Find the locus of a point whose distance from x-axis is equal the distance from the point (1, -1, 2) :

Options

A. `y^(2)+2x-2y-4z+6=0`
B. `x^(2)+2x-2y-4z+6=0`
C. `x^(2)-2x+2y-4z+6=0`
D. `z^(2)-2x+2y-4z+6=0`

Answer

To find the locus of a point whose distance from the x-axis is equal to its distance from the point (1, -1, 2), we can follow these steps:

### Step 1: Define the Point
Let the point be \( P(h, k, l) \).

### Step 2: Distance from the x-axis
The distance of the point \( P(h, k, l) \) from the x-axis is given by the formula:
\[ \text{Distance from x-axis} = \sqrt{k^2 + l^2} \]

### Step 3: Distance from the Point (1, -1, 2)
Next, we calculate the distance from the point \( P(h, k, l) \) to the point \( A(1, -1, 2) \) using the distance formula:
\[ PA = \sqrt{(h - 1)^2 + (k + 1)^2 + (l - 2)^2} \]

### Step 4: Set the Distances Equal
According to the problem, the distance from the x-axis is equal to the distance from the point \( A \):
\[ \sqrt{k^2 + l^2} = \sqrt{(h - 1)^2 + (k + 1)^2 + (l - 2)^2} \]

### Step 5: Square Both Sides
To eliminate the square roots, we square both sides:
\[ k^2 + l^2 = (h - 1)^2 + (k + 1)^2 + (l - 2)^2 \]

### Step 6: Expand the Right Side
Expanding the right side gives:
\[ k^2 + l^2 = (h^2 - 2h + 1) + (k^2 + 2k + 1) + (l^2 - 4l + 4) \]
Combining like terms, we have:
\[ k^2 + l^2 = h^2 - 2h + k^2 + 2k + l^2 + 6 - 4l \]

### Step 7: Simplify the Equation
Cancelling \( k^2 \) and \( l^2 \) from both sides, we get:
\[ 0 = h^2 - 2h + 2k - 4l + 6 \]

### Step 8: Rearranging the Equation
Rearranging gives us the equation of the locus:
\[ h^2 - 2h + 2k - 4l + 6 = 0 \]

### Step 9: Change Variables
To express in standard form, let \( x = h \), \( y = k \), \( z = l \):
\[ x^2 - 2x + 2y - 4z + 6 = 0 \]

### Final Answer
Thus, the equation of the required locus is:
\[ x^2 - 2x + 2y - 4z + 6 = 0 \]

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