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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'