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 of a line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$.

Step2: Find the slope of line \(AB\)

For points \(A(-m,p)\) and \(B(w,z)\), the slope of line \(AB\), \(m_{AB}=\frac{z - p}{w + m}\).

Step3: Find the slope of line \(A'B'\) for each option

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

eq - 1\) (unless specific values of \(m,z,w,p\) which is not general).

• Option 2: For \(A'(p,m)\) and \(B'(z,-w)\), \(m_{A'B'}=\frac{-w - m}{z - p}\). Then \(m_{AB}\times m_{A'B'}=\frac{(z - p)(-w - m)}{(w + m)(z - p)}=- 1\) only if \(z - p

eq0\) and \(w + m
eq0\). But we can check the general perpendicular condition.

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

eq-1\) (unless \(z - p = 0\) which is a degenerate case).

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

eq-1\) (in general).

• The correct condition for two lines with slopes \(m_1\) and \(m_2\) to be perpendicular is \(m_1\times m_2=-1\). If we assume the transformation from \(A(-m,p)\) to \(A'(p,-m)\) (a rotation of \(90^{\circ}\) about the origin, which changes \((x,y)\) to \((y,-x)\)) and from \(B(w,z)\) to \(B'(z,-w)\) (same rotation rule \((x,y)\to(y,-x)\)). The slope of \(AB\) is \(m_{AB}=\frac{z - p}{w + m}\) and the slope of \(A'B'\) is \(m_{A'B'}=\frac{-w + m}{z - p}\), and \(m_{AB}\times m_{A'B'}=\frac{(z - p)(-w + m)}{(w + m)(z - p)}=-1\) (assuming \(z

eq p\) and \(w
eq - m\), non - vertical and non - horizontal lines).

So the coordinate for points \(A'\) and \(B'\) that would help prove that lines \(AB\) and \(A'B'\) are perpendicular is \(A'(p, -m)\) and \(B'(z, -w)\)

Answer:

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