Question
Line A B has a negative slope and goes through points (negative m, p) and (w, z). Line A prime B prime has a positive slope and intersects with line A B.
Which coordinate for points A' and B' would help prove that lines AB and A'B' are perpendicular?
A': (p, m) and B': (z, w)
A': (p, m) and B': (z, −w)
A': (p, −m) and B': (z, w)
A': (p, −m) and B': (z, −w)
Which coordinate for points A' and B' would help prove that lines AB and A'B' are perpendicular?
A': (p, m) and B': (z, w)
A': (p, m) and B': (z, −w)
A': (p, −m) and B': (z, w)
A': (p, −m) and B': (z, −w)
Answers
GPT-4o mini
To determine which coordinate pair for points A' and B' would help prove that lines AB and A'B' are perpendicular, we should consider the slopes of the lines.
- Line AB has a negative slope, which we can denote as \( m_{AB} < 0 \).
- Line A'B' has a positive slope, \( m_{A'B'} > 0 \).
For two lines to be perpendicular, the product of their slopes must equal -1:
\[
m_{AB} \times m_{A'B'} = -1
\]
Given that the slopes you're dealing with need to be negative and positive respectively, it follows that the slopes must be such that they are negative reciprocals of each other.
We want to analyze the slopes based on the suggested points:
1. **A': (p, m) and B': (z, w)**
- The slope \( m_{A'B'} = \frac{w - m}{z - p} \)
2. **A': (p, m) and B': (z, -w)**
- The slope \( m_{A'B'} = \frac{-w - m}{z - p} = \frac{-w - m}{z - p} \) (this slope is dependent on the sign change of w)
3. **A': (p, -m) and B': (z, w)**
- The slope \( m_{A'B'} = \frac{w + m}{z - p} \)
4. **A': (p, -m) and B': (z, -w)**
- The slope \( m_{A'B'} = \frac{-w + m}{z - p} \)
To conclude, we must ensure that the slopes \( m_{AB} \) and \( m_{A'B'} \) yield a product of -1.
Given that \( AB \) has a negative slope, and \( A'B' \) must yield a positive slope that is the negative reciprocal.
To find a suitable pair, consider the signs in the pairs:
- When y-coordinates of one line are the negative counterparts of the other, it implies the lines' slopes will lead to a product of -1.
Out of the options provided, **A': (p, −m) and B': (z, w)** will yield the scenario from the coordinate transformations (i.e., switching the m component negates the original slope/magnitude, endorsing perpendicularity).
Thus, the correct answer is:
**A': (p, −m) and B': (z, w)**.
- Line AB has a negative slope, which we can denote as \( m_{AB} < 0 \).
- Line A'B' has a positive slope, \( m_{A'B'} > 0 \).
For two lines to be perpendicular, the product of their slopes must equal -1:
\[
m_{AB} \times m_{A'B'} = -1
\]
Given that the slopes you're dealing with need to be negative and positive respectively, it follows that the slopes must be such that they are negative reciprocals of each other.
We want to analyze the slopes based on the suggested points:
1. **A': (p, m) and B': (z, w)**
- The slope \( m_{A'B'} = \frac{w - m}{z - p} \)
2. **A': (p, m) and B': (z, -w)**
- The slope \( m_{A'B'} = \frac{-w - m}{z - p} = \frac{-w - m}{z - p} \) (this slope is dependent on the sign change of w)
3. **A': (p, -m) and B': (z, w)**
- The slope \( m_{A'B'} = \frac{w + m}{z - p} \)
4. **A': (p, -m) and B': (z, -w)**
- The slope \( m_{A'B'} = \frac{-w + m}{z - p} \)
To conclude, we must ensure that the slopes \( m_{AB} \) and \( m_{A'B'} \) yield a product of -1.
Given that \( AB \) has a negative slope, and \( A'B' \) must yield a positive slope that is the negative reciprocal.
To find a suitable pair, consider the signs in the pairs:
- When y-coordinates of one line are the negative counterparts of the other, it implies the lines' slopes will lead to a product of -1.
Out of the options provided, **A': (p, −m) and B': (z, w)** will yield the scenario from the coordinate transformations (i.e., switching the m component negates the original slope/magnitude, endorsing perpendicularity).
Thus, the correct answer is:
**A': (p, −m) and B': (z, w)**.