Question

$A_1,A_2,A_3,...A_n$ are $n$ points in a plane whose coordinates are $(x_1,y_1)$, $(x_{2}, y_

{2})$,$(x_{3}, y_{3})$...,$(x_{n}, y_{n})$ respectively.

If $P_1$ is the centroid of the triangle $A_1,A_2,A_3,P_2$ is the centroid of the triangle $A_2,A_3,A_4,P_3$ is that of the triangle $A_3,A_4,A_5$ and so on $P_{n-2}$ is the centroid
of the points $A_{n-2},A_{n-1},A_n$. The coordinates of the centre of mean position of $P_1,P_2,...P_{n-2}$ is

• A. $(\frac{x_1+2x_2+3x_3+...+n{x}_n}{n-2},\frac{y_1+2y_2+3y_3+...+n{y}_n}{n-2})$
• B. $(\frac{x_1+2x_2+3x_3+...+3x_{n-2}+2x_{n-1}+x_n}{(n-2)},\frac{y_1+2y_2+3y_3+...+3_{y}n-2+2y_{n-1}+y_n}{(n-2)})$
• C. $(\frac{3(x_1+x_2+...+x_n)}{n-2},\frac{3(y_1+y_2+...+y_n)}{n-2})$
• D. at the centre of mean position of the points
  $A_1,A_2,...A_n$

Answer 1

Answer: (B)

$(\frac{x_1+2x_2+3x_3+...+3x_{n-2}+2x_{n-1}+x_n}{(n-2)},\frac{y_1+2y_2+3y_3+...+3_{y}n-2+2y_{n-1}+y_n}{(n-2)})$

$P_1(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$
$P_2(\frac{x_2+x_3+x_4}{3},\frac{y_2+y_3+y_4}{3})$
$P_3(\frac{x_3+x_4+x_5}{3},\frac{y_3+y_4+y_5}{3})$...
$P_{n-2}(\frac{x_{n-2}+x_{n-1}+x_n}{3},\frac{y_{n-2}+y_{n-1}+y_n}{3})$
So, the coordinates of the centre of mean position of $P_1,P_2...P_{n-2}$ is $(\overline{x},\overline{y})$
where
$\overline{x}=\frac{x_1+2x_2+3x_3+...+3x_{n-2}+2x_{n-1}+x_n}{n-2}$
$\overline{y}=\frac{y_1+2y_2+3y_3+...+3y_{n-2}+2y_{n-1}+y_n}{n-2}$