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').
mid point of AB is
$\Big(\frac{\text{a}+\text{a}'}{2},\frac{\text{b}+\text{b}'}{2}\Big)$mid point of CD is
$\Big(\frac{-\text{a}+\text{a}'}{2},\frac{\text{b}-\text{b}'}{2}\Big)$Equation is
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$$\text{y}-\text{y}_1=\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1}(\text{x}-\text{x}_1)$
$\text{y}-\Big(\frac{\text{b}+\text{b}'}{2}\Big)=\frac{\Big(\frac{\text{b}-\text{b}'}{2}\Big)-\Big(\frac{\text{b}+\text{b}'}{2}\Big)}{\Big(\frac{-\text{a}+\text{a}'}{2}-\frac{\text{a}+\text{a}'}{2}\Big)}\Big(\text{x}-\Big(\frac{\text{a}+\text{a}'}{2}\Big)\Big)$
$\text{y}-\Big(\frac{\text{b}-\text{b}'}{2}\Big)=\frac{\frac{\text{b}}{2}-\frac{\text{b}'}{2}-\frac{\text{b}}{2}-\frac{\text{b}'}{2}}{-\frac{\text{a}}{2}+\frac{\text{a}'}{2}-\frac{\text{a}}{2}-\frac{\text{a}'}{2}}\Big(\text{x}-\Big(\frac{\text{a}+\text{a}'}{2}\Big)\Big)$
$\text{y}-\Big(\frac{\text{b}+\text{b}'}{2}\Big)=\frac{+\text{b}'}{\text{a}}\Big(\text{x}-\Big(\frac{\text{a}+\text{a}'}{2}\Big)\Big) $
$2\text{ay}-2\text{b}'\text{x}=\text{ab}-\text{a}'\text{b}'$
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