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And so based on just the information given, we actually can't prove congruency. Now, let me ask you a slightly different question. Let's say that we did give you a little bit more information. Let's say we told you that the measure of this angle right over here is 31 degrees, and the measure of this angle right over here is 31 degrees. Can you now prove that triangle DCA is congruent to triangle BAC? So let's see what we can deduce now. Well, we know that AC is in both triangles, so it's going to be congruent to itself.
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Proving triangle congruence Congruence High school geometry Khan Academy.mp3
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Let's say we told you that the measure of this angle right over here is 31 degrees, and the measure of this angle right over here is 31 degrees. Can you now prove that triangle DCA is congruent to triangle BAC? So let's see what we can deduce now. Well, we know that AC is in both triangles, so it's going to be congruent to itself. And let me write that down. We know that segment AC is congruent to segment AC. It sits in both triangles, and this is by reflexivity, which is a fancy way of saying that something is going to be congruent to itself.
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Proving triangle congruence Congruence High school geometry Khan Academy.mp3
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Well, we know that AC is in both triangles, so it's going to be congruent to itself. And let me write that down. We know that segment AC is congruent to segment AC. It sits in both triangles, and this is by reflexivity, which is a fancy way of saying that something is going to be congruent to itself. Now, we also see that AB is parallel to DC just like before, and AC can be viewed as part of a transversal. So we can deduce that angle CAB, let me write this down, actually I'm gonna do it in a different color. We can deduce that angle CAB, CAB is congruent to angle ACD, angle ACD, because they are alternate interior angles where a transversal intersects two parallel lines.
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Proving triangle congruence Congruence High school geometry Khan Academy.mp3
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It sits in both triangles, and this is by reflexivity, which is a fancy way of saying that something is going to be congruent to itself. Now, we also see that AB is parallel to DC just like before, and AC can be viewed as part of a transversal. So we can deduce that angle CAB, let me write this down, actually I'm gonna do it in a different color. We can deduce that angle CAB, CAB is congruent to angle ACD, angle ACD, because they are alternate interior angles where a transversal intersects two parallel lines. So just to be clear, this angle, which is CAB, is congruent to this angle, which is ACD. And so now we have two angles and a side, two angles and a side that are congruent. So we can now deduce by angle-angle-side postulate that the triangles are indeed congruent.
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Proving triangle congruence Congruence High school geometry Khan Academy.mp3
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We can deduce that angle CAB, CAB is congruent to angle ACD, angle ACD, because they are alternate interior angles where a transversal intersects two parallel lines. So just to be clear, this angle, which is CAB, is congruent to this angle, which is ACD. And so now we have two angles and a side, two angles and a side that are congruent. So we can now deduce by angle-angle-side postulate that the triangles are indeed congruent. So we now know that triangle DCA is indeed congruent to triangle BAC because of angle-angle-side congruency, which we've talked about in previous videos. And just to be clear, sometimes people like the two-column proofs, I can make this look a little bit more like a two-column proof by saying these are my statements, statement, and this is my rationale right over here. And we're done.
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Proving triangle congruence Congruence High school geometry Khan Academy.mp3
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And like how all of these proofs start, let's construct ourselves a right triangle. So what I'm going to do, I'm going to construct it so this hypotenuse sits on the bottom. So that's the hypotenuse of my right triangle. Try to draw it as big as possible so that we have space to work with. So that's going to be my hypotenuse. And then let's say that this is the longer side that's not the hypotenuse. We could have two sides that are equal, but I'll just draw it so it looks a little bit longer.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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Try to draw it as big as possible so that we have space to work with. So that's going to be my hypotenuse. And then let's say that this is the longer side that's not the hypotenuse. We could have two sides that are equal, but I'll just draw it so it looks a little bit longer. Let's call that side length a. And then let's draw this side right over here. It has to be a right triangle, so maybe it goes right over there, that side of length b.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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We could have two sides that are equal, but I'll just draw it so it looks a little bit longer. Let's call that side length a. And then let's draw this side right over here. It has to be a right triangle, so maybe it goes right over there, that side of length b. Let me extend the length a a little bit. So it definitely looks like a right triangle. And this is our 90 degree angle.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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It has to be a right triangle, so maybe it goes right over there, that side of length b. Let me extend the length a a little bit. So it definitely looks like a right triangle. And this is our 90 degree angle. So the first thing that I'm going to do is take this triangle and then rotate it counterclockwise by 90 degrees. So if I rotate it counterclockwise by 90 degrees, I'm literally just going to rotate it like that and draw another completely congruent version of this one. So I'm going to rotate it by 90 degrees.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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And this is our 90 degree angle. So the first thing that I'm going to do is take this triangle and then rotate it counterclockwise by 90 degrees. So if I rotate it counterclockwise by 90 degrees, I'm literally just going to rotate it like that and draw another completely congruent version of this one. So I'm going to rotate it by 90 degrees. And if I did that, the hypotenuse would then sit straight up. So I'm going to do my best attempt to draw it almost to scale, as much as I can eyeball it. This side of length a will now come out, will now look something like this.