If the point P(x, y) be equidistant from the points `A(a+b, a-b)` and `B(a-b, a+b)` then

Options

A. ax = by
B. bx = ay
C. `x^(2)-y^(2)=2(ax+by)`
D. P can be (a, b)

Answer

To solve the problem, we need to find the condition under which the point \( P(x, y) \) is equidistant from the points \( A(a+b, a-b) \) and \( B(a-b, a+b) \).

### Step-by-Step Solution:

1. **Set up the distance equations**:
The distance from point \( P(x, y) \) to point \( A(a+b, a-b) \) is given by:
\[ d(P, A) = \sqrt{(x - (a+b))^2 + (y - (a-b))^2} \]
The distance from point \( P(x, y) \) to point \( B(a-b, a+b) \) is given by:
\[ d(P, B) = \sqrt{(x - (a-b))^2 + (y - (a+b))^2} \]

2. **Set the distances equal**:
Since \( P \) is equidistant from \( A \) and \( B \), we have:
\[ d(P, A) = d(P, B) \]
This leads to the equation:
\[ \sqrt{(x - (a+b))^2 + (y - (a-b))^2} = \sqrt{(x - (a-b))^2 + (y - (a+b))^2} \]

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

4. **Expand both sides**:
Expanding the left side:
\[ (x - (a+b))^2 = x^2 - 2x(a+b) + (a+b)^2 \]
\[ (y - (a-b))^2 = y^2 - 2y(a-b) + (a-b)^2 \]
So the left side becomes:
\[ x^2 - 2x(a+b) + (a+b)^2 + y^2 - 2y(a-b) + (a-b)^2 \]

Expanding the right side:
\[ (x - (a-b))^2 = x^2 - 2x(a-b) + (a-b)^2 \]
\[ (y - (a+b))^2 = y^2 - 2y(a+b) + (a+b)^2 \]
So the right side becomes:
\[ x^2 - 2x(a-b) + (a-b)^2 + y^2 - 2y(a+b) + (a+b)^2 \]

5. **Combine like terms**:
After expanding both sides, we can cancel \( x^2 \) and \( y^2 \) from both sides:
\[ -2x(a+b) + (a+b)^2 - 2y(a-b) + (a-b)^2 = -2x(a-b) + (a-b)^2 - 2y(a+b) + (a+b)^2 \]

6. **Rearranging the equation**:
After simplification, we will have:
\[ -2x(a+b) + 2x(a-b) - 2y(a-b) + 2y(a+b) = 0 \]
This simplifies to:
\[ 2x(b) + 2y(b) = 0 \]
Dividing by 2 and factoring out \( b \):
\[ b(x + y) = 0 \]

7. **Conclusion**:
Since \( b \) cannot be zero (as it is a parameter), we conclude:
\[ x + y = 0 \quad \Rightarrow \quad x = -y \]

### Final Result:
The point \( P(x, y) \) is equidistant from points \( A \) and \( B \) if:
\[ x = -y \]