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Find the image of P(–2, –1) after two reflections; first Ry=-5 and then Rx=1.
A. (–2, –1)
B. (–1, –6)
C. (4, –9)
D. (1, –5)
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e.mccormick
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So the Ry=-5 and Rx=1 are the lines you need to reflect across?
OpenSessame
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Those are the reflections
e.mccormick
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A reflection is usually over a line.... so I want to make sure that is the notation you are using.
OpenSessame
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I think so...
e.mccormick
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http://www.mathopenref.com/reflect.html
Like it says there, "A transformation where each point in a shape appears at an equal distance on the opposite side of a given line - the line of reflection. "
So, start by just doing a rough sketch. Does not need to be exact, but you need to find where the dot and line would be and how far the line is from the dot. If you can do that with math, great! But a sketch can help.
OpenSessame
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Okay, but i dont understand what i have to do...
e.mccormick
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Well, it needs to be an equal distance from the line on the other side of the line.
OpenSessame
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Where did the line come from????
OpenSessame
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You lost me..
e.mccormick
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Ry=-5 \(\leftarrow\) that is the line part.
e.mccormick
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Let me show you an example.
OpenSessame
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Okay...Just show me with this one
e.mccormick
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If I work this that is me giving you an answer. If I work an example you can follow the process and answer your own homework.
Lets say I take this random dot:
|dw:1376455485961:dw|
And lets say I have Ry=1, so the line where every Y is 1.
e.mccormick
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|dw:1376455572591:dw|
OpenSessame
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Well, im really lost man...So i wouldnt even get it
OpenSessame
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That is an example already, im just doing a text review for before i take it. so it doesnt count
e.mccormick
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The important measure is the distance from the dot to the line.
|dw:1376455620753:dw|
Because I need to put the reflection the SAME DISTANCE on the other side of the line.
e.mccormick
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|dw:1376455677255:dw|
OpenSessame
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Do the distance is the number in the equation?
OpenSessame
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Let me try to do mine...one sec
e.mccormick
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The distance is the measure from your point and the line.
e.mccormick
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Yours is done twice.... so see what you get for the first step and I can check it. Then the second step.
OpenSessame
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|dw:1376455841757:dw|
e.mccormick
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OK, that is a good (-2,-1).
OpenSessame
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|dw:1376455914360:dw|
OpenSessame
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Now what do i do exactly?
e.mccormick
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OK, so what is the measure of the distance between your dot and the line?
|dw:1376456012468:dw|
OpenSessame
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-3?
e.mccormick
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Not quite. Tewo problems. This is change in y, not x, and distance is always positive.
OpenSessame
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so 4?
e.mccormick
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Yes. So your point is 4 above the line. That means the reflection must be 4 BELOW the line!
OpenSessame
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so at -5?
OpenSessame
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wait no....-4?
e.mccormick
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The line is at -5. So 4 below -5.... or -5-4=?
e.mccormick
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Above is +, below is -. Similarly, to the left of is -, to the right of is +. Has to do with the Cartesian (xy) plane rules.
OpenSessame
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Ohhhh so -9?
e.mccormick
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Yes. That means the X is the same and the Y is now -9. So, the point after 1 translation is at (-2,-9).
|dw:1376456612998:dw|
e.mccormick
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Now you have the line Rx=1 to do the second reflection over.
OpenSessame
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okay:) i get it!
OpenSessame
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|dw:1376456678313:dw|
OpenSessame
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|dw:1376456695550:dw|
e.mccormick
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So, how far this time?
OpenSessame
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3?
e.mccormick
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\(\large\ddot\smile\)
OpenSessame
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so the point would me three to the right?
OpenSessame
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so (4,-9)????
e.mccormick
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Yes, 3 to the right of 1.
And yep, that is it. (4,-9).
OpenSessame
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wait im gonna see if it matches...
e.mccormick
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Now, once you get good at the math, you should not need a sketch, but if yuou are ever in doubt, the rough sketch is there to help!
OpenSessame
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Okay, the textbook just didnt really explain it
e.mccormick
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The thing I think most people mes up here is the line.... because x=0 is the y axis, and y=0 is the x axis. So they give you a y line but you draw it the same way as the x axis and that can be confusing!
OpenSessame
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yea, thats the right answer! thanks:)
OpenSessame
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Can you help with some other questions that i dont get?
e.mccormick
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Probably. I can explain it the same basic way I always do and hopefully it works like this and you get it!
OpenSessame
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Okay, thanks:)
e.mccormick
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OK. Dilation is going to be the size change. The rigit motion will get it from one place to the other.
OpenSessame
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Okay...
e.mccormick
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It is probably easiest to do the dilation first. That will be some multiplication number. It is a ratio of the sides.
OpenSessame
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Okay...
e.mccormick
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For example, this would be a dialation by 3.....
|dw:1376457753408:dw|
So start by finding how long the sides of both rectangles are.
