Find the equation to the locus of a point which is always equidistant from the points whose coordinates are
`( p+q, p-q) and (p- q, p+q)`

Answer

To find the equation of the locus of a point that is always equidistant from the points \((p+q, p-q)\) and \((p-q, p+q)\), we can follow these steps:

### Step 1: Define the Point
Let the point whose locus we want to find be \((H, K)\).

### Step 2: Use the Distance Formula
According to the problem, the point \((H, K)\) is equidistant from the points \((p+q, p-q)\) and \((p-q, p+q)\). We can express this using the distance formula:
\[ \sqrt{(H - (p+q))^2 + (K - (p-q))^2} = \sqrt{(H - (p-q))^2 + (K - (p+q))^2} \]

### Step 3: Square Both Sides
To eliminate the square roots, we square both sides:
\[ (H - (p+q))^2 + (K - (p-q))^2 = (H - (p-q))^2 + (K - (p+q))^2 \]

### Step 4: Expand Both Sides
Now, we expand both sides of the equation:
- Left Side:
\[ (H - (p+q))^2 = H^2 - 2H(p+q) + (p+q)^2 \]
\[ (K - (p-q))^2 = K^2 - 2K(p-q) + (p-q)^2 \]
Combining these:
\[ H^2 - 2H(p+q) + (p^2 + 2pq + q^2) + K^2 - 2K(p-q) + (p^2 - 2pq + q^2) \]

- Right Side:
\[ (H - (p-q))^2 = H^2 - 2H(p-q) + (p-q)^2 \]
\[ (K - (p+q))^2 = K^2 - 2K(p+q) + (p+q)^2 \]
Combining these:
\[ H^2 - 2H(p-q) + (p^2 - 2pq + q^2) + K^2 - 2K(p+q) + (p^2 + 2pq + q^2) \]

### Step 5: Set the Expanded Equations Equal
Now we set the expanded left side equal to the expanded right side:
\[ H^2 - 2H(p+q) + K^2 - 2K(p-q) + 2p^2 + 2q^2 = H^2 - 2H(p-q) + K^2 - 2K(p+q) + 2p^2 + 2q^2 \]

### Step 6: Simplify the Equation
Cancelling out the common terms and simplifying:
\[ -2H(p+q) - 2K(p-q) = -2H(p-q) - 2K(p+q) \]
Rearranging gives:
\[ 2H(p-q - p-q) = 2K(p+q - p+q) \]
This simplifies to:
\[ -2Hq + 2Kq = 0 \]
Factoring out common terms:
\[ q(H - K) = 0 \]

### Step 7: Final Equation
Since \(q\) cannot be zero (as it is a constant), we conclude:
\[ H = K \]
Replacing \(H\) with \(x\) and \(K\) with \(y\), we get the equation of the locus:
\[ x - y = 0 \quad \text{or} \quad y = x \]