus and directrix Conic sections Algebra II Khan Academy.mp3 |