- anonymous

∆ABC is reflected across the x-axis and translated 2 units up to give the image ∆A'B'C' with coordinates A'(1, 1), B'(2, -1), C'(2, 1). What are the coordinates of the preimage ∆ABC?

- katieb

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- anonymous

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- anonymous

Well for starters if something is reflected across the x-axis, then the y-coordinate of a point will become the opposite.

- anonymous

So in this simple figure note how reflecting it across the x-axis just corresponds to the Y-VALUE being negated

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## More answers

- anonymous

Secondly, if something is translated up (or down) a certain amount then it corresponds to adding (or subtracting) that amount from the Y-VALUE of the point

- anonymous

It's just I have to use one of these answers, and I'm very confused and horrible at this.
A(-1, 1), B(-2, 3), C(-2, 1)
A(1, 1), B(2, 3), C(2, 1)
A(1, -3), B(2, -1), C(2, -3)
A(1, -3), B(2, -5), C(2, -3)

- anonymous

Oh sorry. I didn't mean to interrupt.

- anonymous

Ok so given the order in the problem, first reflect then translate, we must apply those two steps in the process IN THAT ORDER. Note if you reverse the order you may (most likely will in this case) get a different answer.

- anonymous

No problem

- anonymous

Ok so please list the verticies of triangle A'B'C' after you NEGATE (find the opposite) of their y-coordinate please.

- anonymous

In other words, list them like this:
A'(1, 1), B'(2, -1), C'(2, 1)
But insert the opposite of the y-coordinate, which corresponds to reflecting about the x-axis.

- anonymous

Ill give you one:
A'(1, -1)

- anonymous

I'm very bad at this.. I get confused easily. So I'm not completely sure how to do any of this. What am I supposed to do with A'(1, -1)? I apologize for making this complicated.

- anonymous

Its ok, but I cant just give you the answer or you will never get better, just like if you were lifting weights... if I lifted the weight for you you wouldn't gain the muscle.... that being said I want to take you step by step through the process:
I want you to find the opposite (i.e. negate) the y-values in the following points:
A'(1, 1), B'(2, -1), C'(2, 1)
To help you along, I will demonstrate what I am looking for by doing the first point for you:
A'(1, -1) (note I negated the y-value)
Now find the other two points please.

- anonymous

Do you understand what I mean by negate?

- anonymous

Not really... I'm just starting this class and it's not going well. And I understand. I am glad you're not just giving the answers. I really wanna understand this.

- anonymous

So when you negate something it is the same as just making it negative. For example, if I have 5, then the OPPOSITE of that number is -5. If I you are asked to NEGATE 5, then the answer is -5.
If the quantity they give you is already negative, no matter :D. The OPPOSITE of -7 is -(-7)=7. If asked to NEGATE -3, then you have -(-3)=3. Understand?

- anonymous

@emma.farrell still there?

- anonymous

Yes, sorry. Just had to help my mom with something. I think I get it.

- anonymous

Ok so please list the points so I can check to make sure

- anonymous

The points from the problem?

- anonymous

Yes

- anonymous

Find the opposite of the y-coordinate (the second value in the order pairs) for all three points A'B'C' (just copy paste what I posted above and just modify the corresponding y-values)

- anonymous

A'(1, 1), B'(2, -1), C'(2, 1)
A' (-1, -1) B' (-2, 1), C' (-2, -1)
Is that what you mean?

- anonymous

Sorry try again... remember only the y-value is negated because (looking at my picture above) the vertical axis is the y-axis and those were the values that changed. The x-values (the first value in the ordered pair) say the same.

- anonymous

Okay, thanks.
A'(1, 1), B'(2, -1), C'(2, 1)
A'(1, -1), B'(2, 1), C'(2, -1)
Is that better? Or am I looking at/doing the wrong thing?

- anonymous

Ok that is correct.

- anonymous

Now we must translate these points UP by 2. Do you think you can do that? Give it a shot and post your answer and I will let you know.

- anonymous

Wait, what do you mean translate them up? Sorry, I don't mean to be difficult.

- anonymous

Its ok... when something is "translated (up/down/left/right)" that is the mathematical way of saying shifted or moved in that direction.

- anonymous

Imagine moving the whole object up by two units... or pointwise... imagine moving each of the the vertices up by two units

- anonymous

Anyways about your message, I can empathize I am exactly the same. But feel good because you can go on to be amazing at whatever you choose. It is merely a question of getting interested in the boring stuff so you can master it so that when you get to the more interesting stuff you can truely enjoy it.

- anonymous

Ok so as for this problem please do be scared to take a guess. One of the worst things you can do in math is be scared to take a shot at a problem. If you get it wrong no worries Im here to help so what do you think should happen to the points if I shift the figure up by two?

- anonymous

@emma.farrell are you there?

- anonymous

I'm sorry! I don't mean to keep disappearing, but I've gotten a bit busy. Would it be a bother if I came back to this in a bit? I've got some personal things to take care of.

- anonymous

Emma if you focus for a minute or two you can finish this problem

- anonymous

So all you need to do is answer the question what happens to these points when I translate UP by 2:
A'(1, -1), B'(2, 1), C'(2, -1)

- anonymous

I have to go now... In order to translate up simply shift the y-values of your verticies by 2... i.e. add two to the y-values of your responses:
A'(1, -1+2), B'(2, 1+2), C'(2, -1+2)=
A'(1, 1), B'(2, 3), C'(2, 1)

- anonymous

Thank fully before I left, I reread the question.... Unfortunately I misread it the first time. Everything we discussed was correct only I solved the problem backwards. We have the final position A'(1, 1), B'(2, -1), C'(2, 1) and our goal is to find the initial image (aka the preImage) ABC.
This means we must do the opposite of everything, including the steps (i.e. we have to deal with the translation first, then the reflection).
First off, A'(1, 1), B'(2, -1), C'(2, 1) has already been shifted up by two (since its the final image), thus we must undo this shift by shifting it DOWN by 2:
A'(1, 1-2), B'(2, -1-2), C'(2, 1-2)=A'(1, -1), B'(2, -3), C'(2, -1)
Then we reflect about the x-axis (again by negating the y-values):
Therefore, A(1, 1), B(2, 3), C(2, 1)
As it turns out this is the same answer. Note once the process finished I dropped the primes on A, B, & C which I neglected to do on my last post above.

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