Question

If z=x+iy and |z-2+i|=|z-3-i| then locus of z is _________

• A. x-2y+5=0
• B. 2x-4y-5=0
• C. x+2y=0
• D. 2x+4y-5=0

Answer 1

The Correct Option is D

Given that \(z = x + iy\), we have the equation for the locus as \(|z-2+i| = |z-3-i|\). This can be expanded as:
\(|(x+iy)-2+i| = |(x+iy)-3-i|\)
Thus,
\(|(x-2) + i(y+1)| = |(x-3) + i(y-1)|\)
This equation compares the distances between the point \(z\) and the complex numbers \(2-i\) and \(3+i\). It signifies that the locus is the perpendicular bisector of the line segment joining the points \((2,-1)\) and \((3,1)\).
First, calculate the midpoint of the line segment joining \((2,-1)\) and \((3,1)\):
\(((2+3)/2, (-1+1)/2) = (2.5, 0)\)
Next, find the slope of the line connecting these two points:
Slope = \((1 - (-1)) / (3 - 2) = 2\)
The slope of the perpendicular bisector is the negative reciprocal of this, which is \(-1/2\).
Thus, the equation of the line (perpendicular bisector) passing through the midpoint \((2.5, 0)\) is:
\(y - 0 = -1/2(x - 2.5)\)
Simplifying, we have:
\(y = -1/2x + 1.25\)
Rewrite in standard form:
\(2x + 4y - 5 = 0\)
Therefore, the locus of \(z\) is \(2x + 4y - 5 = 0\).