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So I'm going to rotate it by 90 degrees. And if I did that, the hypotenuse would then sit straight up. So I'm going to do my best attempt to draw it almost to scale, as much as I can eyeball it. This side of length a will now come out, will now look something like this. It'll actually be parallel to this over here. So let's see. Let me see how well I could draw it.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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This side of length a will now come out, will now look something like this. It'll actually be parallel to this over here. So let's see. Let me see how well I could draw it. So this is the side of length a. And if we care, this would be 90 degrees. The rotation between the corresponding sides are just going to be 90 degrees in every case.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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Let me see how well I could draw it. So this is the side of length a. And if we care, this would be 90 degrees. The rotation between the corresponding sides are just going to be 90 degrees in every case. This is going to be 90 degrees. That's going to be 90 degrees. Now let me draw side b.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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The rotation between the corresponding sides are just going to be 90 degrees in every case. This is going to be 90 degrees. That's going to be 90 degrees. Now let me draw side b. So it's going to look something like that, or the side that's length b. And the right angle is now here. So all I did is I rotated this by 90 degrees counterclockwise.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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Now let me draw side b. So it's going to look something like that, or the side that's length b. And the right angle is now here. So all I did is I rotated this by 90 degrees counterclockwise. Now what I want to do is construct a parallelogram. I'm going to construct a parallelogram by essentially, and let me label this. So this is height c right over here.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So all I did is I rotated this by 90 degrees counterclockwise. Now what I want to do is construct a parallelogram. I'm going to construct a parallelogram by essentially, and let me label this. So this is height c right over here. Let me do that white color. This is height c. Now what I want to do is go from this point and go up c as well. Now, so this is height c as well.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So this is height c right over here. Let me do that white color. This is height c. Now what I want to do is go from this point and go up c as well. Now, so this is height c as well. And what is this length? What is the length over here from this point to this point going to be? What is this length going to be?
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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Now, so this is height c as well. And what is this length? What is the length over here from this point to this point going to be? What is this length going to be? Well, a little clue is this is a parallelogram. This line right over here is going to be parallel to this line. It's maintained the same distance.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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What is this length going to be? Well, a little clue is this is a parallelogram. This line right over here is going to be parallel to this line. It's maintained the same distance. And since it's traveling the same distance in the x direction or in the horizontal direction and the vertical direction, this is going to be the same length. So this is going to be of length a. Now the next question I have for you is what is the area of this parallelogram that I have just constructed?
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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It's maintained the same distance. And since it's traveling the same distance in the x direction or in the horizontal direction and the vertical direction, this is going to be the same length. So this is going to be of length a. Now the next question I have for you is what is the area of this parallelogram that I have just constructed? Well, to think about that, let's redraw this part of the diagram so that the parallelogram is kind of sitting on the ground. So this is length a. This is length a.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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Now the next question I have for you is what is the area of this parallelogram that I have just constructed? Well, to think about that, let's redraw this part of the diagram so that the parallelogram is kind of sitting on the ground. So this is length a. This is length a. This is length c. This is length c. And if you look at this part right over here, it gives you a clue. The height of the parallelogram is given right over here. This side is perpendicular to the base.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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This is length a. This is length c. This is length c. And if you look at this part right over here, it gives you a clue. The height of the parallelogram is given right over here. This side is perpendicular to the base. So the height of the parallelogram is a as well. So what's the area? Well, the area of a parallelogram is just the base times the height.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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This side is perpendicular to the base. So the height of the parallelogram is a as well. So what's the area? Well, the area of a parallelogram is just the base times the height. So the area of this parallelogram right over here is going to be a squared. Now let's do the same thing, but let's rotate our original right triangle. Let's rotate it the other way.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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Well, the area of a parallelogram is just the base times the height. So the area of this parallelogram right over here is going to be a squared. Now let's do the same thing, but let's rotate our original right triangle. Let's rotate it the other way. So let's rotate it 90 degrees clockwise. And this time, instead of pivoting on this point, we're going to pivot on that point right over there. So what are we going to get?
