If the coordinates of the points `A, B, C` and D be `(a, b), (a', b'), (-a, b)` and `(a',-b')` respectively, then the equation of the line bisecting the line segments AB and CD is

Answer

2ay-2b'x = ab-a'b'

Explanation

Mid point of AB ` = E((a+a')/(2), (b+b')/(2)) " and Midpoint of CD"`
`= F((a'-a)/(2), (b-b')/(2))`
Hence, the equation of line EF is
`y-(b+b')/(2) = (b-b'-b-b')/(a'-a-a-a') (x-(a+a')/(2))`
On simplification, we get
2ay-2b'x = ab-a'b'