Statement-1: If z is a complex number satisfying `(z-1)^(n)=z^n` , `n in N`, then the locus of z is a straight line parallel to imaginary axis.
Statement-2: The locus of a point equidistant from two given points is the perpendicular bisector of the line segment joining them.

Options

A. Statement-1 is True, Statement-2 is True: Statement-2 is a correct explanation for statement-1.
B. Statement-1 is true, statement -2 is true, Statement-2 is not a correct explanation for statement-1.
C. Statement-1 is True, statement-2 is false,
D. statement-1 is False, Statement-2 is true.

Answer

To solve the problem, we need to analyze the condition given for the complex number \( z \) and determine the locus of points that satisfy the equation \( (z - 1)^n = z^n \) for \( n \in \mathbb{N} \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ (z - 1)^n = z^n \]

2. **Rearranging the equation:**
We can rewrite the equation as:
\[ \frac{z - 1}{z} = 1^{\frac{1}{n}} = 1 \]
This implies:
\[ z - 1 = z \]

3. **Divide both sides by \( z \) (assuming \( z \neq 0 \)):**
\[ \frac{z - 1}{z} = 1 \]
This simplifies to:
\[ 1 - \frac{1}{z} = 1 \]

4. **Solving for \( z \):**
Rearranging gives:
\[ -\frac{1}{z} = 0 \implies z = \infty \]
However, this does not help us find the locus. Instead, we can consider the modulus.

5. **Taking modulus on both sides:**
Taking the modulus of both sides of the original equation:
\[ |z - 1|^n = |z|^n \]
Since \( n \) is a natural number, we can take the \( n \)-th root:
\[ |z - 1| = |z| \]

6. **Interpret the modulus condition:**
The equation \( |z - 1| = |z| \) means that the distance from the point \( z \) to the point \( 1 \) (which is \( (1, 0) \) in the complex plane) is equal to the distance from \( z \) to the origin \( 0 \) (which is \( (0, 0) \)).

7. **Finding the locus:**
The set of points \( z \) that are equidistant from the points \( 0 \) and \( 1 \) lies on the perpendicular bisector of the line segment joining \( 0 \) and \( 1 \). The perpendicular bisector of the segment joining \( (0, 0) \) and \( (1, 0) \) is the vertical line:
\[ x = \frac{1}{2} \]
This line is parallel to the imaginary axis.

8. **Conclusion:**
Therefore, the locus of \( z \) is indeed a straight line parallel to the imaginary axis.

### Final Answer:
The statement that the locus of \( z \) is a straight line parallel to the imaginary axis is **true**.