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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Let's rotate it the other way. So let's rotate it 90 degrees clockwise. And this time, instead of pivoting on this point, we're going to pivot on that point right over there. So what are we going to get? So this side of length c, if we rotate it like that, it's going to end up right over here. Try to draw it as close to scale as possible. So that side has length c. Now the side of length b is going to pop out, look something like this.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So what are we going to get? So this side of length c, if we rotate it like that, it's going to end up right over here. Try to draw it as close to scale as possible. So that side has length c. Now the side of length b is going to pop out, look something like this. It's going to be parallel to that. This is going to be a right angle. So let me draw it like that.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So that side has length c. Now the side of length b is going to pop out, look something like this. It's going to be parallel to that. This is going to be a right angle. So let me draw it like that. That looks pretty good. And then the side of length a is going to be out here. So that's a.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So let me draw it like that. That looks pretty good. And then the side of length a is going to be out here. So that's a. And then this right over here is b. And I want to do that b in blue. So let me do the b in blue.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So that's a. And then this right over here is b. And I want to do that b in blue. So let me do the b in blue. And then this right angle, once we've rotated it, is just sitting right over here. Now let's do the same exercise. Let's construct a parallelogram right over here.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So let me do the b in blue. And then this right angle, once we've rotated it, is just sitting right over here. Now let's do the same exercise. Let's construct a parallelogram right over here. So this is height c. This is height c as well. So by the same logic we used over here, if this length is b, this length is b as well. These are parallel lines.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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Let's construct a parallelogram right over here. So this is height c. This is height c as well. So by the same logic we used over here, if this length is b, this length is b as well. These are parallel lines. We're going the same distance in the horizontal direction. We're rising the same in the vertical direction. We know that because they're parallel.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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These are parallel lines. We're going the same distance in the horizontal direction. We're rising the same in the vertical direction. We know that because they're parallel. So this is length b down here. This is length b up there. Now what is the area of this parallelogram right over there?
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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We know that because they're parallel. So this is length b down here. This is length b up there. Now what is the area of this parallelogram right over there? What is the area of that parallelogram going to be? Well, once again, to help us visualize it, we can draw it kind of sitting flat. So this is that side.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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Now what is the area of this parallelogram right over there? What is the area of that parallelogram going to be? Well, once again, to help us visualize it, we can draw it kind of sitting flat. So this is that side. Then you have another side right over here. They both have length b. And you have the sides of length c. So that's c. What is its height?
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So this is that side. Then you have another side right over here. They both have length b. And you have the sides of length c. So that's c. What is its height? Well, you see it right over here. Its height is length b as well. It gives right there.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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And you have the sides of length c. So that's c. What is its height? Well, you see it right over here. Its height is length b as well. It gives right there. We know that this is 90 degrees. We did a 90 degree rotation. So this is how we constructed the thing.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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It gives right there. We know that this is 90 degrees. We did a 90 degree rotation. So this is how we constructed the thing. So given that, the area of a parallelogram is just the base times the height. The area of this parallelogram is b squared. So now things are starting to get interesting.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So this is how we constructed the thing. So given that, the area of a parallelogram is just the base times the height. The area of this parallelogram is b squared. So now things are starting to get interesting. And what I'm going to do is I'm going to copy and paste this part right over here. Because this is, in my mind, the most interesting part of our diagram. Let me see how well I can select it.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So now things are starting to get interesting. And what I'm going to do is I'm going to copy and paste this part right over here. Because this is, in my mind, the most interesting part of our diagram. Let me see how well I can select it. So let me select this part right over here. So let me copy. And then I am going to scroll down.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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Let me see how well I can select it. So let me select this part right over here. So let me copy. And then I am going to scroll down. And then let me paste it. So this diagram that we've constructed right over here, it's pretty clear what the area of it is, the combined diagram. And let me delete a little few parts of it just so that it cleans it up.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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And then I am going to scroll down. And then let me paste it. So this diagram that we've constructed right over here, it's pretty clear what the area of it is, the combined diagram. And let me delete a little few parts of it just so that it cleans it up. So let me clean this thing up so we really get the part that we want to focus on. So cleaning that up, cleaning this up, right over there. So what is, and actually let me delete this right down here as well.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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And let me delete a little few parts of it just so that it cleans it up. So let me clean this thing up so we really get the part that we want to focus on. So cleaning that up, cleaning this up, right over there. So what is, and actually let me delete this right down here as well. Let me delete this right over here. Although we know that this length was c. And actually I'll draw it right over here. This is from our original construction.