The point on x - axis which is equidistant from the points (3,2,2) and (5,5,4) is

Options

A. `((49)/(4),0,0)`
B. `((23)/(2),0,0)`
C. `(-(49)/(4),0,0)`
D. (2,0,0)

Answer

To find the point on the x-axis that is equidistant from the points (3, 2, 2) and (5, 5, 4), we will follow these steps:

### Step 1: Define the point on the x-axis
Let the point on the x-axis be \( P(\lambda, 0, 0) \), where \( \lambda \) is the x-coordinate we need to find.

### Step 2: Calculate the distance from point P to point A (3, 2, 2)
The distance \( PA \) from point \( P(\lambda, 0, 0) \) to point \( A(3, 2, 2) \) can be calculated using the distance formula:
\[ PA = \sqrt{(\lambda - 3)^2 + (0 - 2)^2 + (0 - 2)^2} \]
This simplifies to:
\[ PA = \sqrt{(\lambda - 3)^2 + 4 + 4} = \sqrt{(\lambda - 3)^2 + 8} \]

### Step 3: Calculate the distance from point P to point B (5, 5, 4)
Similarly, the distance \( PB \) from point \( P(\lambda, 0, 0) \) to point \( B(5, 5, 4) \) is:
\[ PB = \sqrt{(\lambda - 5)^2 + (0 - 5)^2 + (0 - 4)^2} \]
This simplifies to:
\[ PB = \sqrt{(\lambda - 5)^2 + 25 + 16} = \sqrt{(\lambda - 5)^2 + 41} \]

### Step 4: Set the distances equal to each other
Since point \( P \) is equidistant from points \( A \) and \( B \), we set \( PA = PB \):
\[ \sqrt{(\lambda - 3)^2 + 8} = \sqrt{(\lambda - 5)^2 + 41} \]

### Step 5: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ (\lambda - 3)^2 + 8 = (\lambda - 5)^2 + 41 \]

### Step 6: Expand both sides
Expanding both sides:
\[ \lambda^2 - 6\lambda + 9 + 8 = \lambda^2 - 10\lambda + 25 + 41 \]
This simplifies to:
\[ \lambda^2 - 6\lambda + 17 = \lambda^2 - 10\lambda + 66 \]

### Step 7: Cancel \( \lambda^2 \) and rearrange the equation
Cancelling \( \lambda^2 \) from both sides:
\[ -6\lambda + 17 = -10\lambda + 66 \]
Rearranging gives:
\[ 10\lambda - 6\lambda = 66 - 17 \]
\[ 4\lambda = 49 \]

### Step 8: Solve for \( \lambda \)
Dividing both sides by 4:
\[ \lambda = \frac{49}{4} \]

### Step 9: Write the coordinates of point P
Thus, the coordinates of point \( P \) on the x-axis are:
\[ P\left(\frac{49}{4}, 0, 0\right) \]

### Final Answer
The point on the x-axis which is equidistant from the points (3, 2, 2) and (5, 5, 4) is:
\[ \left(\frac{49}{4}, 0, 0\right) \]

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