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[-/0.71 Points]DETAILSWANEFMZ 1.3.062.NVAWncar cquation whose graph Through (1, elopcstralaht Ilne with thc gwcn property: HINT (Sce Example 2.]Y(x)Jokana Sred Work...

Question

[-/0.71 Points]DETAILSWANEFMZ 1.3.062.NVAWncar cquation whose graph Through (1, elopcstralaht Ilne with thc gwcn property: HINT (Sce Example 2.]Y(x)Jokana Sred Work RormLart Rerronmn[0/0.71 Points]DETAILSPREvIOUS ANSIVERSWANEFMG 43.084.

[-/0.71 Points] DETAILS WANEFMZ 1.3.062.NVA Wncar cquation whose graph Through (1, elopc stralaht Ilne with thc gwcn property: HINT (Sce Example 2.] Y(x) Jokana Sred Work Rorm Lart Rerronmn [0/0.71 Points] DETAILS PREvIOUS ANSIVERS WANEFMG 43.084.



Answers

Plot the given points on graph paper. Draw $\triangle A B C$ and $\triangle D E F$ . Copy and complete the statement $\triangle A B C \cong \longrightarrow$ .
$$
\begin{array}{lll}{A(-7,-3)} & {B(-2,-3)} & {C(-2,0)} \\ {D(0,1)} & {E(5,1)} & {F(0,-2)}\end{array}
$$

Okay, So for this program, we just need to approach the puns. Um, Ms Coordinate. Um So for example, the had A We had this point. Truthful four. So put him to this. Golden made the point. This should be here. And if we do the projection project to X Y planned and the XP print and the wisely plan, we have three different points on the on three different plants and the wilderness. Anything for the 1 to 5 points. And the this, um minus 205 that for these miners to the or five. Because this point is already on y z plants. So we did a projection onto the third. This punk is only xz plane safer. If we do the projection to the exit plan, you have yourself.

Do you hear? They've given us several points of the first of which ISS the 0.2 to 4. And they ask us to, uh, plot and label the points A, B and C, which are the points projected onto the X Y plane, the X Z plane in the Y Z plane, respectively. So, by the way we do that it was first just plot where PS subpoenas to in our extraction to in our Y direction and then four in our Z direction. So maybe it's somewhere like here. Let me see that this kind of forms a rectangular prism if we consider the origin one vortex of erecting your prism. And this point P is the opposite vertex of this retinue of prison, we form to reckon you're prisons like this. Um, wait for my writing your prison with this with two points. Excuse me on. And now we're interested in finding the points. Yeah, this would be a on the x Z plane E on X Y plane and see. Oh, excuse me. I got this. I got this backwards. Um, they should be on the X y plane. Should be a Z plan. Should be and, um z y plan should be. See eso No way do we go about this? We'll use a is the first example A eyes just gonna be directly below our point P is gonna have the same X and y coordinates. It's just going to have a zero in the Z direction, right? So we've kind of projected our point peeled down to our X y plane by making Z zero. And when you make the zero X and why don't change eso? What we're left with is like if we had a light shining above the point p, its shadow would be a right directly below it. Uh, we haven't changed the values of our X or why we only projected Z predicted the point down to where z zero I second on the same thing for be right, instead of projecting onto the X Y plane we're projecting of the xz plane. So in this case of why zeros, we can fill that in like that, and like before, Accent Z aren't going to change. So you can feel that in just like that, so being is the point 204 which is the point p projected onto the XY plane and then finally see is the point p projected onto the Y Z plane, which is, um, where? X zero. So we fill in a zero for X and then y Z remain the same. So it's 0 to 4, right? So this is this is for the point p, which is what? Which was the first example. Our second example. I'll do that over here, maybe. P is 1 to 5125 And now I'm not gonna change the plot because it's working in the same way, right? We have a point here. That's our point. P a label that p um we're projecting a B and C into the X y plane, the X Z plane in the Y Z plane, respectively. So again, a we're gonna keep our X and Y coordinates 12 and R Z coordinate is gonna be zero for be. We're gonna keep our X and Z coordinate and why corner is gonna be zero. And for C, we do the same thing, but our X coordinate zero. So we keep the wine physique Word on in this way, we kind of are projecting our point p onto these different planes. Eso now, one more example and hopefully you see the pattern and don't really even move. Teoh, draw the, um that figure toe understand it. I'll still draw the figure anyway because I guess now we're in a different, um, a different quadrant often. Excuse me. Eso still These are X, y and Z, although now x value is negative. So we extend this in the negative direction, will try to draw that little bit Streeter standing in the native direction like this. So your negative too Zero in the Y Direction five in the Z direction. So our point is up here just like this. Now, this is tricky because we have a zero in the Y direction already. So we think about projecting this. This point is actually already on, um, the xz plane, right? Eso think about the point A, which remember, was the projection onto the X Y plane. It's going to be have the same X and y coordinates negative to zero, but just projected onto the X Y plants is Z becomes your as well. So we're direction just the point. Negative. 200 directly on the negative X axis right here. Um, the point being is the projection onto the ecstasy plane on. We see. Actually, our point is already on the XY plane are white corn. It is already there. So no change rooted here. We're gonna do negative t 05 And, um lastly, point c is projected onto the y Z plane eso instead of what we thought If I pushing back now, we're gonna bring it up to our Y Z plane here. X is going to become zero, right? This is a projection onto the y Z plains of X zero. But why and z remained the same? I mean unchanged. So why I still zero NZ's to five and we see this is just a point directly on the positive z axis. Eso We're doing the same thing, the zero when we're looking at my trip us up. But we're really doing the same thing. Just when we project onto a particular plane, Uh, we're setting one of those components X, y or Z 20 s so that our new point wise on on that plane

So first off, we're gonna graph this triangle. So excess to five. Why is 31 NZ is 202 when we're going to wrote this 90 degrees about the point Negative to negative one point, Pete. So this is my original triangle and we're rotating it. Nine degrees for point Pete. So my new point for Z will be negative. Five Positive one. My new point for X would be negative 83 And my new cornet for Why will be negative for for So this is going to be my new triangle my new rotated triangle.

As you can see, this one is simply bigger than this one proportionally. But they're both Quadra laterals with four sides.


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