Find the co-ordinates of the point equidistant from three given points `A(5,1),B(-3,-7)andC(7,-1).`

Answer

To find the coordinates of the point \( D(x, y) \) that is equidistant from the three given points \( A(5, 1) \), \( B(-3, -7) \), and \( C(7, -1) \), we can follow these steps:

### Step 1: Set up the equations for distances
Since point \( D \) is equidistant from points \( A \), \( B \), and \( C \), we can set up the following equations based on the distance formula:

1. Distance \( DA = DB \)
2. Distance \( DB = DC \)

### Step 2: Write the distance equations
Using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \), we can express the distances:

1. \( DA = \sqrt{(x - 5)^2 + (y - 1)^2} \)
2. \( DB = \sqrt{(x + 3)^2 + (y + 7)^2} \)
3. \( DC = \sqrt{(x - 7)^2 + (y + 1)^2} \)

### Step 3: Set up the first equation
From \( DA = DB \):

\[ \sqrt{(x - 5)^2 + (y - 1)^2} = \sqrt{(x + 3)^2 + (y + 7)^2} \]

### Step 4: Square both sides
Squaring both sides to eliminate the square roots gives:

\[ (x - 5)^2 + (y - 1)^2 = (x + 3)^2 + (y + 7)^2 \]

### Step 5: Expand both sides
Expanding both sides:

\[ (x^2 - 10x + 25) + (y^2 - 2y + 1) = (x^2 + 6x + 9) + (y^2 + 14y + 49) \]

### Step 6: Simplify the equation
After simplification, we have:

\[ -10x + 25 - 2y + 1 = 6x + 9 + 14y + 49 \]

This simplifies to:

\[ -16x - 16y + 25 + 1 - 9 - 49 = 0 \]

Which simplifies to:

\[ -16x - 16y - 32 = 0 \]

Dividing through by -16 gives us:

\[ x + y + 2 = 0 \quad \text{(Equation 1)} \]

### Step 7: Set up the second equation
Now, using \( DB = DC \):

\[ \sqrt{(x + 3)^2 + (y + 7)^2} = \sqrt{(x - 7)^2 + (y + 1)^2} \]

### Step 8: Square both sides again
Squaring both sides gives:

\[ (x + 3)^2 + (y + 7)^2 = (x - 7)^2 + (y + 1)^2 \]

### Step 9: Expand both sides
Expanding both sides:

\[ (x^2 + 6x + 9) + (y^2 + 14y + 49) = (x^2 - 14x + 49) + (y^2 + 2y + 1) \]

### Step 10: Simplify the equation
After simplification, we have:

\[ 6x + 9 + 14y + 49 = -14x + 49 + 2y + 1 \]

This simplifies to:

\[ 20x + 12y + 8 = 0 \]

Dividing through by 4 gives us:

\[ 5x + 3y + 2 = 0 \quad \text{(Equation 2)} \]

### Step 11: Solve the system of equations
Now we have two equations:

1. \( x + y + 2 = 0 \)
2. \( 5x + 3y + 2 = 0 \)

From Equation 1, we can express \( y \) in terms of \( x \):

\[ y = -x - 2 \]

Substituting \( y \) into Equation 2:

\[ 5x + 3(-x - 2) + 2 = 0 \]

This simplifies to:

\[ 5x - 3x - 6 + 2 = 0 \]

\[ 2x - 4 = 0 \]

So,

\[ x = 2 \]

### Step 12: Find \( y \)
Substituting \( x = 2 \) back into Equation 1:

\[ y = -2 - 2 = -4 \]

### Final Answer
Thus, the coordinates of the point \( D \) that is equidistant from points \( A \), \( B \), and \( C \) are:

\[ \boxed{(2, -4)} \]