To determine the signs of the coordinates of the points in triangle \( PQR \) and their transformations, we need to understand how the transformations affect the signs of the coordinates.
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Rotating 90 degrees counterclockwise about the origin:
- A point \((x, y)\) becomes \((-y, x)\) after this transformation.
- If \(x > 0\) and \(y > 0\), then after the transformation, \((-y, x)\) becomes negative in the x-coordinate and positive in the y-coordinate.
- If \(x < 0\) and \(y > 0\), then after the transformation, it would keep the y positive but make x negative.
- If \(x < 0\) and \(y < 0\), both coordinates will remain negative after rotation.
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Reflecting across the x-axis:
- Any point \((x, y)\) becomes \((x, -y)\) after this transformation.
- This means that the x-coordinate's sign remains the same, and the y-coordinate's sign changes.
Now, let's analyze this for points \(P\) and \(Q\) in parts A and B:
Part A (Point P)
- Before transformations: Assume the possible coordinates for point \(P\).
- After rotation: The new coordinates transformed will be \((-y, x)\).
- After reflection across the x-axis: The new coordinates will be \((-y, -x)\).
- Depending on the original signs of \(x\) and \(y\), we can deduce the final signs.
Assuming \(P\) initially has positive coordinates, after transformation:
- If original \( x \) and \( y \) are both positive, \(P'' = (-y, -x)\) would result in both coordinates being negative.
- If \( P \) has \(x\) positive and \(y\) negative, we can derive a similar conclusion.
Conclusion for Part A:
- The signs of the coordinates for point \(P\) must be C: Both \(x\) and \(y\) are negative.
Part B (Point Q)
- Using similar logic for point \(Q\):
- After rotation, it transforms to \((-y', x')\) and after reflection it becomes \((-y', -x')\).
Conclusion for Part B:
- Under similar transformations, if both original signs for coordinate \(Q\) are assumed positive, then:
- The final coordinates would reflect both \(x'\) and \(y'\) negative.
Therefore, for part A the answer is C, and for part B the answer is also C. If you have specific coordinates for triangle \(PQR\), we can directly assign the exact signs to the points to validate this reasoning!