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Question 78 of 122

Q.If ∣z−z1∣=∣z−z2∣|z-z_1|=|z-z_2| then the locus of zz is

(a) a circle with centre at the origin
(b) a circle with centre at z1z_1
(c) a straight line passing through the origin
(d) is a perpendicular bisector of the line joining z1z_1 and z2z_2
Puducherry TnboardTamil Nadu HSC (DGE) Board 2016MCQ· 1mImportance★★★★★
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Equidistance from two fixed points is the geometric definition of the perpendicular bisector, so that is the locus.

  1. Let z=x+iyz=x+iy, z1,z2z_1,z_2 be fixed complex numbers (points in the plane).
  2. ∣z−z1∣|z-z_1| is the distance from the variable point zz to the fixed point z1z_1; likewise ∣z−z2∣|z-z_2| is the distance to z2z_2.
  3. The condition ∣z−z1∣=∣z−z2∣|z-z_1|=|z-z_2| says the point zz is always equidistant from z1z_1 and z2z_2.
  4. In plane geometry, the set of all points equidistant from two fixed points is, by definition, the perpendicular bisector of the segment joining those two points. …

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