Square [tex]\(ABCD\)[/tex] was translated using the rule [tex]\((x, y) \rightarrow (x-4, y+15)\)[/tex] to form [tex]\(A^{\prime}B^{\prime}C^{\prime}D^{\prime}\)[/tex]. What are the coordinates of point [tex]\(D\)[/tex] in the pre-image if the coordinates of point [tex]\(D^{\prime}\)[/tex] in the image are [tex]\((9,-8)\)[/tex]?

A. [tex]\((13,-23)\)[/tex]
B. [tex]\((5,7)\)[/tex]
C. [tex]\((18, 1)\)[/tex]
D. [tex]\((-6,-4)\)[/tex]



Answer :

To find the coordinates of point [tex]\(D\)[/tex] in the pre-image given the coordinates of point [tex]\(D'\)[/tex] in the image and the translation rule, follow these steps:

1. Understand the translation rule: The translation rule given is [tex]\((x, y) \to (x-4, y+15)\)[/tex]. This means that for every original point [tex]\((x, y)\)[/tex] in the pre-image, the new point in the image is obtained by subtracting 4 from the [tex]\(x\)[/tex]-coordinate and adding 15 to the [tex]\(y\)[/tex]-coordinate.

2. Express given coordinates of [tex]\(D'\)[/tex]: The coordinates of [tex]\(D'\)[/tex] in the image are [tex]\((9, -8)\)[/tex].

3. Setup the equations based on the translation rule:
Since [tex]\(D'\)[/tex] was obtained by translating [tex]\(D\)[/tex],
[tex]\[ D' = (Dx - 4, Dy + 15) \][/tex]
Given that [tex]\(D'\)[/tex] is [tex]\((9, -8)\)[/tex], we have:
[tex]\[ Dx - 4 = 9 \][/tex]
[tex]\[ Dy + 15 = -8 \][/tex]

4. Solve for [tex]\(Dx\)[/tex]:
Add 4 to both sides of the equation [tex]\(Dx - 4 = 9\)[/tex]:
[tex]\[ Dx - 4 + 4 = 9 + 4 \][/tex]
[tex]\[ Dx = 13 \][/tex]

5. Solve for [tex]\(Dy\)[/tex]:
Subtract 15 from both sides of the equation [tex]\(Dy + 15 = -8\)[/tex]:
[tex]\[ Dy + 15 - 15 = -8 - 15 \][/tex]
[tex]\[ Dy = -23 \][/tex]

6. Conclusion:
The coordinates of point [tex]\(D\)[/tex] in the pre-image are [tex]\((13, -23)\)[/tex].

Thus, the coordinates of point [tex]\(D\)[/tex] in the pre-image are [tex]\((13, -23)\)[/tex].