To find the coordinates of point [tex]\( D \)[/tex] in the pre-image, given the translation rule and the coordinates of point [tex]\( D' \)[/tex] in the image, we need to reverse the translation operation.
The translation rule provided is:
[tex]\[
(x, y) \rightarrow (x-4, y+15)
\][/tex]
This means that for any point [tex]\((x, y)\)[/tex] in the original pre-image, its coordinates in the image will be:
[tex]\[
(x-4, y+15)
\][/tex]
Given the coordinates of point [tex]\(D'\)[/tex] in the image are [tex]\((9, -8)\)[/tex], we can set up equations to find the original coordinates of point [tex]\(D\)[/tex].
Let's denote the coordinates of point [tex]\(D\)[/tex] in the pre-image as [tex]\((D_x, D_y)\)[/tex].
According to the translation rule:
[tex]\[
(D_x - 4, D_y + 15) = (9, -8)
\][/tex]
We can break this into two separate equations:
1. [tex]\(D_x - 4 = 9\)[/tex]
2. [tex]\(D_y + 15 = -8\)[/tex]
Now, let's solve these equations step by step.
Step 1: Solve for [tex]\(D_x\)[/tex]:
[tex]\[
D_x - 4 = 9
\][/tex]
To isolate [tex]\(D_x\)[/tex], add 4 to both sides of the equation:
[tex]\[
D_x = 9 + 4
\][/tex]
[tex]\[
D_x = 13
\][/tex]
Step 2: Solve for [tex]\(D_y\)[/tex]:
[tex]\[
D_y + 15 = -8
\][/tex]
To isolate [tex]\(D_y\)[/tex], subtract 15 from both sides of the equation:
[tex]\[
D_y = -8 - 15
\][/tex]
[tex]\[
D_y = -23
\][/tex]
Therefore, the coordinates of point [tex]\(D\)[/tex] in the pre-image are:
[tex]\[
(D_x, D_y) = (13, -23)
\][/tex]
Hence, the correct answer is [tex]\((13, -23)\)[/tex].
