A variable line L is drawn through O(0,0) to meet the line `L_(1) " and " L_(2)` given by y-x-10 =0 and y-x-20=0 at Points A and B, respectively.
Locus of P, if `OP^(2) = OA xx OB`, is a. `(y+x)^(2) = 50` b. `(y-x)^(2) = 200` c. `(y-x)^(2) = 100` d. none of these

Options

A. `(y-x)^(2) = 100`
B. `(y+x)^(2) = 50`
C. `(y-x)^(2) = 200`
D. none of these

Explanation

To solve the problem, we need to find the locus of point P such that \( OP^2 = OA \times OB \), where O is the origin, A is the intersection of line L with line \( L_1 \), and B is the intersection of line L with line \( L_2 \).

### Step 1: Define the variable line L
The variable line L passing through the origin can be expressed as:
\[ y = mx \]
where \( m \) is the slope of the line.

### Step 2: Find the equations of lines \( L_1 \) and \( L_2 \)
The equations of lines \( L_1 \) and \( L_2 \) are given as:
\[ L_1: y - x - 10 = 0 \quad \text{(or } y = x + 10\text{)} \]
\[ L_2: y - x - 20 = 0 \quad \text{(or } y = x + 20\text{)} \]

### Step 3: Find the coordinates of point A (intersection of L and \( L_1 \))
To find point A, we set the equation of line L equal to the equation of line \( L_1 \):
\[ mx = x + 10 \]
Rearranging gives:
\[ mx - x = 10 \implies x(m - 1) = 10 \implies x = \frac{10}{m - 1} \]
Now, substituting \( x \) back into the equation of line L to find \( y \):
\[ y = m \left(\frac{10}{m - 1}\right) = \frac{10m}{m - 1} \]
Thus, the coordinates of point A are:
\[ A\left(\frac{10}{m - 1}, \frac{10m}{m - 1}\right) \]

### Step 4: Find the coordinates of point B (intersection of L and \( L_2 \))
To find point B, we set the equation of line L equal to the equation of line \( L_2 \):
\[ mx = x + 20 \]
Rearranging gives:
\[ mx - x = 20 \implies x(m - 1) = 20 \implies x = \frac{20}{m - 1} \]
Now, substituting \( x \) back into the equation of line L to find \( y \):
\[ y = m \left(\frac{20}{m - 1}\right) = \frac{20m}{m - 1} \]
Thus, the coordinates of point B are:
\[ B\left(\frac{20}{m - 1}, \frac{20m}{m - 1}\right) \]

### Step 5: Calculate \( OP^2 \), \( OA \), and \( OB \)
The distance \( OP \) is given by:
\[ OP^2 = x^2 + (mx)^2 = x^2(1 + m^2) \]

The distance \( OA \) is given by:
\[ OA = \sqrt{\left(0 - \frac{10}{m - 1}\right)^2 + \left(0 - \frac{10m}{m - 1}\right)^2} \]
Calculating \( OA^2 \):
\[ OA^2 = \left(\frac{10}{m - 1}\right)^2 + \left(\frac{10m}{m - 1}\right)^2 = \frac{100(1 + m^2)}{(m - 1)^2} \]

The distance \( OB \) is given by:
\[ OB = \sqrt{\left(0 - \frac{20}{m - 1}\right)^2 + \left(0 - \frac{20m}{m - 1}\right)^2} \]
Calculating \( OB^2 \):
\[ OB^2 = \left(\frac{20}{m - 1}\right)^2 + \left(\frac{20m}{m - 1}\right)^2 = \frac{400(1 + m^2)}{(m - 1)^2} \]

### Step 6: Set up the equation \( OP^2 = OA \times OB \)
From the previous steps, we have:
\[ x^2(1 + m^2) = \sqrt{\frac{100(1 + m^2)}{(m - 1)^2}} \times \sqrt{\frac{400(1 + m^2)}{(m - 1)^2}} \]
This simplifies to:
\[ x^2(1 + m^2) = \frac{20000(1 + m^2)}{(m - 1)^2} \]
Dividing both sides by \( (1 + m^2) \) (assuming \( m \neq 0 \)):
\[ x^2 = \frac{20000}{(m - 1)^2} \]

### Step 7: Substitute \( m = \frac{y}{x} \)
Substituting \( m = \frac{y}{x} \):
\[ x^2 = \frac{20000}{\left(\frac{y}{x} - 1\right)^2} \]
Cross-multiplying gives:
\[ x^4 = 20000 \left(\frac{x}{y - x}\right)^2 \]
Thus, we can express this as:
\[ (y - x)^2 = 200 \]

### Final Answer
The locus of point P is:
\[ (y - x)^2 = 200 \]
Thus, the correct option is:
**b. \((y - x)^2 = 200\)**