Find the locus of a point P If the join of the points (2,3) and (-1,5) subtends a right angle at P.

Answer

To find the locus of a point \( P \) such that the line segment joining the points \( A(2, 3) \) and \( B(-1, 5) \) subtends a right angle at \( P \), we can follow these steps:

### Step 1: Define the points and the point P
Let \( P \) be represented by the coordinates \( (x, y) \). The points \( A \) and \( B \) are given as \( A(2, 3) \) and \( B(-1, 5) \).

### Step 2: Use the Pythagorean theorem
Since \( P \) subtends a right angle at the line segment \( AB \), we can apply the Pythagorean theorem. According to the theorem, for the right triangle \( APB \):
\[ PA^2 + PB^2 = AB^2 \]

### Step 3: Calculate the distance \( AB \)
First, we need to find the distance \( AB \):
\[ AB = \sqrt{(2 - (-1))^2 + (3 - 5)^2} = \sqrt{(2 + 1)^2 + (3 - 5)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \]
Thus, \( AB^2 = 13 \).

### Step 4: Calculate \( PA^2 \) and \( PB^2 \)
Next, we calculate the distances \( PA \) and \( PB \):
\[ PA = \sqrt{(x - 2)^2 + (y - 3)^2} \]
\[ PB = \sqrt{(x + 1)^2 + (y - 5)^2} \]
Now squaring both distances:
\[ PA^2 = (x - 2)^2 + (y - 3)^2 \]
\[ PB^2 = (x + 1)^2 + (y - 5)^2 \]

### Step 5: Set up the equation using the Pythagorean theorem
Now substituting these into the Pythagorean theorem:
\[ (x - 2)^2 + (y - 3)^2 + (x + 1)^2 + (y - 5)^2 = 13 \]

### Step 6: Expand and simplify the equation
Expanding each term:
\[ (x - 2)^2 = x^2 - 4x + 4 \]
\[ (y - 3)^2 = y^2 - 6y + 9 \]
\[ (x + 1)^2 = x^2 + 2x + 1 \]
\[ (y - 5)^2 = y^2 - 10y + 25 \]

Combining these:
\[ (x^2 - 4x + 4) + (y^2 - 6y + 9) + (x^2 + 2x + 1) + (y^2 - 10y + 25) = 13 \]

Combining like terms:
\[ 2x^2 + 2y^2 - 2x - 16y + 39 = 13 \]

### Step 7: Rearranging the equation
Now, rearranging gives:
\[ 2x^2 + 2y^2 - 2x - 16y + 26 = 0 \]
Dividing the entire equation by 2:
\[ x^2 + y^2 - x - 8y + 13 = 0 \]

### Final Result
Thus, the locus of the point \( P \) is given by the equation:
\[ x^2 + y^2 - x - 8y + 13 = 0 \]