To solve this problem, we need to perform two transformations on the vertices of triangle [tex]\(\triangle ABC\)[/tex]:
1. Rotate the triangle [tex]\(180^{\circ}\)[/tex] clockwise about the origin.
2. Reflect the resulting triangle across the line [tex]\(y = -x\)[/tex].
### Step 1: Rotate [tex]\(180^{\circ}\)[/tex] Clockwise
Rotation of [tex]\(180^{\circ}\)[/tex] clockwise about the origin can be achieved by negating both the x and y coordinates of each point.
For the vertices:
- [tex]\(A(-3, 0) \rightarrow A' (3, 0)\)[/tex]
- [tex]\(B(-2, 3) \rightarrow B' (2, -3)\)[/tex]
- [tex]\(C(-1, 1) \rightarrow C' (1, -1)\)[/tex]
### Step 2: Reflect across the Line [tex]\(y = -x\)[/tex]
Reflection of a point [tex]\((x, y)\)[/tex] across the line [tex]\(y = -x\)[/tex] results in the coordinates [tex]\((-y, -x)\)[/tex].
Reflecting the rotated points:
- [tex]\(A'(3, 0) \rightarrow A'' (0, 3)\)[/tex]
- [tex]\(B'(2, -3) \rightarrow B'' (-3, 2)\)[/tex]
- [tex]\(C'(1, -1) \rightarrow C'' (-1, 1)\)[/tex]
The final coordinates of the vertices after both transformations are:
- [tex]\(A'' (0, 3)\)[/tex]
- [tex]\(B'' (-3, 2)\)[/tex]
- [tex]\(C'' (-1, 1)\)[/tex]
Therefore, the correct answer is:
- [tex]\((0, 3)\)[/tex] for [tex]\(A'\)[/tex]
- [tex]\((-3, 2)\)[/tex] for [tex]\(B'\)[/tex]
- [tex]\((-1, 1)\)[/tex] for [tex]\(C'\)[/tex]
Thus, the coordinates of the vertices of the image are:
[tex]\[
\boxed{(0, 3), (-3, 2), (-1, 1)}
\][/tex]
The correct answer is:
A. [tex]\(A^{\prime}(0, 3), B^{\prime}(-3, 2), C^{\prime}(-1, 1)\)[/tex]
