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 will follow these steps:

### Step 1: Set up the distance equations
Since point \( P \) is equidistant from points \( A \) and \( B \), we can write the equation:
\[ AP = PB \]
where \( AP \) is the distance from \( P \) to \( A \) and \( PB \) is the distance from \( P \) to \( B \).

### Step 2: Calculate the distances
Using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):

1. The distance \( AP \) is:
\[ AP = \sqrt{(x - 1)^2 + (y - 4)^2} \]

2. The distance \( PB \) is:
\[ PB = \sqrt{(x + 1)^2 + (y - 2)^2} \]

### Step 3: Set the distances equal to each other
Now we set the two distances equal:
\[ \sqrt{(x - 1)^2 + (y - 4)^2} = \sqrt{(x + 1)^2 + (y - 2)^2} \]

### Step 4: Square both sides
To eliminate the square roots, we square both sides:
\[ (x - 1)^2 + (y - 4)^2 = (x + 1)^2 + (y - 2)^2 \]

### Step 5: Expand both sides
Now, we expand both sides:
1. Left side:
\[ (x - 1)^2 = x^2 - 2x + 1 \]
\[ (y - 4)^2 = y^2 - 8y + 16 \]
So,
\[ x^2 - 2x + 1 + y^2 - 8y + 16 = x^2 + y^2 - 2x - 8y + 17 \]

2. Right side:
\[ (x + 1)^2 = x^2 + 2x + 1 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
So,
\[ x^2 + 2x + 1 + y^2 - 4y + 4 = x^2 + y^2 + 2x - 4y + 5 \]

### Step 6: Set the expanded forms equal
Now we have:
\[ x^2 + y^2 - 2x - 8y + 17 = x^2 + y^2 + 2x - 4y + 5 \]

### Step 7: Simplify the equation
Cancel \( x^2 \) and \( y^2 \) from both sides:
\[ -2x - 8y + 17 = 2x - 4y + 5 \]

### Step 8: Rearrange the equation
Rearranging gives:
\[ -2x - 8y + 4y + 17 - 5 = 2x \]
\[ -4x - 4y + 12 = 0 \]

### Step 9: Factor out common terms
Factoring out \(-4\):
\[ -4(x + y - 3) = 0 \]
This simplifies to:
\[ x + y = 3 \]

### Final Relation
Thus, the relation between \( x \) and \( y \) is:
\[ x + y = 3 \]

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