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So what is, and actually let me delete this right down here as well. Let me delete this right over here. Although we know that this length was c. And actually I'll draw it right over here. This is from our original construction. We know that this length is c. We know this height is c. We know this down here is c. But my question for you is, what is the area of this combined shape? Well, it's just a squared plus b squared. Let me write that down.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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This is from our original construction. We know that this length is c. We know this height is c. We know this down here is c. But my question for you is, what is the area of this combined shape? Well, it's just a squared plus b squared. Let me write that down. The area is just a squared plus b squared, the area of those two parallelograms. Now how can we maybe rearrange pieces of this shape so that we can express it in terms of c? Well, it might have jumped out at you when I drew this line right over here.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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Let me write that down. The area is just a squared plus b squared, the area of those two parallelograms. Now how can we maybe rearrange pieces of this shape so that we can express it in terms of c? Well, it might have jumped out at you when I drew this line right over here. We know that this is of length, I want to do that in white. We know that this part right over here is of length c. This comes from our original construction. Whoops, lost my diagram.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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Well, it might have jumped out at you when I drew this line right over here. We know that this is of length, I want to do that in white. We know that this part right over here is of length c. This comes from our original construction. Whoops, lost my diagram. This is of length c, that's of length c, and then this right over here is of length c. And so what we could do is take this top right triangle, which is completely congruent to our original right triangle, and shift it down. So remember, the entire area, including this top right triangle, is a squared plus b squared. And we're excluding this part down here, which is our original triangle.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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Whoops, lost my diagram. This is of length c, that's of length c, and then this right over here is of length c. And so what we could do is take this top right triangle, which is completely congruent to our original right triangle, and shift it down. So remember, the entire area, including this top right triangle, is a squared plus b squared. And we're excluding this part down here, which is our original triangle. But what happens if we take that? So let me actually cut, and then let me paste it. And all I'm doing is I'm moving that triangle down here.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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And we're excluding this part down here, which is our original triangle. But what happens if we take that? So let me actually cut, and then let me paste it. And all I'm doing is I'm moving that triangle down here. So now it looks like this. So I've just rearranged the area that was a squared b squared. So this entire area of this entire square is still a squared plus b squared.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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And all I'm doing is I'm moving that triangle down here. So now it looks like this. So I've just rearranged the area that was a squared b squared. So this entire area of this entire square is still a squared plus b squared. a squared is this entire area right over here. It was before a parallelogram. I just shifted that top part of the parallelogram down.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So this entire area of this entire square is still a squared plus b squared. a squared is this entire area right over here. It was before a parallelogram. I just shifted that top part of the parallelogram down. b squared is this entire area right over here. Well, what's this going to be in terms of c? Well, we know that this entire thing is a c by c squared.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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I just shifted that top part of the parallelogram down. b squared is this entire area right over here. Well, what's this going to be in terms of c? Well, we know that this entire thing is a c by c squared. So the area in terms of c is just c squared. So a squared plus b squared is equal to c squared. And we have once again proven the Pythagorean theorem.
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Another Pythagorean theorem proof Right triangles and trigonometry Geometry Khan Academy.mp3
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So however far A is from point P, it's going to be three times further under the dilation, three times further in the same direction. So how do we think about that? Well one way to think about it is to go from P to A, you have to go one down and two to the left. So minus one and minus two. And so if you dilate it with a factor of three, then you're gonna want to go three times as far down, so minus three, minus three, and three times as far to the left, so you'll go minus six. So one, or that's, let me do this. So negative one, negative two, negative three, negative four, negative five, negative six.
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Dilating points example.mp3
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So minus one and minus two. And so if you dilate it with a factor of three, then you're gonna want to go three times as far down, so minus three, minus three, and three times as far to the left, so you'll go minus six. So one, or that's, let me do this. So negative one, negative two, negative three, negative four, negative five, negative six. So you will end up right over there. And you can even see it, that this is indeed three times as far from P in the same direction. And so we could call the image of point A, maybe we call that A prime.
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Dilating points example.mp3
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So negative one, negative two, negative three, negative four, negative five, negative six. So you will end up right over there. And you can even see it, that this is indeed three times as far from P in the same direction. And so we could call the image of point A, maybe we call that A prime. And so there you have it, it has been dilated with a scale factor of three. And so you might be saying wait, I'm used to dilation being stretching or scaling, how have I stretched or scaled something? Well imagine a bunch of points here that represents some type of picture.
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Dilating points example.mp3
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And so we could call the image of point A, maybe we call that A prime. And so there you have it, it has been dilated with a scale factor of three. And so you might be saying wait, I'm used to dilation being stretching or scaling, how have I stretched or scaled something? Well imagine a bunch of points here that represents some type of picture. And if you push them all three times further from point P, which you could use your center of dilation, then you would expand the size of your picture by a scale factor of three. Let's do another example with a point. So here we're told, plot the image of point A under a dilation about the origin with a scale factor of 1 3rd.
