Find the relation between x and y such that the point P (x,y) is equidistant from the points `A(1,4)and B(-1,2).`

Answer

To find the relation between \( x \) and \( y \) such that the point \( P(x, y) \) is equidistant from the points \( A(1, 4) \) and \( B(-1, 2) \), we can use the distance formula.

### Step-by-Step Solution:

1. **Distance from Point P to Point A**:
The distance \( PA \) from point \( P(x, y) \) to point \( A(1, 4) \) is given by the distance formula:
\[ PA = \sqrt{(x - 1)^2 + (y - 4)^2} \]

2. **Distance from Point P to Point B**:
The distance \( PB \) from point \( P(x, y) \) to point \( B(-1, 2) \) is:
\[ PB = \sqrt{(x + 1)^2 + (y - 2)^2} \]

3. **Setting the Distances Equal**:
Since point \( P \) is equidistant from points \( A \) and \( B \), we set the two distances equal:
\[ \sqrt{(x - 1)^2 + (y - 4)^2} = \sqrt{(x + 1)^2 + (y - 2)^2} \]

4. **Squaring Both Sides**:
To eliminate the square roots, we square both sides:
\[ (x - 1)^2 + (y - 4)^2 = (x + 1)^2 + (y - 2)^2 \]

5. **Expanding Both Sides**:
Expanding the left side:
\[ (x - 1)^2 = x^2 - 2x + 1 \]
\[ (y - 4)^2 = y^2 - 8y + 16 \]
So,
\[ x^2 - 2x + 1 + y^2 - 8y + 16 = x^2 + y^2 - 2x - 8y + 17 \]

Expanding the right side:
\[ (x + 1)^2 = x^2 + 2x + 1 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
So,
\[ x^2 + 2x + 1 + y^2 - 4y + 4 = x^2 + y^2 + 2x - 4y + 5 \]

6. **Setting the Expanded Equations Equal**:
Now we have:
\[ x^2 - 2x + 1 + y^2 - 8y + 16 = x^2 + y^2 + 2x - 4y + 5 \]

7. **Cancelling \( x^2 \) and \( y^2 \)**:
Cancel \( x^2 \) and \( y^2 \) from both sides:
\[ -2x + 1 - 8y + 16 = 2x - 4y + 5 \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ -2x - 2x + 8y - 4y + 1 + 16 - 5 = 0 \]
\[ -4x + 4y + 12 = 0 \]

9. **Dividing by 4**:
Dividing the entire equation by 4:
\[ -x + y + 3 = 0 \]
or
\[ y = x - 3 \]

### Final Relation:
The relation between \( x \) and \( y \) such that the point \( P(x, y) \) is equidistant from points \( A(1, 4) \) and \( B(-1, 2) \) is:
\[ y = x - 3 \]