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

$\text{The line that bisects the distance between the points A (a, b), B (a', b') and between C (-a, b) and D (a', -b') means a line passing through the mid-point of AB and CD}\text{mid point of AB is}(\frac{a+a^{\prime }}{2},\frac{b+b^{\prime }}{2})\text{mid point of CD is}(\frac{-a+a^{\prime }}{2},\frac{b-b^{\prime }}{2})\text{Equation is}y-y_1=m(x-x_1)y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)y-\left(\frac{b+b^{\prime }}{2}\right)=\frac{\left(\frac{b-b^{\prime }}{2}\right)-\left(\frac{b+b^{\prime }}{2}\right)}{(\frac{-a+a^{\prime }}{2}-\frac{a+a^{\prime }}{2})}(x-\left(\frac{a+a^{\prime }}{2}\right))y-\left(\frac{b+b^{\prime }}{2}\right)=\frac{\frac{b}{2}-\frac{b^{\prime }}{2}-\frac{b}{2}-\frac{b^{\prime }}{2}}{\frac{-a}{2}+\frac{a^{\prime }}{2}-\frac{a}{2}-\frac{a^{\prime }}{2}}(x-\left(\frac{a+a^{\prime }}{2}\right))y-\left(\frac{b+b^{\prime }}{2}\right)=\frac{+b^{\prime }}{a}(x-\left(\frac{a+a^{\prime }}{2}\right))2ay-2b^{\prime }x=ab-a^{\prime }{b}^{\prime }$