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

Answer

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

### Step 1: Define the Coordinates
Let the coordinates of point \( P \) be \( (x, y) \). The coordinates of points \( A \) and \( B \) are given as:
- \( A = (a+b, a-b) \)
- \( B = (a-b, a+b) \)

### Step 2: Set Up the Distance Equation
Since point \( P \) is equidistant from points \( A \) and \( B \), we can write the equation:
\[ PA = PB \]
Using the distance formula, we have:
\[ \sqrt{(x - (a+b))^2 + (y - (a-b))^2} = \sqrt{(x - (a-b))^2 + (y - (a+b))^2} \]

### 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
Now, we will expand both sides of the equation:
1. Left side:
\[ (x - (a+b))^2 + (y - (a-b))^2 = (x^2 - 2x(a+b) + (a+b)^2) + (y^2 - 2y(a-b) + (a-b)^2) \]
2. Right side:
\[ (x - (a-b))^2 + (y - (a+b))^2 = (x^2 - 2x(a-b) + (a-b)^2) + (y^2 - 2y(a+b) + (a+b)^2) \]

### Step 5: Simplify the Equation
After expanding, we can simplify both sides:
- Left side becomes:
\[ x^2 - 2x(a+b) + (a^2 + 2ab + b^2) + y^2 - 2y(a-b) + (a^2 - 2ab + b^2) \]
- Right side becomes:
\[ x^2 - 2x(a-b) + (a^2 - 2ab + b^2) + y^2 - 2y(a+b) + (a^2 + 2ab + b^2) \]

### Step 6: Cancel Common Terms
We can cancel \( x^2 \) and \( y^2 \) from both sides, as well as the \( a^2 \) and \( b^2 \) terms:
\[ -2x(a+b) - 2y(a-b) = -2x(a-b) - 2y(a+b) \]

### Step 7: Rearranging the Equation
Rearranging the equation gives:
\[ -2x(a+b) + 2x(a-b) + 2y(a-b) + 2y(a+b) = 0 \]
This simplifies to:
\[ 2x(-b) + 2y(a) = 0 \]

### Step 8: Final Locus Equation
Dividing through by 2, we get:
\[ -bx + ay = 0 \quad \Rightarrow \quad ax = by \]
This can be rewritten as:
\[ x - y = 0 \]

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