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A figure is transformed by T(x, y) = (x+3, y-1) and then transformed by S(x-3, y + 1). How does the preimage relate to the final image?​ please show work sorry

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Answer: The preimage and final image are the same point

This applies to every point in the xy plane.

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Step-by-step explanation:

Let's say we started with the example point of (5,7)

Put the coordinates of the point into the T(x,y) transformation

We'll replace every x with 5 and every y with 7

So,

T(x, y) = (x+3, y-1)

T(5, 7) = (5+3, 7-1)

T(5,7) = (8, 6)

What does this mean? It means that the original preimage point (5,7) moves to (8,6). This is a shift of 3 units right and 1 unit down.

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Now we'll plug the coordinates of that result, the (8,6), into the transformation of S(x, y) = (x-3, y + 1)

Plug in x = 8 and y = 6 to get...

S(x, y) = (x-3, y + 1)

S(8, 6) = (8-3, 6 + 1)

S(8, 6) = (5, 7)

We have gone from (8,6) to (5,7). We're back where we started.

The second transformation of (x-3, y+1) shifts 3 units left and 1 unit up. Notice both of these undo the original "3 units right and 1 unit down" from earlier. So this is why we get back to the starting point. It's not a coincidence.

Therefore, transformation T and transformation S are inverse transformations of one another. They undo each other. It's like taking a step forward (transformation T) and then taking a step back (transformation S)

In terms of algebra, it's like how multiplication and division are inverses of each other.

As you can probably guess, there's nothing particularly special about (5,7). I chose two random numbers. Pick whatever two favorite numbers you want for x and y. Plug them into T(x,y) and then whatever that result is, plug it into S(x,y) to notice you should get back to your original choice of (x,y). I recommend GeoGebra as a useful tool to help with geometric transformations.

User Alexander Volkov
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