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If $a, b, c$ form an A. P. with common difference $d(eq 0)$ and $x, y, z$ form a G. P. with common ratio $r eq 1$ ), then the area of the triangle with vertices $(a...

Question

If $a, b, c$ form an A. P. with common difference $d(eq 0)$ and $x, y, z$ form a G. P. with common ratio $r eq 1$ ), then the area of the triangle with vertices $(a, x),(b, y)$ and $(c, z)$ is independent of(A) $b$(B) $r$(C) $d$(D) $x$

If $a, b, c$ form an A. P. with common difference $d( eq 0)$ and $x, y, z$ form a G. P. with common ratio $r eq 1$ ), then the area of the triangle with vertices $(a, x),(b, y)$ and $(c, z)$ is independent of (A) $b$ (B) $r$ (C) $d$ (D) $x$



Answers

Triangle area Find a concise formula for the area of a triangle in the $x y$ -plane with vertices $\left(a_{1}, a_{2}\right),\left(b_{1}, b_{2}\right),$ and $\left(c_{1}, c_{2}\right)$.

So we're going to suppose that a B and C form the mercies of some triangle and a B and C of the midpoint of the opposite side. So what we want to show is when we take capital A and little A Capital B B capital seed, little C all of those inspectors and Adam together get zero. All right, so let's first just go ahead and find what the's mid points actually on. So little a so little a should be. So my purpose in green so little a is going to be the midpoint between the point B and C So we just add each of them component wise and then divide though it was gonna be be one plus c one over two be two plus C two over, too, and then be three plus C three over likewise for beat. You can see how in a the only thing we really missing words the a component and you can use similar laundry too. Get that. This should be a one plus C one over two. No, a two plus C two over too. And a three plus c three over. And then lastly, we'll get C is equal to. So the A one plus B one over two a two plus two B two over to you and then a three plus B three over two where so we have bees now, and so we need to find each of the vectors. So the first specter will find we'll just find, what little capital A of little A's? Well, if we think about this, what this is really saying is if we start from the point, eh? And add this vector to it, we should land on little A go ahead and set up a little bit. So if we want to find the factories were gonna attract capital e over and doing that will give us that this vector. Keep it a blue. The specter capital a little a is going to be little a minus capital. So we know how to plug in a and capital A into here, uh, and doing that. Remember, when we subtract these, we can think about the miss vectors, and all we need to do is attract their components. So we're gonna have this component and there was attract a one from it. So it's going to be be one plus C one over too, minus a one. And then be two plus C two over, too. By this, you two there looks like a little bit. And then be three plus C to first. See three over too. Minus a. All right, so we have this. Now, let's go ahead and turn these into just one fraction for each of the components. So we're gonna have to multiply a one a two a three by two. So that's gonna give us be one plus C one by this to a one all over, too. Be two plus two minus to a two. All over, too. And then people plus eatery by this to a all two. And you might notice that if we were to go through the same process of finding, uh, capital Bee Little bee the vector, the only thing that's really going to changes these 1st 2 components should not be whatever letter I have out here And this second component, we multiply it by two, and then we go ahead and add a one Herb, it's attractive. Like how we have it here. So this next one, if you were to give you the same process of subtracting it just to kind of, uh, keep video. Shoot. This is going to be a one plus C one. Since we're doing with B of n minus to be won all over, too. And then it's gonna be a one plus C one minus two B two all over too. And a three section last one should be to not want this pregnancy and then a three plus secret bias to be three all over, too. And then lastly, see Capital Seed, lower case C is going to be so 1st 2 terms do not need to include sees it was gonna be a one plus one, and then the second term is gonna be two times. See component, and then it's going to be a two plus be to minus to see to all over, too. And then it's going to be a three plus B three minus two C three all over two. All right, now, remember, our whole goal for this is to show that capital a little, eh? Plus capital bee, little bee plus capital seed. But we'll see some Studio zero vector. So now all we need to do is add to these component wise. So, uh, first people will question mark on top of that because of what we want to show. So we're gonna add this this and this. So doing that will give us full. We have to a ones there, so it should give us to a one, and then we also have to be ones of them. We have to see one's there. And then we also have all of the parts of her subtracting. There's B negative to a one minus negative or minus to be one minus two C one and this is all going to be over two. And you might just notice a ones cancel out to be one's counsel to see one's cancel out. So this term actually just becomes zero. So let me just go ahead and write that below, so it's gonna be zero, and then the next term, we're going to do the same thing. So we're going to take these three components here, add them all together. So again we're gonna have to a two plus two b. Choose plus to see twos and in minus two twos, minus two twos minus to see to me a little bit more, actually. Let me just go ahead and speak this over. Did not. So this is a little bit so we'll have that. And then all this is still divided by two. And once again, the two way two's comes out to be choose counselor to see twos Council out. So that term also goes to zero. And then, lastly, we're going to add our last components for all these together, and it'll look pretty much just like, Well, we got for the 1st 2 components except with subscript three. So it's gonna be too a green plus two B three plus two secretly minus to a three last 23 minus two See? And then we close the come back here and we divide that last component. And just like with the other three components, a three's cancel out the B three scouts on the C three camps ops that also becomes zero. So when we add those up, each of the components become zero. And this here is our zero. So we've shown that when we add those three factors together, we always end up getting the zero vector

