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

Options

A. `ax=by`
B. `bx=ay`
C. `xy=ab`
D. none

Answer

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

### Step-by-step Solution:

1. **Set Up the Distance Equations**:
Since point \( P \) is equidistant from points \( A \) and \( B \), we can express this mathematically as:
\[ PA = PB \]
This means:
\[ \sqrt{(x - (a+b))^2 + (y - (b-a))^2} = \sqrt{(x - (a-b))^2 + (y - (a+b))^2} \]

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

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

Now, 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:
\[ x^2 - 2x(a-b) + (a-b)^2 + y^2 - 2y(a+b) + (a+b)^2 \]

4. **Combine and Simplify**:
Now we can set both sides equal to each other and simplify:
\[ x^2 - 2x(a+b) + y^2 - 2y(b-a) + (a+b)^2 + (b-a)^2 = x^2 - 2x(a-b) + y^2 - 2y(a+b) + (a-b)^2 + (a+b)^2 \]

Cancel \( x^2 \) and \( y^2 \) from both sides:
\[ -2x(a+b) - 2y(b-a) + (a+b)^2 + (b-a)^2 = -2x(a-b) - 2y(a+b) + (a-b)^2 + (a+b)^2 \]

5. **Rearranging Terms**:
Rearranging gives us:
\[ -2x(a+b) + 2x(a-b) + 2y(a+b) - 2y(b-a) = (a-b)^2 - (b-a)^2 \]

6. **Factor and Solve**:
This simplifies to:
\[ 2x(b-a) + 2y(a+b) = 0 \]
Dividing through by 2, we get:
\[ x(b-a) + y(a+b) = 0 \]

### Final Condition:
Thus, the condition for point \( P(x, y) \) to be equidistant from points \( A \) and \( B \) is:
\[ ay = -bx \]