Let `3-i` and `2+i` be affixes of two points A and B in the Argand plane and P represents the complex number `z=x+iy`. Then, the locus of the P if `|z-3+i|=|z-2-i|`, is

Options

A. circle on AB as diameter
B. the line AB
C. the perpendicular bisector of AB
D. none of these

Answer

To solve the problem, we need to find the locus of the point \( P \) represented by the complex number \( z = x + iy \) such that the distance from \( P \) to point \( A \) (with affix \( 3 - i \)) is equal to the distance from \( P \) to point \( B \) (with affix \( 2 + i \)).

### Step-by-step Solution:

1. **Set up the equation for distances**:
We start with the condition given in the problem:
\[ |z - (3 - i)| = |z - (2 + i)| \]
This translates to:
\[ |z - 3 + i| = |z - 2 - i| \]

2. **Substitute \( z \)**:
Substitute \( z = x + iy \) into the equation:
\[ |(x + iy) - (3 - i)| = |(x + iy) - (2 + i)| \]
This simplifies to:
\[ |(x - 3) + (y + 1)i| = |(x - 2) + (y - 1)i| \]

3. **Calculate the moduli**:
The modulus of a complex number \( a + bi \) is given by \( \sqrt{a^2 + b^2} \). Therefore, we have:
\[ \sqrt{(x - 3)^2 + (y + 1)^2} = \sqrt{(x - 2)^2 + (y - 1)^2} \]

4. **Square both sides**:
To eliminate the square roots, we square both sides:
\[ (x - 3)^2 + (y + 1)^2 = (x - 2)^2 + (y - 1)^2 \]

5. **Expand both sides**:
Expanding both sides, we get:
\[ (x^2 - 6x + 9) + (y^2 + 2y + 1) = (x^2 - 4x + 4) + (y^2 - 2y + 1) \]
This simplifies to:
\[ x^2 - 6x + 9 + y^2 + 2y + 1 = x^2 - 4x + 4 + y^2 - 2y + 1 \]

6. **Combine like terms**:
Cancel \( x^2 \) and \( y^2 \) from both sides:
\[ -6x + 9 + 2y + 1 = -4x + 4 - 2y + 1 \]
This simplifies to:
\[ -6x + 2y + 10 = -4x + 4 - 2y + 1 \]

7. **Rearranging the equation**:
Rearranging gives:
\[ -6x + 4x + 2y + 2y + 10 - 5 = 0 \]
Which simplifies to:
\[ -2x + 4y + 5 = 0 \]

8. **Final equation of the locus**:
Rearranging gives the equation of the line:
\[ 2x - 4y = 5 \]
or
\[ x - 2y = \frac{5}{2} \]

### Conclusion:
The locus of the point \( P \) is a straight line given by the equation:
\[ x - 2y = \frac{5}{2} \]