Let A(3-i) and B(2+i) be two points in the Ar | vedclass

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(C) Given the complex number $z=x+iy$ and the condition $|z-3+i|=|z-2-i|$.This can be written as $|z-(3-i)|=|z-(2+i)|$.Let $A$ be the point representi

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the perpendicular bisector of $AB$

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(C) Given the complex number $z=x+iy$ and the condition $|z-3+i|=|z-2-i|$.This can be written as $|z-(3-i)|=|z-(2+i)|$.Let $A$ be the point representing $3-i$,i.e.,$A(3, -1)$,and $B$ be the point representing $2+i$,i.e.,$B(2, 1)$.The equation $|z-z_A|=|z-z_B|$ represents the set of all points $P(z)$ that are equidistant from points $A$ and $B$.By definition,the locus of points equidistant from two fixed points is the perpendicular bisector of the line segment joining them.Algebraically:$|x+iy-3+i|=|x+iy-2-i|$$| (x-3) + i(y+1) | = | (x-2) + i(y-1) |$$(x-3)^2 + (y+1)^2 = (x-2)^2 + (y-1)^2$$x^2 - 6x + 9 + y^2 + 2y + 1 = x^2 - 4x + 4 + y^2 - 2y + 1$$-6x + 2y + 10 = -4x - 2y + 5$$-2x + 4y + 5 = 0$This is the equation of a line,which is the perpendicular bisector of $AB$.

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

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