We want to find a two by two determinant formula for the area of the triangle in the X Y plane with mercy 00 a one a two a and B one, B two. All right, so if we were to go ahead and first think about the sides of this triangle as maybe oh, vectors. So going from 00 to a one a two. I'll call this factor a going from 002 He won't be too. Oh, call this Victor be well, we know if we were to add these together, we end up with this parallelogram. And if we were to take the cross product of these two vectors, the bagged A tube off that will give us what we're interested in for the area of the parallelogram. And if we just want the triangle won't looking at that there, it's gonna be half. So if we were to find the magnitude Oh, a cross be that there should tell us our magnitude of the parallelogram. They would just divided by two to find the area of a triangle. Let's go ahead and start that. So since both of these East vectors air centered at the origin. We know A is just gonna be able to and B is going to be the one. So when we go ahead and we take a cross, be so remember we can write it as this determined that we have in green one of the equations to give us in the structure. Nothing. Just fill everything out. So said the I J. K. And then we could go ahead and put Dr A first. So it's gonna be a one, a two. And remember, since we're in the X Y plane, that means easy is it to zero. And then same thing with B B one B two and zero for the Z Courtney. Now, if we want to go ahead and find this, determine it so we could just go straight across. But one of the things ah that we can also do is so starting in this top corner here to determine whether it's gonna be adding or subtracting, we can go around like this. So in the top left, it's positive that alternates just going too negative, positive, negative, positive like that. So we can instead come over here and start in this cake column and move downwards. Just alternate with adding and subtracting like that with our coefficients being whatever we're crossing out at the time. All right, so let's do that cause it'll help simplify this a little bit. So first we're going to have. So it's gonna be plus que And then we have the determinant of a one. A two being one. Me too. And then we can go ahead and move this down one. So then is going to be now minus zero and then times to determine of what's left behind. So I j b one b two and then we can go ahead and move this down one more time and it's going to be adding and then zero again for the co Fisher. And then the determinant left behind will be I j a one a two. Now, notice that it doesn't even matter what the determined of the two baker sees on the right are going to be, since you're multiplying it by zero. So those descent be the zero vector. All right, now remember, this here is going to give us our magnitude of this is the just across product of them. So some perpendicular vector. Now, if we were to go ahead and find the magnitude of this, well, that would be the magnitude of K and then determine of a one a two b one, b two. And remember, one of the nice properties of magnitude is we can pull Constance and it paid out look like this. But this is really just a constant or I should say scaler since a one a two be wannabe to her just numbers so we can go ahead and factor that out and get a one a two b one b two and then magnitude of our vector K end. This is a unit vector, so unit. So we know that that's just going to be ableto one. So we end up with the magnitude of a cross be or the area of the parallelogram is going to be the determination of a one a two be one beach. But remember, we didn't want the whole area. We just wanted the area of the triangle. So we're going to do now is divide this by two and then this year and actually go ahead and multiply it by 1/2 just to make it look a little bit nicer. As of now, this here will be the area of the triangle given that its points are at the origin. A one, a two and be one too.

So for this problem, we're just given a general triangle in the X y plane and were asked to find the formula expressing its area as a determinant. So we know the area of the triangle is going to involve taking across product of two vectors so we can take, for instance, a vector going from a to B and a vector going from A to C so we can represent that as so from a to see that's going to be C one minus 81 C two minus a two and then zero, cause it's in the X Y plane. And then the vector from A to B is similarly given by this formula. Okay? And so then we know the area is 1/2 the cross product magnitude of the cross product of, say, a be cross a C, which if we work that out well, it's just going to be 1/2 the magnitude of Okay, so the think about it, the 1st 2 components in the cross product are going to be zero because of this zeros in the last column. And so the only non zero element of the cross product is the last component. So the magnitude is just the absolute value of the last component. And the last component is given by C one minus a one. Sorry. If I'm doing a be cross a C, it should really be the one minus a one C two minus a two minus be to minus a two C one minus 81 And if we wanted to write this as a determinant, we could write it as B one line of, say, one c two minus a two. The two minus a two C one minus 81 This gives us a general formula for the area of a triangle in the

So Triangle X y Z has to be in a sauce Elise triangle. So let's take a look at it. We have X, we have X and Y down here and then we have Z at the top. So X and Y are both mid points there at the same level. There are always going to be at the same level. Z is the only thing that's changing. So while de changes, these sides are actually going to change at the same rate of each other while this side is gonna be in a separate in the separate little plane of its own, so it has to be a sauces.


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