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Dilating points example.mp3
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Well imagine a bunch of points here that represents some type of picture. And if you push them all three times further from point P, which you could use your center of dilation, then you would expand the size of your picture by a scale factor of three. Let's do another example with a point. So here we're told, plot the image of point A under a dilation about the origin with a scale factor of 1 3rd. So first of all, we don't even see the point A here, so it's actually below the fold. So let's see, there we go, that's our point A. We want it to be about the origin, so about the point zero, zero.
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Dilating points example.mp3
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So here we're told, plot the image of point A under a dilation about the origin with a scale factor of 1 3rd. So first of all, we don't even see the point A here, so it's actually below the fold. So let's see, there we go, that's our point A. We want it to be about the origin, so about the point zero, zero. This is what we want to, the dilation about the origin with a scale factor of 1 3rd. Scale is 1 3rd, scale factor I should say. So how do we do this?
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Dilating points example.mp3
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We want it to be about the origin, so about the point zero, zero. This is what we want to, the dilation about the origin with a scale factor of 1 3rd. Scale is 1 3rd, scale factor I should say. So how do we do this? Well here, however far A is from the origin, we now want to be in the same direction, but 1 3rd as far. So one way to think about it, to go from the origin to A, you have to go six down and three to the right. So 1 3rd of that would be two down and one to the right.
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Dilating points example.mp3
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So how do we do this? Well here, however far A is from the origin, we now want to be in the same direction, but 1 3rd as far. So one way to think about it, to go from the origin to A, you have to go six down and three to the right. So 1 3rd of that would be two down and one to the right. Two is 1 3rd of six and one is 1 3rd of three. So you will end up right over here. That would be our A prime.
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Dilating points example.mp3
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One is the idea of things being perpendicular. And usually people are talking about perpendicular, actually let me, I'm misspelling it, perpendicular, perpendicular lines, normally perpendicular lines, and the idea of parallel lines, parallel, parallel lines. So perpendicular lines are two lines that intersect at a right angle. So what am I talking about? So let's say that this is one line right over here, and that this is another line right over here. We would say these two lines are perpendicular if they intersect at a right angle. So they clearly intersect.
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Parallel and perpendicular lines intro Analytic geometry Geometry Khan Academy.mp3
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So what am I talking about? So let's say that this is one line right over here, and that this is another line right over here. We would say these two lines are perpendicular if they intersect at a right angle. So they clearly intersect. In order for them to intersect at a right angle, the angle formed between these two lines needs to be 90 degrees. And if any one of these angles is 90 degrees, the rest of them are going to be 90 degrees. So if this is 90 degrees then these are perpendicular lines.
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Parallel and perpendicular lines intro Analytic geometry Geometry Khan Academy.mp3
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So they clearly intersect. In order for them to intersect at a right angle, the angle formed between these two lines needs to be 90 degrees. And if any one of these angles is 90 degrees, the rest of them are going to be 90 degrees. So if this is 90 degrees then these are perpendicular lines. And if that's 90 degrees, then that's going to be 90 degrees, that's going to be 90 degrees, and that's going to be 90 degrees. So if any of them are 90 degrees, the rest of them are 90 degrees, and we have perpendicular lines. If you have two lines that on a two-dimensional surface like your paper or like the screen never intersect, they stay the same distance apart, then we are talking about parallel lines.
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Parallel and perpendicular lines intro Analytic geometry Geometry Khan Academy.mp3
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So if this is 90 degrees then these are perpendicular lines. And if that's 90 degrees, then that's going to be 90 degrees, that's going to be 90 degrees, and that's going to be 90 degrees. So if any of them are 90 degrees, the rest of them are 90 degrees, and we have perpendicular lines. If you have two lines that on a two-dimensional surface like your paper or like the screen never intersect, they stay the same distance apart, then we are talking about parallel lines. So this line right over here, and this line right over here, the way I've drawn them, are parallel lines, they aren't intersecting. They're both kind of going in the same direction, but they're kind of shifted versions of each other. They will never intersect with each other.
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Parallel and perpendicular lines intro Analytic geometry Geometry Khan Academy.mp3
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If you have two lines that on a two-dimensional surface like your paper or like the screen never intersect, they stay the same distance apart, then we are talking about parallel lines. So this line right over here, and this line right over here, the way I've drawn them, are parallel lines, they aren't intersecting. They're both kind of going in the same direction, but they're kind of shifted versions of each other. They will never intersect with each other. So these two are parallel. If we have two lines that, let's say they intersect, but they don't intersect at a right angle, so let's say we have that line, and we have this line right over here, and they're clearly not intersecting at a right angle, then we call these neither perpendicular nor parallel lines. These lines just intersect.
