Dear Pranay,

Solution:- To prove-  (z1 - z2)/(z¯1 - z¯2)  +  (z1 - z3)/(z¯1 - z¯3) = 0

or, (z1 - z2)/(z1 - z3)  +  (z¯1 - z¯2)/(z¯1 - z¯3) = 0.......(1)

From Rotation Th. of complex no.'s (Refer ?http://www.askiitians.com/iit-jee-algebra/complex-numbers/rotation.aspx), (z1 - z2)/(z1 - z3) = [|z1 - z2|/|z1 - z3|]exp(i∏/2) = i[|z1 - z2|/|z1 - z3|]

Also, (z¯1 - z¯2) = conjugate of (z1 - z2) and |z¯1 - z¯2| =  |z1 - z2|

So, (z¯1 - z¯2)/(z¯1 - z¯3) = [|z1 - z2|/|z1 - z3|]exp(-i∏/2) = -i[|z1 - z2|/|z1 - z3|]

Here, negative sign comes because of the conjugate (Refer ?http://www.askiitians.com/iit-jee-algebra/complex-numbers/conjugate-of-a-complex-number.aspx)

Putting ?these values in eq. (1), we get-

i[|z1 - z2|/|z1 - z3|] + (-i[|z1 - z2|/|z1 - z3|]) = 0

0 = 0

Hence Proved


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