anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
e.mccormick
  • e.mccormick
So the Ry=-5 and Rx=1 are the lines you need to reflect across?
anonymous
  • anonymous
Those are the reflections
e.mccormick
  • e.mccormick
A reflection is usually over a line.... so I want to make sure that is the notation you are using.

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anonymous
  • anonymous
I think so...
e.mccormick
  • e.mccormick
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.
anonymous
  • anonymous
Okay, but i dont understand what i have to do...
e.mccormick
  • e.mccormick
Well, it needs to be an equal distance from the line on the other side of the line.
anonymous
  • anonymous
Where did the line come from????
anonymous
  • anonymous
You lost me..
e.mccormick
  • e.mccormick
Ry=-5 \(\leftarrow\) that is the line part.
e.mccormick
  • e.mccormick
Let me show you an example.
anonymous
  • anonymous
Okay...Just show me with this one
e.mccormick
  • e.mccormick
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
  • e.mccormick
|dw:1376455572591:dw|
anonymous
  • anonymous
Well, im really lost man...So i wouldnt even get it
anonymous
  • anonymous
That is an example already, im just doing a text review for before i take it. so it doesnt count
e.mccormick
  • e.mccormick
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
  • e.mccormick
|dw:1376455677255:dw|
anonymous
  • anonymous
Do the distance is the number in the equation?
anonymous
  • anonymous
Let me try to do mine...one sec
e.mccormick
  • e.mccormick
The distance is the measure from your point and the line.
e.mccormick
  • e.mccormick
Yours is done twice.... so see what you get for the first step and I can check it. Then the second step.
anonymous
  • anonymous
|dw:1376455841757:dw|
e.mccormick
  • e.mccormick
OK, that is a good (-2,-1).
anonymous
  • anonymous
|dw:1376455914360:dw|
anonymous
  • anonymous
Now what do i do exactly?
e.mccormick
  • e.mccormick
OK, so what is the measure of the distance between your dot and the line? |dw:1376456012468:dw|
anonymous
  • anonymous
-3?
e.mccormick
  • e.mccormick
Not quite. Tewo problems. This is change in y, not x, and distance is always positive.
anonymous
  • anonymous
so 4?
e.mccormick
  • e.mccormick
Yes. So your point is 4 above the line. That means the reflection must be 4 BELOW the line!
anonymous
  • anonymous
so at -5?
anonymous
  • anonymous
wait no....-4?
e.mccormick
  • e.mccormick
The line is at -5. So 4 below -5.... or -5-4=?
e.mccormick
  • e.mccormick
Above is +, below is -. Similarly, to the left of is -, to the right of is +. Has to do with the Cartesian (xy) plane rules.
anonymous
  • anonymous
Ohhhh so -9?
e.mccormick
  • e.mccormick
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
  • e.mccormick
Now you have the line Rx=1 to do the second reflection over.
anonymous
  • anonymous
okay:) i get it!
anonymous
  • anonymous
|dw:1376456678313:dw|
anonymous
  • anonymous
|dw:1376456695550:dw|
e.mccormick
  • e.mccormick
So, how far this time?
anonymous
  • anonymous
3?
e.mccormick
  • e.mccormick
\(\large\ddot\smile\)
anonymous
  • anonymous
so the point would me three to the right?
anonymous
  • anonymous
so (4,-9)????
e.mccormick
  • e.mccormick
Yes, 3 to the right of 1. And yep, that is it. (4,-9).
anonymous
  • anonymous
wait im gonna see if it matches...
e.mccormick
  • e.mccormick
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!
anonymous
  • anonymous
Okay, the textbook just didnt really explain it
e.mccormick
  • e.mccormick
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!
anonymous
  • anonymous
yea, thats the right answer! thanks:)
anonymous
  • anonymous
Can you help with some other questions that i dont get?
e.mccormick
  • e.mccormick
Probably. I can explain it the same basic way I always do and hopefully it works like this and you get it!
anonymous
  • anonymous
Okay, thanks:)
anonymous
  • anonymous
What composition of rigid motions and a dilation maps EFGH to the dashed figure? https://study.ashworthcollege.edu/access/content/group/45b8c516-1008-46d7-aa1d-bb9b62c786ff/geometry_exam_9_files/mc020-1.jpg
e.mccormick
  • e.mccormick
OK. Dilation is going to be the size change. The rigit motion will get it from one place to the other.
anonymous
  • anonymous
Okay...
e.mccormick
  • e.mccormick
It is probably easiest to do the dilation first. That will be some multiplication number. It is a ratio of the sides.
anonymous
  • anonymous
Okay...
e.mccormick
  • e.mccormick
For example, this would be a dialation by 3..... |dw:1376457753408:dw| So start by finding how long the sides of both rectangles are.
anonymous
  • anonymous
wait is the answer D2*T<0,-6>
e.mccormick
  • e.mccormick
2x the size, so yah, a dialation of 2. I didn't do the rest yet, but that would probably do it.
anonymous
  • anonymous
Really??? woooohoooo
e.mccormick
  • e.mccormick
Hmmm.... -6.... is that enough on the translation?
anonymous
  • anonymous
Idk...
e.mccormick
  • e.mccormick
Well, lets look at this real quick: |dw:1376458028692:dw| Now you need to double the size.
