Find a point on X-axis which is equidistant from both the points (1, 2, 3) and (3, 5, -2) .

Answer

To find a point on the X-axis that is equidistant from the points \( A(1, 2, 3) \) and \( B(3, 5, -2) \), we can follow these steps:

### Step 1: Define the point on the X-axis
Let the point on the X-axis be \( P(x, 0, 0) \).

### Step 2: Calculate the distance from point P to point A
Using the distance formula in three dimensions, the distance \( PA \) from point \( P \) to point \( A(1, 2, 3) \) is given by:

\[ PA = \sqrt{(x - 1)^2 + (0 - 2)^2 + (0 - 3)^2} \]

This simplifies to:

\[ PA = \sqrt{(x - 1)^2 + 4 + 9} = \sqrt{(x - 1)^2 + 13} \]

### Step 3: Calculate the distance from point P to point B
Similarly, the distance \( PB \) from point \( P \) to point \( B(3, 5, -2) \) is:

\[ PB = \sqrt{(x - 3)^2 + (0 - 5)^2 + (0 + 2)^2} \]

This simplifies to:

\[ PB = \sqrt{(x - 3)^2 + 25 + 4} = \sqrt{(x - 3)^2 + 29} \]

### Step 4: Set the distances equal
Since point \( P \) is equidistant from points \( A \) and \( B \), we set the distances equal to each other:

\[ \sqrt{(x - 1)^2 + 13} = \sqrt{(x - 3)^2 + 29} \]

### Step 5: Square both sides to eliminate the square roots
Squaring both sides gives:

\[ (x - 1)^2 + 13 = (x - 3)^2 + 29 \]

### Step 6: Expand both sides
Expanding both sides:

\[ (x^2 - 2x + 1) + 13 = (x^2 - 6x + 9) + 29 \]

This simplifies to:

\[ x^2 - 2x + 14 = x^2 - 6x + 38 \]

### Step 7: Simplify the equation
Subtract \( x^2 \) from both sides:

\[ -2x + 14 = -6x + 38 \]

### Step 8: Rearrange the equation
Rearranging gives:

\[ 4x = 24 \]

### Step 9: Solve for x
Dividing both sides by 4:

\[ x = 6 \]

### Step 10: Write the final point
Thus, the point on the X-axis that is equidistant from both points \( A \) and \( B \) is:

\[ P(6, 0, 0) \]