OpenSessame
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wait is the answer D2*T<0,-6>
e.mccormick
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2x the size, so yah, a dialation of 2. I didn't do the rest yet, but that would probably do it.
OpenSessame
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Really??? woooohoooo
e.mccormick
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Hmmm.... -6.... is that enough on the translation?
OpenSessame
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Idk...
e.mccormick
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Well, lets look at this real quick:
|dw:1376458028692:dw|
Now you need to double the size.
OpenSessame
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The dialation is times 2?
e.mccormick
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|dw:1376458060944:dw|
OK, that makes them reflections of each other.
e.mccormick
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So now, one of two things needs to happne.... It either needs to be reflected over the x axis, OR, corner H needs to go all the way down to the bottom corner.
|dw:1376458143764:dw|
OpenSessame
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so its shifted down -6
e.mccormick
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Well, is that 6? That bottom corner is at -9....
OpenSessame
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-10?
e.mccormick
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or -10, actually.... counted wrong.
e.mccormick
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So it has to go from 2 above to -10 below... so....
OpenSessame
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so uhm...It would be 11?
OpenSessame
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12
e.mccormick
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Yep!
So 2D, T\(\langle 0,-12\rangle\) is more like it.
OpenSessame
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that isnt one of my options
e.mccormick
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Hmmm.... AH.... OK... the dialaion is probably also changing the translation! \(2\langle 0,-6\rangle = \langle 0,-12\rangle\)
e.mccormick
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So my bad. Looks like you did have it right.
OpenSessame
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YUS:) thanks though!
OpenSessame
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The hexagon GIKMPR and FJN are regular. The dashed line segments
form 30° angles. What is r(240drg,0)(G)
e.mccormick
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Well, the good news is you should understand the why a bit better! By exploring these, it lets you understand the why of it all....
OpenSessame
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okay!
e.mccormick
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Hmmm..... I have not seen that form of question.... my geometry was a while back, and they have changed some of the things they do... Do you know what it is looking for, any terms? An angle?
OpenSessame
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240 degrees
OpenSessame
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Thats really all i understand...
e.mccormick
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r is usually a radius.... but there is no distance....
OpenSessame
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I think r is the point?
e.mccormick
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Is it R and not r?
OpenSessame
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Thats the problem..
e.mccormick
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hmmm..... Well, 240/30=8 segments.
OpenSessame
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so 8 to the left or right?
e.mccormick
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Well, if r is right... but that would be a guess on my part. Right 8 of G? I would need to see the reference to know they type of peoblem to get this one.
OpenSessame
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So would it be K???
e.mccormick
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I would only be guessing without the book and chapter.
OpenSessame
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Okay...Well for me thats better then nothing!
OpenSessame
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The vertices of a triangle are P(–8, 6), Q(1, –3), and R(–6, –3). Name the vertices of \[R _{y=x}(PQR)\]
e.mccormick
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Is that a reflection over the y axis?
OpenSessame
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I believe it is.
OpenSessame
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You did??
OpenSessame
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OOOOO
e.mccormick
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The composition part does not matter here.... the \(R_y\) part is reflection over the y axis!
OpenSessame
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So just make it the other sign?
OpenSessame
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Would the first point be (-6,8)
OpenSessame
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OHHHH!!! MAKES SENSE!
OpenSessame
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so (6,-8)
e.mccormick
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Amazing what finding a reference to the symbols can do!
OpenSessame
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YEA! Thanks so much man!
OpenSessame
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Write a sequence of rigid motions that maps ab and xy
OpenSessame
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LAST ONE AND I WILL UNDERSTAND GEOMETRY!
e.mccormick
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That composition part also confirms what we talked about with the <0,-6> becoming <0,-12>. The SECOND part of the compositon happens first!
OpenSessame
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the second part is (AB=XY)
e.mccormick
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Are those seperate calculations for those two?
OpenSessame
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The top doesnt matter just look at the lines
e.mccormick
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Yah, well, then that is some sort of what? Rotation, reflection etc?
OpenSessame
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Reflection?
e.mccormick
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Remember that A is going to X and B is going to Y..... should be a huge clue....
YES! Some sort of reflection.
e.mccormick
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So, you just need to find the line they are reflected over.
e.mccormick
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It would be half way between them.
OpenSessame
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0?
e.mccormick
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Are they both the same distance from x=0?
OpenSessame
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yES...
e.mccormick
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Did you count the distance from 0 to X and 0 to A?
OpenSessame
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no...
e.mccormick
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Take a second look.... they are not the same.
OpenSessame
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Oh...
e.mccormick
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You want the line half way between. So take the distance between the two points and divide by 2. It will be that far to the left of the right most line, and right of the left most line.
OpenSessame
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okay...