`A(2,3) , B (1,5), C(-1,2) ` are the three points . If P is a point such that `PA^(2)+PB^(2)=2PC^2`, then find locus of P.

Answer

To find the locus of the point \( P(x, y) \) such that \( PA^2 + PB^2 = 2PC^2 \) for the points \( A(2, 3) \), \( B(1, 5) \), and \( C(-1, 2) \), we will follow these steps:

### Step 1: Express \( PA^2 \), \( PB^2 \), and \( PC^2 \) using the distance formula.

Using the distance formula, we can express the squares of the distances as follows:

1. **For \( PA^2 \)**:
\[ PA^2 = (x - 2)^2 + (y - 3)^2 \]

2. **For \( PB^2 \)**:
\[ PB^2 = (x - 1)^2 + (y - 5)^2 \]

3. **For \( PC^2 \)**:
\[ PC^2 = (x + 1)^2 + (y - 2)^2 \]

### Step 2: Substitute these expressions into the given equation.

Now we substitute these expressions into the equation \( PA^2 + PB^2 = 2PC^2 \):
\[ (x - 2)^2 + (y - 3)^2 + (x - 1)^2 + (y - 5)^2 = 2((x + 1)^2 + (y - 2)^2) \]

### Step 3: Expand each term.

Now we will expand each term on both sides:

1. **Left Side**:
\[ (x - 2)^2 = x^2 - 4x + 4 \]
\[ (y - 3)^2 = y^2 - 6y + 9 \]
\[ (x - 1)^2 = x^2 - 2x + 1 \]
\[ (y - 5)^2 = y^2 - 10y + 25 \]

Combining these:
\[ PA^2 + PB^2 = (x^2 - 4x + 4) + (y^2 - 6y + 9) + (x^2 - 2x + 1) + (y^2 - 10y + 25) \]
\[ = 2x^2 + 2y^2 - 6x - 16y + 39 \]

2. **Right Side**:
\[ (x + 1)^2 = x^2 + 2x + 1 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
Thus,
\[ 2PC^2 = 2((x^2 + 2x + 1) + (y^2 - 4y + 4)) \]
\[ = 2x^2 + 4x + 2 + 2y^2 - 8y + 8 \]
\[ = 2x^2 + 2y^2 + 4x - 8y + 10 \]

### Step 4: Set the left side equal to the right side and simplify.

Now we set the left side equal to the right side:
\[ 2x^2 + 2y^2 - 6x - 16y + 39 = 2x^2 + 2y^2 + 4x - 8y + 10 \]

Subtract \( 2x^2 + 2y^2 \) from both sides:
\[ -6x - 16y + 39 = 4x - 8y + 10 \]

Rearranging gives:
\[ -6x - 4x - 16y + 8y + 39 - 10 = 0 \]
\[ -10x - 8y + 29 = 0 \]

### Step 5: Multiply through by -1 to simplify.

Multiplying through by -1:
\[ 10x + 8y - 29 = 0 \]

### Final Result

Thus, the locus of the point \( P \) is given by the equation:
\[ 10x + 8y - 29 = 0 \]