Question

Find the equation to the straight line which bisects the distance between the points (a, b), (a', b') and also bisects the distance between the points (−a, b) and (a', −b').

Answer 1

Let the given points be A (a, b), B (a', b'), C (−a, b) and D (a', −b').
Let P and Q be the midpoints of AB and CD, respectively.

$$\begin{gathered} \therefore P\equiv \left(\frac{a+a^{\text{'}}}{2},\frac{b+b^{\text{'}}}{2}\right) \\ Q\equiv \left(\frac{a^{\text{'}}-a}{2},\frac{b-b^{\text{'}}}{2}\right) \end{gathered}$$

The equation of the line passing through P and Q is

$$\begin{gathered} y-\frac{b+b^{\text{'}}}{2}=\frac{{\displaystyle \frac{b-b\text{'}}{2}}-{\displaystyle \frac{b+b\text{'}}{2}}}{{\displaystyle \frac{a\text{'}-a}{2}}-{\displaystyle \frac{a\text{'}+a}{2}}}\left(x-\frac{a+a^{\text{'}}}{2}\right) \\ \Rightarrow 2y-b-b^{\text{'}}=\frac{b^{\text{'}}}{a}\left(2x-a-a^{\text{'}}\right) \\ \Rightarrow 2ay-2b^{\text{'}}x=ab-a^{\text{'}}{b}^{\text{'}} \end{gathered}$$

Hence, the equation of the required straight line is $2ay-2b^{\text{'}}x=ab-a^{\text{'}}{b}^{\text{'}}$