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Hey, we're here today to talk about some examples of coordinate geometry and how to use coordinate proof to prove different things, in this case about a triangle, whether it's classifying triangle by its sides by its angles and so forth. Here are a few examples of how that might work so we're dealing with a triangle. Tr I It's in the first quadrant. You could tell by the points 15, 651 in 5. 11 are all first quadrant point. So the first thing I wanna do always in any kind of coordinate proof is I'm gonna plot my points. Um, so keep in mind, we're dealing with all first chord, first quadrant, and so 15, 612345 There's 15 6. Okay, we'll call that point. T 15, 651 There we go. We're gonna call that point are, And 5, 11. Okay, we'll go ahead and connect those pieces. That's how we can see our triangle, even though it's not technically necessary. Okay, What about triangle T? All right, question is, is that I saw Seles recalling isosceles triangles, the triangle with exactly to congruent sides and recall that our main tools for doing any kind of coordinate proof is to use either the distance formula, the slope formula or the midpoint formula in this case is we're talking about classifying a triangle by its sides. I'm going to use the distance formula. So we got lucky. We know one side. We know our I we know that this side right here is 10 units long because we're going from 1 to 11. So really, it comes down to finding the distance of I t vs rt. So the distance of I t is equal to the square root of 15 minus five. Quantity squared plus six minus 11 squared well, five minus five is 10 squared is 100 six minus 11 is negative five but squared is 25. And so I left with a radical 1 25 the distance of rt square root. And that's gonna be 15 minus five again squared plus six minus one square again. 15 minus five is 10 10 square. It is 100. You might see where we're going with this. Six minus one is five. Hey, look, five squares 25 we're left with once again radical. 1 25 So this is an isosceles triangle. Andi, we've proven that by showing that two of the sides have the same length Now, there should have been some intuition here. The fact that these points are along the same vertical and at this point right here is exactly in the middle. Um, would indicate that the two blue sides have to be equal. We learned a the're, um, about that before. If you have the perpendicular by sector in this case, the perpendicular by sector of IR than any point on that line or segment or Ray, what have you is going to be equidistant to the end point of that segment? So it would be equidistant I and are so that should make sense that what we have here is true. Okay, here's another example. Okay, this is similar to what we saw, um, in the notes, A previous video and that is proof that we have a triangle show that it is, or prove that it is a right triangle. Two ways to do that. You can show that it's got a right angle, which is gonna make it a right triangle, or show that the three sides, um work with or yeah, show that the Pythagorean theorem holds by showing that the three sides work in terms of a scripless B squared equals C squared. I'm gonna go with, I think, a little less work. And that's showing the slopes and looking for two sides that air have opposite reciprocal slope, which would make the two lines perpendicular and therefore have a right angle. Lets go ahead first, label our points. Okay, so we have negative 10 4, which would be here. Okay, so we're gonna call that point right there. D negative. 10 4 that we have. Negative 41 That point right there is a and then we have negative to five, which is point in and again, we'll connect those points. Now, keep in mind, a visual might give us an indication of something, but it is no proof. So we're gonna go ahead, maybe use it as if anything is gonna be perpendicular. It looks like D A and n a. So let's look for the slopes of D A and A and let's see if their opposite receptacles um, so the slope of D A would be four minus one over negative 10 minus negative four. It gets us a negative three over a negative six. Or, in other words, a pardon me, the top. There is a positive three over a negative six, which is going to give us a value of a negative one half. Now, if we look close enough at our picture, you can see that if I go down 1/2 down, 1/2, down, one over to that is confirmed. And by looking at our pictures, well, we see that you're over one up to over one up too. So it does look like here. We have a slope of negative a half, and here we have a slope of 2/1. So, yeah, we're looking at a right triangle here. We're looking at perpendicular, assuming my points are drawn accurately. My lines were drawn nice and straight, but if we're really gonna prove it, let's use the math. And let's show that we have obvious reciprocal. So the proof of a would be five minus one over negative two minus negative four. That's going to get us a four over to, because negative two plus four, which ends up Equalling to and sure enough. We have opposite reciprocal. So we do indeed have a right triangle. The math shows it, and our picture shows it. There you go. Question three is asking about Show that we have a triangle g e N c r e J E n And it wants just to prove that it's isosceles right triangle, folks. What you would do is essentially both of these steps combined. You'd have to show that not on Lee, Do you have two sides congruent but that those two sides are also have slopes that are opposite reciprocal. So these two steps would have to be shown. So essentially, it's the same thing, so I'm not gonna necessarily repeat that. But you would use distance, you would use slope. Um, and both conditions would have to be met.

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