Question

which coordinate for points a’ and b’ would help prove that lines ab and a’b’ are perpendicular? a(-m, p) b(w, z) options: 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)

Explanation:

Step1: Find slope of AB

Slope of \( AB \): \( m_{AB} = \frac{z - p}{w - (-m)} = \frac{z - p}{w + m} \)

Step2: Analyze slope of A'B' for perpendicularity

For lines to be perpendicular, product of slopes \( m_{AB} \times m_{A'B'} = -1 \). Let's check each option.

• Option 1: \( A'(p, m) \), \( B'(z, w) \). Slope \( m_{A'B'} = \frac{w - m}{z - p} \). Product: \( \frac{z - p}{w + m} \times \frac{w - m}{z - p} = \frac{w - m}{w + m}

eq -1 \) (generally).

• Option 2: \( A'(p, m) \), \( B'(z, -w) \). Slope \( m_{A'B'} = \frac{-w - m}{z - p} \). Product: \( \frac{z - p}{w + m} \times \frac{-w - m}{z - p} = \frac{-(w + m)}{w + m} = -1 \). This satisfies perpendicularity.
• Option 3: \( A'(p, -m) \), \( B'(z, w) \). Slope \( m_{A'B'} = \frac{w - (-m)}{z - p} = \frac{w + m}{z - p} \). Product: \( \frac{z - p}{w + m} \times \frac{w + m}{z - p} = 1

eq -1 \).

• Option 4: \( A'(p, -m) \), \( B'(z, -w) \). Slope \( m_{A'B'} = \frac{-w - (-m)}{z - p} = \frac{m - w}{z - p} \). Product: \( \frac{z - p}{w + m} \times \frac{m - w}{z - p} = \frac{m - w}{w + m}

eq -1 \) (generally).

Answer:

A': \((p, m)\) and B': \((z, -w)\)