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Parallel and perpendicular lines intro Analytic geometry Geometry Khan Academy.mp3
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What is the central angle of the arc in degrees? So they're asking for this one. So this is the arc that they're talking about. That's 221 over 18 pi long. And they want to know this angle that it subtends, this central angle right over here. So we just have to remind ourselves that the ratio of this arc length to the entire circumference, let me write that down, the ratio of this arc length, which is 221 over 18 pi, to the entire circumference, which is 20 pi, is going to be equal to the ratio of this central angle, which we can call theta, the ratio of theta to 360 degrees if we were to go all the way around the circle, to 360 degrees. This will give us our theta in degrees.
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Finding central angle measure given arc length Circles Geometry Khan Academy.mp3
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That's 221 over 18 pi long. And they want to know this angle that it subtends, this central angle right over here. So we just have to remind ourselves that the ratio of this arc length to the entire circumference, let me write that down, the ratio of this arc length, which is 221 over 18 pi, to the entire circumference, which is 20 pi, is going to be equal to the ratio of this central angle, which we can call theta, the ratio of theta to 360 degrees if we were to go all the way around the circle, to 360 degrees. This will give us our theta in degrees. If we wanted it in radians, we would think of it in terms of 2 pi radians around the circle. But it was 360 degrees since we're in degrees. Now we just have to simplify.
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Finding central angle measure given arc length Circles Geometry Khan Academy.mp3
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This will give us our theta in degrees. If we wanted it in radians, we would think of it in terms of 2 pi radians around the circle. But it was 360 degrees since we're in degrees. Now we just have to simplify. Now the easiest thing is just to multiply both sides times 360 degrees. So let's do that. So this is 300.
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Finding central angle measure given arc length Circles Geometry Khan Academy.mp3
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Now we just have to simplify. Now the easiest thing is just to multiply both sides times 360 degrees. So let's do that. So this is 300. If we multiply the left-hand side by 360 degrees, we get 360 times 221 times pi over, see we have 18 times 20 times 20 times pi times pi. And on the right-hand side, if we multiply it by 360, we are just left with theta. So we really just have to simplify this now.
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Finding central angle measure given arc length Circles Geometry Khan Academy.mp3
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So this is 300. If we multiply the left-hand side by 360 degrees, we get 360 times 221 times pi over, see we have 18 times 20 times 20 times pi times pi. And on the right-hand side, if we multiply it by 360, we are just left with theta. So we really just have to simplify this now. Let's see, pi divided by pi is going to be 1. 360 divided by 20, well, it's going to be the same thing as 36 over 2, which is the same thing as 18. And 18 divided by 18 is 1.
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Finding central angle measure given arc length Circles Geometry Khan Academy.mp3
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We are told the triangle N prime is the image of triangle N under a dilation. So this is N prime in this red color and then the original N is in this blue color. What is the center of dilation? And they give us some choices here. Choice A, B, C, or D is the center of dilation. So pause this video and see if you can figure it out on your own. So there's a couple of ways to think about it.
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Example identifying the center of dilation.mp3
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And they give us some choices here. Choice A, B, C, or D is the center of dilation. So pause this video and see if you can figure it out on your own. So there's a couple of ways to think about it. One way, I like to just first think about well what is the scale factor here? So in our original N, we have this side here, it has a length of two. And then once we dilated it by, and used that scale factor, that corresponding side has a length of four.
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Example identifying the center of dilation.mp3
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So there's a couple of ways to think about it. One way, I like to just first think about well what is the scale factor here? So in our original N, we have this side here, it has a length of two. And then once we dilated it by, and used that scale factor, that corresponding side has a length of four. So we went from two to four. So we can figure out our scale factor. Scale factor is equal to two.
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Example identifying the center of dilation.mp3
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And then once we dilated it by, and used that scale factor, that corresponding side has a length of four. So we went from two to four. So we can figure out our scale factor. Scale factor is equal to two. Two times two is equal to four. Now what about our center of dilation? So one way to think about it is pick two corresponding points.
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Example identifying the center of dilation.mp3
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Scale factor is equal to two. Two times two is equal to four. Now what about our center of dilation? So one way to think about it is pick two corresponding points. So let's say we were to pick this point and this point. So the image, the corresponding point on N prime is going to be the scale factor as far away from our center of dilation as the original point. So in this example, we know the scale factor is two.
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Example identifying the center of dilation.mp3
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So one way to think about it is pick two corresponding points. So let's say we were to pick this point and this point. So the image, the corresponding point on N prime is going to be the scale factor as far away from our center of dilation as the original point. So in this example, we know the scale factor is two. So this is going to be twice as far from our center of dilation as the corresponding point. Well you can immediately see, and it's going to be in the same direction. So actually, if you just draw a line connecting these two, there's actually only one choice that sits on that line, and that is choice D right over here as being the center of dilation.
