`p_(1),p_(2),p_(3)` are the distances of points (1,1),(2,0) and (0,2) from a variable line L such that `p_(1)+p_(2)+p_(3)=0`. The line L passes through a fixed point

Options

A. (1,2)
B. (1,1)
C. (2,1)
D. none

Answer

To solve the problem, we need to find the equation of the variable line \( L \) that passes through a fixed point and satisfies the condition that the sum of the distances from the given points to the line is zero.

### Step-by-step Solution:

1. **Define the Variable Line**:
Let the equation of the variable line \( L \) be given by:
\[ Ax + By + C = 0 \]

2. **Calculate Distances**:
The distances \( p_1, p_2, p_3 \) from the points \( (1, 1) \), \( (2, 0) \), and \( (0, 2) \) to the line are given by the formula for the distance from a point to a line:
\[ p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Thus, we can write:
- For point \( (1, 1) \):
\[ p_1 = \frac{|A(1) + B(1) + C|}{\sqrt{A^2 + B^2}} = \frac{|A + B + C|}{\sqrt{A^2 + B^2}} \]
- For point \( (2, 0) \):
\[ p_2 = \frac{|A(2) + B(0) + C|}{\sqrt{A^2 + B^2}} = \frac{|2A + C|}{\sqrt{A^2 + B^2}} \]
- For point \( (0, 2) \):
\[ p_3 = \frac{|A(0) + B(2) + C|}{\sqrt{A^2 + B^2}} = \frac{|2B + C|}{\sqrt{A^2 + B^2}} \]

3. **Set Up the Equation**:
According to the problem, we have:
\[ p_1 + p_2 + p_3 = 0 \]
Substituting the distances:
\[ \frac{|A + B + C|}{\sqrt{A^2 + B^2}} + \frac{|2A + C|}{\sqrt{A^2 + B^2}} + \frac{|2B + C|}{\sqrt{A^2 + B^2}} = 0 \]
Since \( \sqrt{A^2 + B^2} \) is a positive quantity, we can multiply through by \( \sqrt{A^2 + B^2} \):
\[ |A + B + C| + |2A + C| + |2B + C| = 0 \]

4. **Analyze the Absolute Values**:
The only way for the sum of absolute values to equal zero is if each absolute value is individually zero:
\[ |A + B + C| = 0, \quad |2A + C| = 0, \quad |2B + C| = 0 \]
This leads to the following equations:
\[ A + B + C = 0 \quad (1) \]
\[ 2A + C = 0 \quad (2) \]
\[ 2B + C = 0 \quad (3) \]

5. **Solve the System of Equations**:
From equation (2), we can express \( C \) in terms of \( A \):
\[ C = -2A \]
From equation (3), we can express \( C \) in terms of \( B \):
\[ C = -2B \]
Setting these equal gives:
\[ -2A = -2B \implies A = B \]
Substituting \( C = -2A \) into equation (1):
\[ A + A - 2A = 0 \implies 0 = 0 \]
This confirms that \( A \) and \( B \) can take any value as long as they are equal.

6. **Final Equation of the Line**:
Therefore, the equation of the line can be expressed as:
\[ A(x + y) - 2A = 0 \implies x + y - 2 = 0 \]
This line passes through the point \( (1, 1) \).

### Conclusion:
The variable line \( L \) that satisfies the given conditions is:
\[ x + y - 2 = 0 \]