If the points (5, 4) and (x, y) are equidistant from the point (4, 5) , prove that `x^(2)+y^(2)-8x-10y+39=0`

Answer

To prove that the points (5, 4) and (x, y) are equidistant from the point (4, 5), we will use the distance formula and set the distances equal to each other.

### Step 1: Identify the points
Let:
- Point A = (5, 4)
- Point B = (x, y)
- Point C = (4, 5)

### Step 2: Use the distance formula
The distance between two points (x1, y1) and (x2, y2) is given by the formula:
\[ d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} \]

### Step 3: Calculate the distance AB
Using the distance formula, the distance AB (from point A to point C) is:
\[ AB = \sqrt{(4 - 5)^2 + (5 - 4)^2} \]
Calculating this gives:
\[ AB = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 4: Calculate the distance BC
Now, we calculate the distance BC (from point B to point C):
\[ BC = \sqrt{(4 - x)^2 + (5 - y)^2} \]

### Step 5: Set the distances equal
Since the points are equidistant from point C, we set the distances equal:
\[ \sqrt{2} = \sqrt{(4 - x)^2 + (5 - y)^2} \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides gives:
\[ 2 = (4 - x)^2 + (5 - y)^2 \]

### Step 7: Expand the right-hand side
Expanding the right-hand side:
\[ (4 - x)^2 = 16 - 8x + x^2 \]
\[ (5 - y)^2 = 25 - 10y + y^2 \]
So,
\[ 2 = (16 - 8x + x^2) + (25 - 10y + y^2) \]
Combining these gives:
\[ 2 = x^2 + y^2 - 8x - 10y + 41 \]

### Step 8: Rearranging the equation
Rearranging the equation to bring all terms to one side:
\[ x^2 + y^2 - 8x - 10y + 41 - 2 = 0 \]
This simplifies to:
\[ x^2 + y^2 - 8x - 10y + 39 = 0 \]

### Conclusion
Thus, we have proved that:
\[ x^2 + y^2 - 8x - 10y + 39 = 0 \]