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Example identifying the center of dilation.mp3
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So in this example, we know the scale factor is two. So this is going to be twice as far from our center of dilation as the corresponding point. Well you can immediately see, and it's going to be in the same direction. So actually, if you just draw a line connecting these two, there's actually only one choice that sits on that line, and that is choice D right over here as being the center of dilation. And you can also verify that notice, this first point on the original triangle, its change in X is two and its change in Y is three, two, three, to go from point D to point to that point. And then if you want to go to point D to its image, well now you gotta go twice as far. Your change in X is four and your change in Y is six.
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Example identifying the center of dilation.mp3
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So actually, if you just draw a line connecting these two, there's actually only one choice that sits on that line, and that is choice D right over here as being the center of dilation. And you can also verify that notice, this first point on the original triangle, its change in X is two and its change in Y is three, two, three, to go from point D to point to that point. And then if you want to go to point D to its image, well now you gotta go twice as far. Your change in X is four and your change in Y is six. You could use the Pythagorean theorem to calculate this distance and then the longer distance. But what you see is is that the corresponding point is now twice as far from your center of dilation. So there's a couple of ways to think about it.
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Example identifying the center of dilation.mp3
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Find the equation of a line perpendicular to this line that passes through the point 2, 8. So this first piece of information, that it's perpendicular to that line right over there. What does that tell us? Well, if it's perpendicular to this line, its slope has to be the negative inverse of 2 5ths. So its slope, the negative inverse of 2 5ths, the inverse of 2 5ths is 5. Let me do it in a better color, a nicer green. If this line's slope is negative 2 5ths, the equation of the line we have to figure out that's perpendicular, its slope is going to be the inverse.
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Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3
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Well, if it's perpendicular to this line, its slope has to be the negative inverse of 2 5ths. So its slope, the negative inverse of 2 5ths, the inverse of 2 5ths is 5. Let me do it in a better color, a nicer green. If this line's slope is negative 2 5ths, the equation of the line we have to figure out that's perpendicular, its slope is going to be the inverse. So instead of 2 5ths, it's going to be 5 halves. Instead of being a negative, it's going to be a positive. So this is the negative inverse of negative 2 5ths.
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Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3
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If this line's slope is negative 2 5ths, the equation of the line we have to figure out that's perpendicular, its slope is going to be the inverse. So instead of 2 5ths, it's going to be 5 halves. Instead of being a negative, it's going to be a positive. So this is the negative inverse of negative 2 5ths. You take the negative sign, it becomes positive. You swap the 5 and the 2, you get 5 halves. So that is going to have to be our slope.
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Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3
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So this is the negative inverse of negative 2 5ths. You take the negative sign, it becomes positive. You swap the 5 and the 2, you get 5 halves. So that is going to have to be our slope. And we can actually use the point slope form right here. It goes through this point right there. So let's use point slope form.
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Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3
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So that is going to have to be our slope. And we can actually use the point slope form right here. It goes through this point right there. So let's use point slope form. y minus this y value, which has to be on the line, is equal to our slope, 5 halves, times x minus this x value, the x value when y is equal to 8. And this is the equation of the line in point slope form. If you want to put it in slope intercept form, you can just do a little bit of algebraic manipulation.
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Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3
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So let's use point slope form. y minus this y value, which has to be on the line, is equal to our slope, 5 halves, times x minus this x value, the x value when y is equal to 8. And this is the equation of the line in point slope form. If you want to put it in slope intercept form, you can just do a little bit of algebraic manipulation. y minus 8 is equal to, let's distribute the 5 halves. So 5 halves x minus 5 halves times 2 is just 5. And then add 8 to both sides.
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Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3
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If you want to put it in slope intercept form, you can just do a little bit of algebraic manipulation. y minus 8 is equal to, let's distribute the 5 halves. So 5 halves x minus 5 halves times 2 is just 5. And then add 8 to both sides. You get y is equal to 5 halves x. Add 8 to negative 5, so plus 3. And we are done.
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Writing equations of perpendicular lines (example 2) High School Math Khan Academy.mp3
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So pause this video and try to think about this on your own before we work through it together. All right, now let's work through it together. So in general, if you know that something is already a parallelogram and you wanna determine whether it's a rectangle, it's really a question of whether the adjacent sides intersect at a right angle. So for example, a parallelogram might look something like this. What we know about a parallelogram is that the opposite sides are parallel. So this side is parallel to that side and that this side is parallel to this side. And all rectangles are parallelograms, but not all parallelograms are rectangles.
