Answer:
See Explanation
Step-by-step explanation:
The question has missing details. So, I'll solve using a general assumption.
Let one of the coordinates of the figure be
[tex]A = (x,y)[/tex]
Let the scale factor be n
When dilated, the new figure is:
[tex]A' = n * A[/tex]
[tex]A' = n * (x,y)[/tex]
To return the image back to the original figure, the new scale factor mus be a reciprocal of the previous scale factor i.e. [tex]\frac{1}{n}[/tex]
So:
[tex]A = A' * \frac{1}{n}[/tex]
Substitute n(x,y) for A';
[tex]A = n(x,y) * \frac{1}{n}[/tex]
[tex]A = \frac{n(x,y)}{n}[/tex]
[tex]A = (x,y)[/tex]
Take for instance:
[tex]A = (2,3)[/tex]
[tex]n = 2[/tex] -- scale factor
[tex]A' = n * A[/tex]
[tex]A' = 2 * (2,3)[/tex]
[tex]A' = (4,6)[/tex]
To get A from A', using the analysis above
[tex]A = \frac{1}{2} * (4,6)[/tex]
[tex]A = (2,3)[/tex]