Question

which coordinate for points a and b would help prove that lines ab and ab 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: Recall slope formula

The slope of a line passing through two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(m = \frac{y_2 - y_1}{x_2 - x_1}\). For two lines to be perpendicular, the product of their slopes should be \(- 1\) (or one slope is the negative reciprocal of the other).

Step2: Find slope of AB

Points \(A(-m,p)\) and \(B(w,z)\). Slope of \(AB\), \(m_{AB}=\frac{z - p}{w-(-m)}=\frac{z - p}{w + m}\).

Step3: Analyze each option for slope of A'B'

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

eq - 1\) (in general).

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

eq - 1\) (in general).

Answer:

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