Let the algebric sum of the perpendicular distance from the points (2,0),(0,2),(1,1) to a variable straight line be zero then the line passes through a fixed point whose co ordinates are

Options

A. (1,1)
B. (2,2)
C. (0,0)
D. None of these

Answer

To solve the problem, we need to find the fixed point through which the variable straight line passes, given that the algebraic sum of the perpendicular distances from the points (2,0), (0,2), and (1,1) to the line is zero.

### Step-by-Step Solution:

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

2. **Calculate the Perpendicular Distances**:
The perpendicular distance \( P \) from a point \( (x_1, y_1) \) to the line \( Ax + By + C = 0 \) is given by:
\[ P = \frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}} \]

3. **Calculate Distances for Each Point**:
- For point \( P_1 = (1, 1) \):
\[ P_1 = \frac{A(1) + B(1) + C}{\sqrt{A^2 + B^2}} = \frac{A + B + C}{\sqrt{A^2 + B^2}} \]
- For point \( P_2 = (0, 2) \):
\[ P_2 = \frac{A(0) + B(2) + C}{\sqrt{A^2 + B^2}} = \frac{2B + C}{\sqrt{A^2 + B^2}} \]
- For point \( P_3 = (2, 0) \):
\[ P_3 = \frac{A(2) + B(0) + C}{\sqrt{A^2 + B^2}} = \frac{2A + C}{\sqrt{A^2 + B^2}} \]

4. **Set Up the Equation**:
According to the problem, the algebraic sum of these distances is zero:
\[ P_1 + P_2 + P_3 = 0 \]
Substituting the distances calculated:
\[ \frac{A + B + C}{\sqrt{A^2 + B^2}} + \frac{2B + C}{\sqrt{A^2 + B^2}} + \frac{2A + C}{\sqrt{A^2 + B^2}} = 0 \]

5. **Combine the Terms**:
Combine the numerators:
\[ \frac{(A + B + C) + (2B + C) + (2A + C)}{\sqrt{A^2 + B^2}} = 0 \]
This simplifies to:
\[ \frac{3A + 3B + 3C}{\sqrt{A^2 + B^2}} = 0 \]
Since \( \sqrt{A^2 + B^2} \) is non-zero, we can set the numerator to zero:
\[ 3A + 3B + 3C = 0 \implies A + B + C = 0 \]

6. **Find the Fixed Point**:
The line \( Ax + By + C = 0 \) can be rewritten using \( C = -A - B \):
\[ Ax + By - (A + B) = 0 \implies Ax + By + A + B = 0 \]
Rearranging gives:
\[ A(x + 1) + B(y + 1) = 0 \]
This indicates that the line passes through the point \( (-1, -1) \).

### Final Answer:
The fixed point through which the line passes is:
\[ \text{Coordinates: } (-1, -1) \]