Question

which coordinate for points a and b would help prove that lines ab and ab 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)

Explanation:

Step1: Recall the slope formula

The slope formula is \(m = \frac{y_2 - y_1}{x_2 - x_1}\). For two perpendicular lines with slopes \(m_1\) and \(m_2\), \(m_1\times m_2=- 1\).
The slope of line \(AB\) with \(A(-m,p)\) and \(B(w,z)\) is \(m_{AB}=\frac{z - p}{w + m}\).

Step2: Calculate slopes for each option

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

eq - 1\) (in general).

• Option 2: If \(A'(p,m)\) and \(B'(z,-w)\), then \(m_{A'B'}=\frac{-w - m}{z - p}\). \(m_{AB}\times m_{A'B'}=\frac{(z - p)(-w - m)}{(w + m)(z - p)}=-1\).
• Option 3: If \(A'(p,-m)\) and \(B'(z,w)\), then \(m_{A'B'}=\frac{w + m}{z - p}\). \(m_{AB}\times m_{A'B'}=\frac{(z - p)(w + m)}{(w + m)(z - p)} = 1

eq-1\).

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

eq - 1\) (in general).

Answer:

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