anonymous
  • anonymous
The dialation is times 2?
e.mccormick
  • e.mccormick
|dw:1376458060944:dw| OK, that makes them reflections of each other.
e.mccormick
  • e.mccormick
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|
anonymous
  • anonymous
so its shifted down -6
e.mccormick
  • e.mccormick
Well, is that 6? That bottom corner is at -9....
anonymous
  • anonymous
-10?
e.mccormick
  • e.mccormick
or -10, actually.... counted wrong.
e.mccormick
  • e.mccormick
So it has to go from 2 above to -10 below... so....
anonymous
  • anonymous
so uhm...It would be 11?
anonymous
  • anonymous
12
e.mccormick
  • e.mccormick
Yep! So 2D, T\(\langle 0,-12\rangle\) is more like it.
anonymous
  • anonymous
that isnt one of my options
e.mccormick
  • e.mccormick
Hmmm.... AH.... OK... the dialaion is probably also changing the translation! \(2\langle 0,-6\rangle = \langle 0,-12\rangle\)
e.mccormick
  • e.mccormick
So my bad. Looks like you did have it right.
anonymous
  • anonymous
YUS:) thanks though!
anonymous
  • anonymous
The hexagon GIKMPR and FJN are regular. The dashed line segments form 30° angles. What is r(240drg,0)(G)
anonymous
  • anonymous
https://study.ashworthcollege.edu/access/content/group/45b8c516-1008-46d7-aa1d-bb9b62c786ff/geometry_exam_9_files/nar001-2.jpg
e.mccormick
  • e.mccormick
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....
anonymous
  • anonymous
okay!
e.mccormick
  • e.mccormick
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?
anonymous
  • anonymous
240 degrees
anonymous
  • anonymous
Thats really all i understand...
e.mccormick
  • e.mccormick
r is usually a radius.... but there is no distance....
anonymous
  • anonymous
I think r is the point?
e.mccormick
  • e.mccormick
Is it R and not r?
anonymous
  • anonymous
https://study.ashworthcollege.edu/access/content/group/45b8c516-1008-46d7-aa1d-bb9b62c786ff/geometry_exam_9_files/mc011-1.jpg
anonymous
  • anonymous
Thats the problem..
e.mccormick
  • e.mccormick
hmmm..... Well, 240/30=8 segments.
anonymous
  • anonymous
so 8 to the left or right?
e.mccormick
  • e.mccormick
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.
anonymous
  • anonymous
So would it be K???
e.mccormick
  • e.mccormick
I would only be guessing without the book and chapter.
anonymous
  • anonymous
Okay...Well for me thats better then nothing!
anonymous
  • anonymous
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
  • e.mccormick
Is that a reflection over the y axis?
anonymous
  • anonymous
I believe it is.
e.mccormick
  • e.mccormick
AH HA! Found a reference! http://www.regentsprep.org/Regents/math/geometry/GT6/composition.htm
anonymous
  • anonymous
You did??
anonymous
  • anonymous
OOOOO
e.mccormick
  • e.mccormick
The composition part does not matter here.... the \(R_y\) part is reflection over the y axis!
anonymous
  • anonymous
So just make it the other sign?
e.mccormick
  • e.mccormick
And even better! http://www.regentsprep.org/Regents/math/geometry/GT5/reviewTranformations.htm
anonymous
  • anonymous
Would the first point be (-6,8)
anonymous
  • anonymous
OHHHH!!! MAKES SENSE!
anonymous
  • anonymous
so (6,-8)
e.mccormick
  • e.mccormick
Amazing what finding a reference to the symbols can do!
anonymous
  • anonymous
YEA! Thanks so much man!
anonymous
  • anonymous
Write a sequence of rigid motions that maps ab and xy
anonymous
  • anonymous
https://study.ashworthcollege.edu/access/content/group/45b8c516-1008-46d7-aa1d-bb9b62c786ff/geometry_exam_9_files/nar003-1.jpg
anonymous
  • anonymous
LAST ONE AND I WILL UNDERSTAND GEOMETRY!
e.mccormick
  • e.mccormick
That composition part also confirms what we talked about with the <0,-6> becoming <0,-12>. The SECOND part of the compositon happens first!
anonymous
  • anonymous
the second part is (AB=XY)
e.mccormick
  • e.mccormick
Are those seperate calculations for those two?
anonymous
  • anonymous
The top doesnt matter just look at the lines
e.mccormick
  • e.mccormick
Yah, well, then that is some sort of what? Rotation, reflection etc?
anonymous
  • anonymous
Reflection?
e.mccormick
  • e.mccormick
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
  • e.mccormick
So, you just need to find the line they are reflected over.
e.mccormick
  • e.mccormick
It would be half way between them.
anonymous
  • anonymous
0?
e.mccormick
  • e.mccormick
Are they both the same distance from x=0?
anonymous
  • anonymous
yES...
e.mccormick
  • e.mccormick
Did you count the distance from 0 to X and 0 to A?
anonymous
  • anonymous
no...
e.mccormick
  • e.mccormick
Take a second look.... they are not the same.
anonymous
  • anonymous
Oh...
e.mccormick
  • e.mccormick
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.
anonymous
  • anonymous
okay...

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