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)

- anonymous

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- e.mccormick

So the Ry=-5 and Rx=1 are the lines you need to reflect across?

- anonymous

Those are the reflections

- 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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## More answers

- anonymous

I think so...

- 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

Okay, but i dont understand what i have to do...

- e.mccormick

Well, it needs to be an equal distance from the line on the other side of the line.

- anonymous

Where did the line come from????

- anonymous

You lost me..

- e.mccormick

Ry=-5 \(\leftarrow\) that is the line part.

- e.mccormick

Let me show you an example.

- anonymous

Okay...Just show me with this one

- 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

|dw:1376455572591:dw|

- anonymous

Well, im really lost man...So i wouldnt even get it

- anonymous

That is an example already, im just doing a text review for before i take it. so it doesnt count

- 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

|dw:1376455677255:dw|

- anonymous

Do the distance is the number in the equation?

- anonymous

Let me try to do mine...one sec

- e.mccormick

The distance is the measure from your point and the line.

- 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

|dw:1376455841757:dw|

- e.mccormick

OK, that is a good (-2,-1).

- anonymous

|dw:1376455914360:dw|

- anonymous

Now what do i do exactly?

- e.mccormick

OK, so what is the measure of the distance between your dot and the line?
|dw:1376456012468:dw|

- anonymous

-3?

- e.mccormick

Not quite. Tewo problems. This is change in y, not x, and distance is always positive.

- anonymous

so 4?

- e.mccormick

Yes. So your point is 4 above the line. That means the reflection must be 4 BELOW the line!

- anonymous

so at -5?

- anonymous

wait no....-4?

- e.mccormick

The line is at -5. So 4 below -5.... or -5-4=?

- 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

Ohhhh so -9?

- 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

Now you have the line Rx=1 to do the second reflection over.

- anonymous

okay:) i get it!

- anonymous

|dw:1376456678313:dw|

- anonymous

|dw:1376456695550:dw|

- e.mccormick

So, how far this time?

- anonymous

3?

- e.mccormick

\(\large\ddot\smile\)

- anonymous

so the point would me three to the right?

- anonymous

so (4,-9)????

- e.mccormick

Yes, 3 to the right of 1.
And yep, that is it. (4,-9).

- anonymous

wait im gonna see if it matches...

- 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

Okay, the textbook just didnt really explain it

- 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

yea, thats the right answer! thanks:)

- anonymous

Can you help with some other questions that i dont get?

- 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

Okay, thanks:)

- 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

OK. Dilation is going to be the size change. The rigit motion will get it from one place to the other.

- anonymous

Okay...

- 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

Okay...

- 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

wait is the answer D2*T<0,-6>

- 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

Really??? woooohoooo

- e.mccormick

Hmmm.... -6.... is that enough on the translation?

- anonymous

Idk...

- e.mccormick

Well, lets look at this real quick:
|dw:1376458028692:dw|
Now you need to double the size.

- anonymous

The dialation is times 2?

- e.mccormick

|dw:1376458060944:dw|
OK, that makes them reflections of each other.

- 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

so its shifted down -6

- e.mccormick

Well, is that 6? That bottom corner is at -9....

- anonymous

-10?

- e.mccormick

or -10, actually.... counted wrong.

- e.mccormick

So it has to go from 2 above to -10 below... so....

- anonymous

so uhm...It would be 11?

- anonymous

12

- e.mccormick

Yep!
So 2D, T\(\langle 0,-12\rangle\) is more like it.

- anonymous

that isnt one of my options

- e.mccormick

Hmmm.... AH.... OK... the dialaion is probably also changing the translation! \(2\langle 0,-6\rangle = \langle 0,-12\rangle\)

- e.mccormick

So my bad. Looks like you did have it right.

- anonymous

YUS:) thanks though!

- anonymous

The hexagon GIKMPR and FJN are regular. The dashed line segments
form 30° angles. What is r(240drg,0)(G)

- anonymous

https://study.ashworthcollege.edu/access/content/group/45b8c516-1008-46d7-aa1d-bb9b62c786ff/geometry_exam_9_files/nar001-2.jpg

- 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

okay!

- 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

240 degrees

- anonymous

Thats really all i understand...

- e.mccormick

r is usually a radius.... but there is no distance....

- anonymous

I think r is the point?

- e.mccormick

Is it R and not r?

- anonymous

https://study.ashworthcollege.edu/access/content/group/45b8c516-1008-46d7-aa1d-bb9b62c786ff/geometry_exam_9_files/mc011-1.jpg

- anonymous

Thats the problem..

- e.mccormick

hmmm..... Well, 240/30=8 segments.

- anonymous

so 8 to the left or right?

- 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

So would it be K???

- e.mccormick

I would only be guessing without the book and chapter.

- anonymous

Okay...Well for me thats better then nothing!

- 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

Is that a reflection over the y axis?

- anonymous

I believe it is.

- e.mccormick

AH HA! Found a reference!
http://www.regentsprep.org/Regents/math/geometry/GT6/composition.htm

- anonymous

You did??

- anonymous

OOOOO

- e.mccormick

The composition part does not matter here.... the \(R_y\) part is reflection over the y axis!

- anonymous

So just make it the other sign?

- e.mccormick

And even better!
http://www.regentsprep.org/Regents/math/geometry/GT5/reviewTranformations.htm

- anonymous

Would the first point be (-6,8)

- anonymous

OHHHH!!! MAKES SENSE!

- anonymous

so (6,-8)

- e.mccormick

Amazing what finding a reference to the symbols can do!

- anonymous

YEA! Thanks so much man!

- anonymous

Write a sequence of rigid motions that maps ab and xy

- anonymous

https://study.ashworthcollege.edu/access/content/group/45b8c516-1008-46d7-aa1d-bb9b62c786ff/geometry_exam_9_files/nar003-1.jpg

- anonymous

LAST ONE AND I WILL UNDERSTAND GEOMETRY!

- 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

the second part is (AB=XY)

- e.mccormick

Are those seperate calculations for those two?

- anonymous

The top doesnt matter just look at the lines

- e.mccormick

Yah, well, then that is some sort of what? Rotation, reflection etc?

- anonymous

Reflection?

- 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

So, you just need to find the line they are reflected over.

- e.mccormick

It would be half way between them.

- anonymous

0?

- e.mccormick

Are they both the same distance from x=0?

- anonymous

yES...

- e.mccormick

Did you count the distance from 0 to X and 0 to A?

- anonymous

no...

- e.mccormick

Take a second look.... they are not the same.

- anonymous

Oh...

- 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

okay...

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