Question
A coordinate plane's axes range from negative 6 to 6, both by 1-unit increments. Points upper L, upper O, and upper W are plotted and connected via lines that form a triangle.
Which quadrant will ΔLOW
be in when it is reflected across the y-axis and then reflected across the x-axis?
(1 point)
Responses
Quadrant IV
Quadrant IV
Quadrant I
Quadrant I
Quadrant II
Quadrant II
Quadrant III
Which quadrant will ΔLOW
be in when it is reflected across the y-axis and then reflected across the x-axis?
(1 point)
Responses
Quadrant IV
Quadrant IV
Quadrant I
Quadrant I
Quadrant II
Quadrant II
Quadrant III
Answers
GPT-4o mini
To determine which quadrant triangle \( \Delta LOW \) will be in after the given reflections, we can follow these steps:
1. **Reflect across the y-axis**: Reflecting a point \((x, y)\) across the y-axis changes the sign of its x-coordinate, resulting in the point \((-x, y)\).
2. **Reflect across the x-axis**: Reflecting a point \((-x, y)\) across the x-axis changes the sign of its y-coordinate, resulting in the point \((-x, -y)\).
So, reflecting the triangle \( \Delta LOW \) first across the y-axis and then across the x-axis would transform the vertices as follows:
- \((x, y)\) becomes \((-x, y)\),
- then \((-x, y)\) becomes \((-x, -y)\).
If the original points \(L\), \(O\), and \(W\) are in any of the four quadrants, after both reflections:
- Quadrant I points \((+,+)\) will move to Quadrant III \((-,-)\).
- Quadrant II points \((-,+)\) will move to Quadrant IV \((-, -)\).
- Quadrant III points \((-,-)\) will move to Quadrant I \((+,+)\).
- Quadrant IV points \((+,-)\) will move to Quadrant II \((-,+)\).
Because the triangle's vertices will effectively shift to their corresponding positions in opposite quadrants, if you start from Quadrant I (positive x and positive y), you will end up in Quadrant III (negative x and negative y) after both reflections.
So, **the answer depends on the original position of the triangle in the coordinate plane.**
Assuming triangle \( \Delta LOW \) lies in **any quadrant**, after reflecting it first across the y-axis and then across the x-axis, it will always end up in Quadrant III \((-,-)\).
Thus, if the options are only Quadrants I, II, III, and IV, the final transformed triangle \(\Delta LOW\) will be in:
**Quadrant III.**
1. **Reflect across the y-axis**: Reflecting a point \((x, y)\) across the y-axis changes the sign of its x-coordinate, resulting in the point \((-x, y)\).
2. **Reflect across the x-axis**: Reflecting a point \((-x, y)\) across the x-axis changes the sign of its y-coordinate, resulting in the point \((-x, -y)\).
So, reflecting the triangle \( \Delta LOW \) first across the y-axis and then across the x-axis would transform the vertices as follows:
- \((x, y)\) becomes \((-x, y)\),
- then \((-x, y)\) becomes \((-x, -y)\).
If the original points \(L\), \(O\), and \(W\) are in any of the four quadrants, after both reflections:
- Quadrant I points \((+,+)\) will move to Quadrant III \((-,-)\).
- Quadrant II points \((-,+)\) will move to Quadrant IV \((-, -)\).
- Quadrant III points \((-,-)\) will move to Quadrant I \((+,+)\).
- Quadrant IV points \((+,-)\) will move to Quadrant II \((-,+)\).
Because the triangle's vertices will effectively shift to their corresponding positions in opposite quadrants, if you start from Quadrant I (positive x and positive y), you will end up in Quadrant III (negative x and negative y) after both reflections.
So, **the answer depends on the original position of the triangle in the coordinate plane.**
Assuming triangle \( \Delta LOW \) lies in **any quadrant**, after reflecting it first across the y-axis and then across the x-axis, it will always end up in Quadrant III \((-,-)\).
Thus, if the options are only Quadrants I, II, III, and IV, the final transformed triangle \(\Delta LOW\) will be in:
**Quadrant III.**