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Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3
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So for example, a parallelogram might look something like this. What we know about a parallelogram is that the opposite sides are parallel. So this side is parallel to that side and that this side is parallel to this side. And all rectangles are parallelograms, but not all parallelograms are rectangles. In order for a parallelogram to be a rectangle, these sides need to intersect at right angles. And clearly the way I drew this one, it doesn't look like that. But let's see if we can figure that out based on the coordinates that they have given us.
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Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3
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And all rectangles are parallelograms, but not all parallelograms are rectangles. In order for a parallelogram to be a rectangle, these sides need to intersect at right angles. And clearly the way I drew this one, it doesn't look like that. But let's see if we can figure that out based on the coordinates that they have given us. And to help us visualize, let me just put some coordinates. Let me draw some axes here. So that's my X axis, and then this is my Y axis.
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Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3
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But let's see if we can figure that out based on the coordinates that they have given us. And to help us visualize, let me just put some coordinates. Let me draw some axes here. So that's my X axis, and then this is my Y axis. Let's see the coordinates. Let's see, we have twos, fours, sixes. Let me actually count by an eight.
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Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3
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So that's my X axis, and then this is my Y axis. Let's see the coordinates. Let's see, we have twos, fours, sixes. Let me actually count by an eight. So let me count by twos here. So we have two, four, six, and eight. And then we have negative two, negative four, negative six, negative eight.
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Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3
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Let me actually count by an eight. So let me count by twos here. So we have two, four, six, and eight. And then we have negative two, negative four, negative six, negative eight. We have two, four, six, and eight. And then we'd have negative two, negative four, negative six, and negative eight. So each hash mark is another two.
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Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3
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And then we have negative two, negative four, negative six, negative eight. We have two, four, six, and eight. And then we'd have negative two, negative four, negative six, and negative eight. So each hash mark is another two. I'm counting by twos here. And so let's plot these points. And I'll do it in different colors so we can keep track.
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Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3
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So each hash mark is another two. I'm counting by twos here. And so let's plot these points. And I'll do it in different colors so we can keep track. So A is negative six, negative four. So negative two, negative four, negative six, and then negative four would go right over here. That is point A.
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Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3
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And I'll do it in different colors so we can keep track. So A is negative six, negative four. So negative two, negative four, negative six, and then negative four would go right over here. That is point A. Then we have point B, which is negative two, six. So negative two, six. So that's going to go up to four and six.
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Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3
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That is point A. Then we have point B, which is negative two, six. So negative two, six. So that's going to go up to four and six. So that is point B right over there. Then we have point C, which is at eight, two. So eight, comma, two right over there.
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Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3
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So that's going to go up to four and six. So that is point B right over there. Then we have point C, which is at eight, two. So eight, comma, two right over there. That is point C. And then last but not least, we have point D, which is at four, comma, negative eight. Four, comma, negative eight right over there, point D. And so our quadrilateral, or we actually know it's a parallelogram, looks like this. So you have segment AB like that.
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Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3
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So eight, comma, two right over there. That is point C. And then last but not least, we have point D, which is at four, comma, negative eight. Four, comma, negative eight right over there, point D. And so our quadrilateral, or we actually know it's a parallelogram, looks like this. So you have segment AB like that. You have segment BC that looks like that. Segment CD looks like this. And segment AD looks like this.
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Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3
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So you have segment AB like that. You have segment BC that looks like that. Segment CD looks like this. And segment AD looks like this. And we know already that it's a parallelogram. So we know that segment AB is parallel to segment DC, and segment BC is parallel to segment AD. But what we really need to do is figure out whether they are intersecting at right angles.
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Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3
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And segment AD looks like this. And we know already that it's a parallelogram. So we know that segment AB is parallel to segment DC, and segment BC is parallel to segment AD. But what we really need to do is figure out whether they are intersecting at right angles. And to do that, using the coordinates to figure that out, we have to figure out the slopes of these different line segments. And so let's figure out first the slope of AB. So the slope of segment AB is going to be equal to our change in y over change in x.
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Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3
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But what we really need to do is figure out whether they are intersecting at right angles. And to do that, using the coordinates to figure that out, we have to figure out the slopes of these different line segments. And so let's figure out first the slope of AB. So the slope of segment AB is going to be equal to our change in y over change in x. So our change in y is going to be six minus negative four, six minus negative four, over negative two minus negative six. Negative two minus negative six. And so this is going to be equal to six plus four, which is 10, over negative two minus negative six.
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Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3
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So the slope of segment AB is going to be equal to our change in y over change in x. So our change in y is going to be six minus negative four, six minus negative four, over negative two minus negative six. Negative two minus negative six. And so this is going to be equal to six plus four, which is 10, over negative two minus negative six. That's the same thing as negative two plus six. So that's going to be over four, which is the same thing as 5 1⁄2. All right, that's interesting.
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Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3
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