Question
A four-sided figure W(quadrant II), X(quadrant III), Y(quadrant IV), Z(quadrant I) is shown on the coordinate plane. The figure is then reflected across the y-axis. Which point of the figure will end up in quadrant I.
Answers
GPT-4o mini
To determine which point of the figure will end up in quadrant I after reflection across the y-axis, we need to understand how the reflection affects the coordinates of each point.
1. **Reflection across the y-axis** changes the sign of the x-coordinate while keeping the y-coordinate the same. Thus, for any point \((x, y)\), its reflected point will be \((-x, y)\).
2. The points are given as follows:
- W is in quadrant II, which means its coordinates are \((x_W, y_W)\) where \(x_W < 0\) and \(y_W > 0\).
- X is in quadrant III, so its coordinates are \((x_X, y_X)\) where \(x_X < 0\) and \(y_X < 0\).
- Y is in quadrant IV, so its coordinates are \((x_Y, y_Y)\) where \(x_Y > 0\) and \(y_Y < 0\).
- Z is in quadrant I, where its coordinates are \((x_Z, y_Z)\) such that \(x_Z > 0\) and \(y_Z > 0\).
Now, we will analyze each point after reflection:
- **Point W**: After reflection, it becomes \((-x_W, y_W)\). Since \(x_W < 0\), \(-x_W > 0\), placing it in quadrant I (positive x and positive y).
- **Point X**: After reflection, it becomes \((-x_X, y_X)\). Since both \(x_X < 0\) and \(y_X < 0\), this point remains in quadrant II or III.
- **Point Y**: After reflection, it becomes \((-x_Y, y_Y)\). Here, \(x_Y > 0\) implies \(-x_Y < 0\) and since \(y_Y < 0\), it remains in quadrant II or III.
- **Point Z**: After reflection, it becomes \((-x_Z, y_Z)\). While \(x_Z > 0\), thus \(-x_Z < 0\) and \(y_Z > 0\), this point also remains in quadrant II.
Based on the above analysis, only point W (from quadrant II) will end up in quadrant I after the reflection across the y-axis.
**Answer:** Point W (from quadrant II) will end up in quadrant I after reflection.
1. **Reflection across the y-axis** changes the sign of the x-coordinate while keeping the y-coordinate the same. Thus, for any point \((x, y)\), its reflected point will be \((-x, y)\).
2. The points are given as follows:
- W is in quadrant II, which means its coordinates are \((x_W, y_W)\) where \(x_W < 0\) and \(y_W > 0\).
- X is in quadrant III, so its coordinates are \((x_X, y_X)\) where \(x_X < 0\) and \(y_X < 0\).
- Y is in quadrant IV, so its coordinates are \((x_Y, y_Y)\) where \(x_Y > 0\) and \(y_Y < 0\).
- Z is in quadrant I, where its coordinates are \((x_Z, y_Z)\) such that \(x_Z > 0\) and \(y_Z > 0\).
Now, we will analyze each point after reflection:
- **Point W**: After reflection, it becomes \((-x_W, y_W)\). Since \(x_W < 0\), \(-x_W > 0\), placing it in quadrant I (positive x and positive y).
- **Point X**: After reflection, it becomes \((-x_X, y_X)\). Since both \(x_X < 0\) and \(y_X < 0\), this point remains in quadrant II or III.
- **Point Y**: After reflection, it becomes \((-x_Y, y_Y)\). Here, \(x_Y > 0\) implies \(-x_Y < 0\) and since \(y_Y < 0\), it remains in quadrant II or III.
- **Point Z**: After reflection, it becomes \((-x_Z, y_Z)\). While \(x_Z > 0\), thus \(-x_Z < 0\) and \(y_Z > 0\), this point also remains in quadrant II.
Based on the above analysis, only point W (from quadrant II) will end up in quadrant I after the reflection across the y-axis.
**Answer:** Point W (from quadrant II) will end up in quadrant I after reflection.