The locus of the point which is equidistant from the points (2, -2, 1) and (0, 2, 3) is

Options

A. 2x - y-z=1
B. x - 2y-z=1
C. x - 2y - 2+1 = 0
D. None of these

Answer

To find the locus of the point which is equidistant from the points \( A(2, -2, 1) \) and \( B(0, 2, 3) \), we can follow these steps:

### Step 1: Define the point
Let the point \( P(h, k, r) \) be the point that is equidistant from points \( A \) and \( B \).

### Step 2: Write the distance formula
The distance from point \( P \) to point \( A \) is given by:
\[ d_A = \sqrt{(h - 2)^2 + (k + 2)^2 + (r - 1)^2} \]
The distance from point \( P \) to point \( B \) is given by:
\[ d_B = \sqrt{(h - 0)^2 + (k - 2)^2 + (r - 3)^2} \]

### Step 3: Set the distances equal
Since point \( P \) is equidistant from points \( A \) and \( B \), we have:
\[ d_A = d_B \]
Squaring both sides to eliminate the square roots gives:
\[ (h - 2)^2 + (k + 2)^2 + (r - 1)^2 = h^2 + (k - 2)^2 + (r - 3)^2 \]

### Step 4: Expand both sides
Expanding the left-hand side:
\[ (h - 2)^2 = h^2 - 4h + 4 \]
\[ (k + 2)^2 = k^2 + 4k + 4 \]
\[ (r - 1)^2 = r^2 - 2r + 1 \]
So, the left-hand side becomes:
\[ h^2 - 4h + 4 + k^2 + 4k + 4 + r^2 - 2r + 1 = h^2 + k^2 + r^2 - 4h + 4k + 9 \]

Now expanding the right-hand side:
\[ h^2 + (k - 2)^2 + (r - 3)^2 = h^2 + (k^2 - 4k + 4) + (r^2 - 6r + 9) \]
So, the right-hand side becomes:
\[ h^2 + k^2 - 4k + 4 + r^2 - 6r + 9 = h^2 + k^2 + r^2 - 4k - 6r + 13 \]

### Step 5: Set the expanded forms equal
Setting the two expanded forms equal gives:
\[ h^2 + k^2 + r^2 - 4h + 4k + 9 = h^2 + k^2 + r^2 - 4k - 6r + 13 \]

### Step 6: Simplify the equation
Cancelling \( h^2, k^2, r^2 \) from both sides:
\[ -4h + 4k + 9 = -4k - 6r + 13 \]
Rearranging gives:
\[ -4h + 8k + 6r - 4 = 0 \]

### Step 7: Divide by -2
Dividing the entire equation by -2 gives:
\[ 2h - 4k - 3r + 2 = 0 \]

### Step 8: Replace variables
Now, replace \( h \) with \( x \), \( k \) with \( y \), and \( r \) with \( z \):
\[ 2x - 4y - 3z + 2 = 0 \]

### Final Equation
Thus, the locus of the point which is equidistant from the points \( (2, -2, 1) \) and \( (0, 2, 3) \) is:
\[ 2x - 4y - 3z + 2 = 0 \]