If the point P(x, y) is equidistant from A(5, 1) and B(-1, 5) then find the relation between x and y.

Answer

To find the relation between \( x \) and \( y \) for the point \( P(x, y) \) that is equidistant from points \( A(5, 1) \) and \( B(-1, 5) \), we can use the distance formula.

### Step 1: Write the distance formula
The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

### Step 2: Set up the distances from point P to points A and B
The distance from point \( P(x, y) \) to point \( A(5, 1) \) is:
\[ d_{PA} = \sqrt{(x - 5)^2 + (y - 1)^2} \]
The distance from point \( P(x, y) \) to point \( B(-1, 5) \) is:
\[ d_{PB} = \sqrt{(x + 1)^2 + (y - 5)^2} \]

### Step 3: Set the distances equal to each other
Since point \( P \) is equidistant from points \( A \) and \( B \), we can set the distances equal:
\[ \sqrt{(x - 5)^2 + (y - 1)^2} = \sqrt{(x + 1)^2 + (y - 5)^2} \]

### Step 4: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ (x - 5)^2 + (y - 1)^2 = (x + 1)^2 + (y - 5)^2 \]

### Step 5: Expand both sides
Expanding the left side:
\[ (x - 5)^2 = x^2 - 10x + 25 \]
\[ (y - 1)^2 = y^2 - 2y + 1 \]
So,
\[ x^2 - 10x + 25 + y^2 - 2y + 1 = x^2 + y^2 - 10x - 2y + 26 \]

Expanding the right side:
\[ (x + 1)^2 = x^2 + 2x + 1 \]
\[ (y - 5)^2 = y^2 - 10y + 25 \]
So,
\[ x^2 + 2x + 1 + y^2 - 10y + 25 = x^2 + y^2 + 2x - 10y + 26 \]

### Step 6: Set the expanded equations equal to each other
Now we have:
\[ x^2 - 10x + 25 + y^2 - 2y + 1 = x^2 + y^2 + 2x - 10y + 26 \]

### Step 7: Simplify the equation
Cancel \( x^2 \) and \( y^2 \) from both sides:
\[ -10x + 25 - 2y + 1 = 2x - 10y + 26 \]
This simplifies to:
\[ -10x - 2y + 26 = 2x - 10y + 26 \]

### Step 8: Rearranging the equation
Rearranging gives:
\[ -10x + 10y = 2x + 2y \]
Combining like terms:
\[ -12x + 8y = 0 \]

### Step 9: Final relation
Dividing through by 4 gives:
\[ -3x + 2y = 0 \]
or
\[ 2y = 3x \]
Thus, the relation between \( x \) and \( y \) is:
\[ y = \frac{3}{2}x \]