Square ABCD was translated using the rule (x, y) → (x – 4, y + 15) to form A'B'C'D'. What are the coordinates of point D in the pre-image if the coordinates of point D’ in the image are (9, –8)? (13, –23) (5, 7) (18, 1) (–6, –4)

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Square ABCD was translated using the rule (x, y) → (x – 4, y + 15) to form A'B'C'D'. What are the coordinates of point D in the pre-image if the coordinates of point D’ in the image are (9, –8)? (13, –23) (5, 7) (18, 1) (–6, –4)

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i was once told subtract the original from the spicy... to me it didnt make sense but here how it work\[\left(\begin{matrix}9 \\ -8\end{matrix}\right)-\left(\begin{matrix}-4 \\ 15\end{matrix}\right)=\left(\begin{matrix}13 \\ -23\end{matrix}\right)\]
As an alternative to @lynfran's method, you could also look at it as follows: The transformation is T: (x,y) -> (x-4, y+15) to go from ABCD to A'B'C'D'. The inverse transformation, \(T^{-1}\) is given by: \(T^{-1}\) : (x,y) -> (x+4, y-15) So apply the inverse transformation to D' to get back D.

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so, it's A?
yea

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