Find the co ordinates of the point equidistant form the four points with ordinates (2,0,0) ,(0,-1,0) ,(0,0,5) and (0,0,0) find also the distance of the point from the four points

Answer

To find the coordinates of the point that is equidistant from the four given points \( A(2, 0, 0) \), \( B(0, -1, 0) \), \( C(0, 0, 5) \), and \( D(0, 0, 0) \), we will follow these steps:

### Step 1: Set up the distance equations
Let the point \( P \) have coordinates \( (x, y, z) \). The distances from point \( P \) to each of the points \( A \), \( B \), \( C \), and \( D \) must be equal. Therefore, we can write:

1. \( AP^2 = BP^2 \)
2. \( BP^2 = CP^2 \)
3. \( CP^2 = DP^2 \)

### Step 2: Write the distance formulas
Using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \), we can express the squared distances:

1. \( AP^2 = (x - 2)^2 + (y - 0)^2 + (z - 0)^2 \)
2. \( BP^2 = (x - 0)^2 + (y + 1)^2 + (z - 0)^2 \)
3. \( CP^2 = (x - 0)^2 + (y - 0)^2 + (z - 5)^2 \)
4. \( DP^2 = (x - 0)^2 + (y - 0)^2 + (z - 0)^2 \)

### Step 3: Set up the equations
From \( AP^2 = BP^2 \):

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

Expanding both sides:

\[ (x^2 - 4x + 4) + y^2 + z^2 = x^2 + (y^2 + 2y + 1) + z^2 \]

Cancelling \( x^2 \), \( y^2 \), and \( z^2 \):

\[ -4x + 4 = 2y + 1 \]

Rearranging gives:

\[ 4x + 2y = 3 \quad \text{(Equation 1)} \]

### Step 4: Set up the second equation
From \( BP^2 = DP^2 \):

\[ x^2 + (y + 1)^2 + z^2 = x^2 + y^2 + z^2 \]

Cancelling \( x^2 \), \( y^2 \), and \( z^2 \):

\[ (y + 1)^2 = y^2 \]

Expanding gives:

\[ y^2 + 2y + 1 = y^2 \]

Cancelling \( y^2 \):

\[ 2y + 1 = 0 \]

Solving gives:

\[ y = -\frac{1}{2} \quad \text{(Equation 2)} \]

### Step 5: Set up the third equation
From \( CP^2 = DP^2 \):

\[ (x^2 + y^2 + (z - 5)^2) = (x^2 + y^2 + z^2) \]

Cancelling \( x^2 \) and \( y^2 \):

\[ (z - 5)^2 = z^2 \]

Expanding gives:

\[ z^2 - 10z + 25 = z^2 \]

Cancelling \( z^2 \):

\[ -10z + 25 = 0 \]

Solving gives:

\[ z = \frac{5}{2} \quad \text{(Equation 3)} \]

### Step 6: Solve for \( x \)
Substituting \( y = -\frac{1}{2} \) into Equation 1:

\[ 4x + 2\left(-\frac{1}{2}\right) = 3 \]

This simplifies to:

\[ 4x - 1 = 3 \]

So,

\[ 4x = 4 \implies x = 1 \]

### Step 7: Final coordinates
Thus, the coordinates of the point \( P \) are:

\[ P(1, -\frac{1}{2}, \frac{5}{2}) \]

### Step 8: Calculate the distance from point \( P \) to each of the four points
To find the distance from \( P \) to any of the points (let's calculate \( DP \)):

\[ DP = \sqrt{(1 - 0)^2 + \left(-\frac{1}{2} - 0\right)^2 + \left(\frac{5}{2} - 0\right)^2} \]

Calculating each term:

\[ = \sqrt{1^2 + \left(-\frac{1}{2}\right)^2 + \left(\frac{5}{2}\right)^2} \]

Calculating:

\[ = \sqrt{1 + \frac{1}{4} + \frac{25}{4}} = \sqrt{1 + \frac{26}{4}} = \sqrt{\frac{4}{4} + \frac{26}{4}} = \sqrt{\frac{30}{4}} = \sqrt{\frac{15}{2}} \]

### Final Result
The coordinates of the point \( P \) are \( (1, -\frac{1}{2}, \frac{5}{2}) \) and the distance from \( P \) to each of the four points is \( \sqrt{\frac{15}{2}} \).