The correct option is A Straight line
a¯¯¯z+b¯¯¯z+c=0 (1)
or ¯¯¯a¯¯¯z+¯¯bz+¯¯c=0 (2)
Eliminating ¯¯¯z from (1) and (2), we geet
z=c¯¯¯a−b¯¯c|b|2−|a|2
If |a|≠|b|, then z represents one point on the Argand plane. If |a|=|b| and ¯¯¯ac≠b¯¯c, then no such z exists. Adding (1) and (2),
(¯¯¯a+b)¯¯¯z+(a+¯¯b)z+(c+¯¯c)=0
This is of the form A¯¯¯z+¯¯¯¯Az+B=0, where B=c+¯¯c=0 is real.
Hence, locus of z is a straight line.