If the distances of p (x, y) from A (-1, 5) and B (5, 1) are equal, then

Options

A. `2x=y`
B. `3x=2y`
C. `3x=y`
D. `2x=3y`

Answer

To solve the problem where the distances of point P (x, y) from points A (-1, 5) and B (5, 1) are equal, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to find the relationship between x and y such that the distance from point P to point A is equal to the distance from point P to point B.

2. **Using the Distance Formula**: The distance between two points (x1, y1) and (x2, y2) is given by the formula:
\[ D = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} \]

3. **Setting Up the Distances**:
- The distance from P (x, y) to A (-1, 5) is:
\[ PA = \sqrt{(x - (-1))^2 + (y - 5)^2} = \sqrt{(x + 1)^2 + (y - 5)^2} \]
- The distance from P (x, y) to B (5, 1) is:
\[ PB = \sqrt{(x - 5)^2 + (y - 1)^2} \]

4. **Equating the Distances**: Since the distances are equal, we can set them equal to each other:
\[ \sqrt{(x + 1)^2 + (y - 5)^2} = \sqrt{(x - 5)^2 + (y - 1)^2} \]

5. **Squaring Both Sides**: To eliminate the square roots, we square both sides:
\[ (x + 1)^2 + (y - 5)^2 = (x - 5)^2 + (y - 1)^2 \]

6. **Expanding Both Sides**:
- Left Side:
\[ (x + 1)^2 = x^2 + 2x + 1 \]
\[ (y - 5)^2 = y^2 - 10y + 25 \]
So,
\[ x^2 + 2x + 1 + y^2 - 10y + 25 = x^2 + y^2 + 2x - 10y + 26 \]

- Right Side:
\[ (x - 5)^2 = x^2 - 10x + 25 \]
\[ (y - 1)^2 = y^2 - 2y + 1 \]
So,
\[ x^2 - 10x + 25 + y^2 - 2y + 1 = x^2 + y^2 - 10x - 2y + 26 \]

7. **Setting the Expanded Equations Equal**:
\[ x^2 + y^2 + 2x - 10y + 26 = x^2 + y^2 - 10x - 2y + 26 \]

8. **Canceling Common Terms**: Both sides have \(x^2\), \(y^2\), and \(26\), so we can cancel them out:
\[ 2x - 10y = -10x - 2y \]

9. **Rearranging the Equation**: Move all terms involving x to one side and y to the other:
\[ 2x + 10x = -2y + 10y \]
\[ 12x = 8y \]

10. **Simplifying the Relationship**: Divide both sides by 4:
\[ 3x = 2y \]

### Final Answer:
The relationship between x and y is:
\[ 3x = 2y \]