If any point P is at the equal distances from points A(a+b,a-b) and B(a-b,a+b), then locus of a point is

Options

A. x-y=0
B. ax+by=0
C. bx+ay=0
D. x+y=0

Explanation

To find the locus of a point \( P(x, y) \) that is equidistant from the points \( A(a+b, a-b) \) and \( B(a-b, a+b) \), we can follow these steps:

### Step 1: Set Up the Distance Equations
The point \( P \) is equidistant from points \( A \) and \( B \). Therefore, we can set the distances equal to each other:
\[ PA = PB \]

### Step 2: Apply the Distance Formula
Using the distance formula, we can express \( PA \) and \( PB \):
\[ PA = \sqrt{(x - (a+b))^2 + (y - (a-b))^2} \]
\[ PB = \sqrt{(x - (a-b))^2 + (y - (a+b))^2} \]

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

### Step 4: Expand Both Sides
Expanding both sides gives:
\[ (x^2 - 2x(a+b) + (a+b)^2) + (y^2 - 2y(a-b) + (a-b)^2) = (x^2 - 2x(a-b) + (a-b)^2) + (y^2 - 2y(a+b) + (a+b)^2) \]

### Step 5: Simplify the Equation
Now, we can simplify the equation by canceling out \( x^2 \) and \( y^2 \) from both sides:
\[ -2x(a+b) - 2y(a-b) + (a+b)^2 + (a-b)^2 = -2x(a-b) - 2y(a+b) + (a-b)^2 + (a+b)^2 \]

### Step 6: Rearranging Terms
Rearranging the terms leads to:
\[ -2x(a+b) + 2x(a-b) - 2y(a-b) + 2y(a+b) = 0 \]

### Step 7: Factor Out Common Terms
Factoring out common terms gives:
\[ 2x(-b + b) + 2y(a + b - a + b) = 0 \]
This simplifies to:
\[ 2x(b - b) + 2y(2b) = 0 \]

### Step 8: Final Equation
This leads to:
\[ y = x \]

### Conclusion
Thus, the locus of the point \( P \) is given by the equation:
\[ x - y = 0 \]