To find the coordinates of the vertices of the pre-image of a triangle after a 90° counterclockwise rotation about the origin, we need to reverse the transformation. A 90° counterclockwise rotation involves the transformation of a point [tex]\((x, y)\)[/tex] to [tex]\((-y, x)\)[/tex]. To find the pre-image, we use the inverse transformation [tex]\((x, y) \rightarrow (y, -x)\)[/tex].
Given the coordinates of the vertices of the image after rotation:
- [tex]\(A' (6, 3)\)[/tex]
- [tex]\(B' (-2, 1)\)[/tex]
- [tex]\(C' (1, 7)\)[/tex]
We will now apply the inverse transformation [tex]\((x, y) \rightarrow (y, -x)\)[/tex] to each of these vertices.
### Step 1: Finding the pre-image of [tex]\(A'\)[/tex]
For [tex]\(A' (6, 3)\)[/tex]:
- [tex]\(x = 6\)[/tex]
- [tex]\(y = 3\)[/tex]
Applying the inverse transformation:
[tex]\[ (6, 3) \rightarrow (3, -6) \][/tex]
So, the coordinates of [tex]\(A\)[/tex] are [tex]\((3, -6)\)[/tex].
### Step 2: Finding the pre-image of [tex]\(B'\)[/tex]
For [tex]\(B' (-2, 1)\)[/tex]:
- [tex]\(x = -2\)[/tex]
- [tex]\(y = 1\)[/tex]
Applying the inverse transformation:
[tex]\[ (-2, 1) \rightarrow (1, 2) \][/tex]
So, the coordinates of [tex]\(B\)[/tex] are [tex]\((1, 2)\)[/tex].
### Step 3: Finding the pre-image of [tex]\(C'\)[/tex]
For [tex]\(C' (1, 7)\)[/tex]:
- [tex]\(x = 1\)[/tex]
- [tex]\(y = 7\)[/tex]
Applying the inverse transformation:
[tex]\[ (1, 7) \rightarrow (7, -1) \][/tex]
So, the coordinates of [tex]\(C\)[/tex] are [tex]\((7, -1)\)[/tex].
### Conclusion
The coordinates of the vertices of the pre-image of the triangle are:
- [tex]\(A (3, -6)\)[/tex]
- [tex]\(B (1, 2)\)[/tex]
- [tex]\(C (7, -1)\)[/tex]
Thus, after reversing the 90° counterclockwise rotation, the original coordinates of the vertices of the pre-image are:
[tex]\[ \boxed{(3, -6), (1, 2), (7, -1)} \][/tex]
