If the point P(x, y) is equidistant from the points A(a + b, b - a) and B(a - b, a + b). Prove that bx = ay.

Answer

To prove that \( bx = ay \) given that point \( P(x, y) \) is equidistant from points \( A(a + b, b - a) \) and \( B(a - b, a + b) \), we can follow these steps:

### Step 1: Write the distance formulas
The distance from point \( P(x, y) \) to point \( A(a + b, b - a) \) is given by:
\[ PA = \sqrt{(x - (a + b))^2 + (y - (b - a))^2} \]
The distance from point \( P(x, y) \) to point \( B(a - b, a + b) \) is given by:
\[ PB = \sqrt{(x - (a - b))^2 + (y - (a + b))^2} \]

### Step 2: Set the distances equal
Since \( P \) is equidistant from \( A \) and \( B \), we have:
\[ PA = PB \]

### Step 3: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (x - (a + b))^2 + (y - (b - a))^2 = (x - (a - b))^2 + (y - (a + b))^2 \]

### Step 4: 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,
\[ x^2 - 2x(a + b) + (a + b)^2 + y^2 - 2y(b - a) + (b - a)^2 \]

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

### Step 5: Set the expanded forms equal
Now we have:
\[ x^2 - 2x(a + b) + (a + b)^2 + y^2 - 2y(b - a) + (b - a)^2 = x^2 - 2x(a - b) + (a - b)^2 + y^2 - 2y(a + b) + (a + b)^2 \]

### Step 6: Cancel common terms
Cancel \( x^2 \) and \( y^2 \) from both sides:
\[ -2x(a + b) + (a + b)^2 - 2y(b - a) + (b - a)^2 = -2x(a - b) + (a - b)^2 - 2y(a + b) + (a + b)^2 \]

### Step 7: Rearrange and simplify
Rearranging gives:
\[ -2x(a + b) + 2x(a - b) = 2y(a + b) - 2y(b - a) \]
This simplifies to:
\[ 2x(-b) = 2y(a) \]

### Step 8: Divide by 2
Dividing both sides by 2 gives:
\[ -xb = ay \]

### Step 9: Rearranging gives the final result
Thus, we can rearrange this to show:
\[ bx = ay \]

### Conclusion
We have proved that if point \( P(x, y) \) is equidistant from points \( A \) and \( B \), then \( bx = ay \).

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