If `P(1,0),Q(-1,0)` and `R(2,0)` are three given points, then the locus of point S satisfying the relation `SQ^(2)+SR^(2)=2sP^(2)` is

Options

A. a straight line parallel to x-axis
B. circle through origin
C. circle with centre at the origin
D. a straight line|| to y-axis

Answer

To solve the problem, we need to find the locus of the point \( S(h, k) \) that satisfies the equation \( SQ^2 + SR^2 = 2SP^2 \), where \( P(1, 0) \), \( Q(-1, 0) \), and \( R(2, 0) \) are given points.

### Step 1: Write the distances \( SQ \), \( SR \), and \( SP \)

1. **Distance \( SQ \)**:
\[ SQ = \sqrt{(h + 1)^2 + k^2} \]
Therefore,
\[ SQ^2 = (h + 1)^2 + k^2 \]

2. **Distance \( SR \)**:
\[ SR = \sqrt{(h - 2)^2 + k^2} \]
Therefore,
\[ SR^2 = (h - 2)^2 + k^2 \]

3. **Distance \( SP \)**:
\[ SP = \sqrt{(h - 1)^2 + k^2} \]
Therefore,
\[ SP^2 = (h - 1)^2 + k^2 \]

### Step 2: Substitute into the equation

Now, substituting these distances into the given equation \( SQ^2 + SR^2 = 2SP^2 \):

\[ (h + 1)^2 + k^2 + (h - 2)^2 + k^2 = 2[(h - 1)^2 + k^2] \]

### Step 3: Simplify the equation

1. Combine the left-hand side:
\[ (h + 1)^2 + (h - 2)^2 + 2k^2 = 2[(h - 1)^2 + k^2] \]

2. Expanding both sides:
- Left-hand side:
\[ (h^2 + 2h + 1) + (h^2 - 4h + 4) + 2k^2 = 2h^2 - 2h + 5 + 2k^2 \]
- Right-hand side:
\[ 2(h^2 - 2h + 1 + k^2) = 2h^2 - 4h + 2 + 2k^2 \]

3. Equating both sides:
\[ 2h^2 - 2h + 5 + 2k^2 = 2h^2 - 4h + 2 + 2k^2 \]

### Step 4: Cancel and simplify

1. Cancel \( 2h^2 \) and \( 2k^2 \) from both sides:
\[ -2h + 5 = -4h + 2 \]

2. Rearranging gives:
\[ 2h = 3 \implies h = \frac{3}{2} \]

### Step 5: Determine the locus

Since \( h \) is constant at \( \frac{3}{2} \), the locus of point \( S \) is a vertical line given by:
\[ x = \frac{3}{2} \]

### Conclusion

The locus of point \( S \) is a vertical line parallel to the y-axis at \( x = \frac{3}{2} \).