$A$ straight line $L$ is such that the algebraic sum of perpendicular distances of vertices of a triangle

from $L$is zero . Then $L$passes through

$L:ax+by+c=0$ be the line and $A\left(x_1,y_1\right)$,

$B\left(x_2,y_2\right),C\left(x_3,y_3\right)$ be the vertices of the triangle. The

sum of algebraic distances $A,B,C$ from $L$ is zero

$\begin{array}{r}\therefore \sum _{i=1}^3 \frac{a{x}_i+b{y}_i+c}{\sqrt{a^2+b^2}}=0\\ \Rightarrow a\sum _{i=1}^3 x_i+b\sum _{i=1}^3 y_i+3c=0\end{array}$

or $a\left(\frac{\mathrm{\Sigma }{x}_i}{3}\right)+b\left(\frac{\mathrm{\Sigma }{y}_i}{3}\right)+c=0$

$\therefore$ The point $\left(\frac{\mathrm{\Sigma }{x}_i}{3},\frac{\mathrm{\Sigma }{y}_i}{3}\right)$ lies on the line $ax+by+c=0$

the line passes through the centroid.