Let A(3-i), B(2+i) be two points in the argand plane. If the point P represents the complex number z=x+i y , which satisfies |z-3+i|=|z-2-i| , then the locus of the point P is

Step-by-step solution

We have a complex number z=x+i y aligned & and |z-3+i|=|z-2-i| & |x+i y-3+i|=|x+i y-2-i| & (x-3)^2+(y+1)^2=(x-2)^2+(y-1)^2 & x^3-6 x+9+y^2+2 y+1 & =x^2-4 x+4+y^2-2 y+1 aligned So, it represent a line Point A(3,-1) and B(2,1) So, mid-point of A B= ( 5 2 , 0 ) m₁= slope of A B= 1-(-1) 2-3 =-2 Point ( 5 2 , 0 ) satisfies the equation -2 x+4 y+5=0 and slope of line =m₂= 1 2 Now, m₁ m₂=-2 1 2 =-1 So, line -2 x+4 y+5 is perpendicular to A B . Hence, locus of point p is the perpendicular bisector of A B .