Question

proving perpendicular lines
which coordinate for points ( a ) and ( b ) would help prove that lines ( ab ) and ( ab ) are perpendicular?
( \bigcirc a:(p, m) ) and ( b:(z, w) )
( \bigcirc a:(p, m) ) and ( b:(z,-w) )
( \bigcirc a:(p,-m) ) and ( b:(z, w) )
( \bigcirc 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 lines to be perpendicular, the product of their slopes \(m_1\times m_2=- 1\).
Let \(A(-m,p)\) and \(B(w,z)\), the slope of line \(AB\) is \(m_{AB}=\frac{z - p}{w + m}\).

Step2: Calculate slopes for each option

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