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

Options

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

Answer

To find the locus of point \( S \) satisfying the relation \( SQ^2 + SR^2 = 2SP^2 \), where \( P(1,0) \), \( Q(-1,0) \), and \( R(2,0) \) are given points, we will follow these steps:

### Step 1: Define the coordinates of point \( S \)
Let the coordinates of point \( S \) be \( S(h, k) \).

### Step 2: Calculate the distances \( SQ \), \( SR \), and \( SP \)
Using the distance formula, we can express the distances as follows:
- \( SQ = \sqrt{(h + 1)^2 + k^2} \)
- \( SR = \sqrt{(h - 2)^2 + k^2} \)
- \( SP = \sqrt{(h - 1)^2 + k^2} \)

### Step 3: Square the distances
Now, we will square each distance:
- \( SQ^2 = (h + 1)^2 + k^2 \)
- \( SR^2 = (h - 2)^2 + k^2 \)
- \( SP^2 = (h - 1)^2 + k^2 \)

### Step 4: Substitute the squared distances into the given equation
Substituting into the equation \( SQ^2 + SR^2 = 2SP^2 \):
\[ (h + 1)^2 + k^2 + (h - 2)^2 + k^2 = 2((h - 1)^2 + k^2) \]

### Step 5: Simplify the equation
Combine like terms:
\[ (h + 1)^2 + (h - 2)^2 + 2k^2 = 2((h - 1)^2 + k^2) \]
Expanding both sides:
\[ (h^2 + 2h + 1) + (h^2 - 4h + 4) + 2k^2 = 2(h^2 - 2h + 1 + k^2) \]
This simplifies to:
\[ 2h^2 - 2h + 5 + 2k^2 = 2h^2 - 4h + 2 + 2k^2 \]

### Step 6: Cancel out common terms
Cancel \( 2h^2 \) and \( 2k^2 \) from both sides:
\[ -2h + 5 = -4h + 2 \]

### Step 7: Solve for \( h \)
Rearranging gives:
\[ -2h + 4h = 2 - 5 \]
\[ 2h = -3 \quad \Rightarrow \quad h = -\frac{3}{2} \]

### Step 8: Determine the locus
Since \( h = -\frac{3}{2} \) is a constant, the locus of point \( S \) is a vertical line given by:
\[ x = -\frac{3}{2} \]

### Final Answer
The locus of point \( S \) is the vertical line \( x = -\frac{3}{2} \).

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