The algebraic sum of the perpendicular distances from `A(x_1, y_1)`, `B(x_2, y_2)` and `C(x_3, y_3)` to a variable line is zero. Then the line passes through (A) the orthocentre of `triangleABC` (B) centroid of `triangleABC` (C) incentre of `triangleABC` (D) circumcentre of `triangleABC`

Options

A. the orthocentre of `DeltaABC`
B. the centroid of `DeltaABC`
C. the circumcentre of `DeltaABC`
D. the incentre of `DeltaABC`

Answer

To solve the problem, we need to analyze the given condition that the algebraic sum of the perpendicular distances from points A, B, and C to a variable line is zero.

### Step-by-Step Solution:

1. **Define the Variable Line**:
Let the equation of the variable line be given by:
\[ ax + by + c = 0 \]

2. **Calculate the Perpendicular Distances**:
The perpendicular distance \(d\) from a point \((x_0, y_0)\) to the line \(ax + by + c = 0\) is given by:
\[ d = \frac{ax_0 + by_0 + c}{\sqrt{a^2 + b^2}} \]
Therefore, the distances from points A, B, and C to the line are:
- From point A \((x_1, y_1)\):
\[ d_1 = \frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}} \]
- From point B \((x_2, y_2)\):
\[ d_2 = \frac{ax_2 + by_2 + c}{\sqrt{a^2 + b^2}} \]
- From point C \((x_3, y_3)\):
\[ d_3 = \frac{ax_3 + by_3 + c}{\sqrt{a^2 + b^2}} \]

3. **Set Up the Equation**:
According to the problem, the algebraic sum of these distances is zero:
\[ d_1 + d_2 + d_3 = 0 \]
Substituting the distances:
\[ \frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}} + \frac{ax_2 + by_2 + c}{\sqrt{a^2 + b^2}} + \frac{ax_3 + by_3 + c}{\sqrt{a^2 + b^2}} = 0 \]

4. **Combine the Terms**:
Taking a common denominator:
\[ \frac{(ax_1 + by_1 + c) + (ax_2 + by_2 + c) + (ax_3 + by_3 + c)}{\sqrt{a^2 + b^2}} = 0 \]
This implies:
\[ ax_1 + ax_2 + ax_3 + by_1 + by_2 + by_3 + 3c = 0 \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ a(x_1 + x_2 + x_3) + b(y_1 + y_2 + y_3) + 3c = 0 \]

6. **Divide by 3**:
Dividing the entire equation by 3:
\[ \frac{a}{3}(x_1 + x_2 + x_3) + \frac{b}{3}(y_1 + y_2 + y_3) + c = 0 \]

7. **Identify the Centroid**:
The centroid \(G\) of triangle ABC is given by:
\[ G\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]
Thus, substituting the coordinates of the centroid into the equation shows that the line passes through the centroid.

### Conclusion:
The line passes through the centroid of triangle ABC. Therefore, the correct answer is:
**(B) centroid of triangle ABC.**