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Domain

A, B, C, D......

Question

A, B, C, D... are n points in a plane whose coordinates are $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}), . . . . . . A B$ is bisected in the point $G_{1}; G_{2} C$ is divided at $G_{3}$ in the ratio $1 : 2; G_{3} D$ is divided at $G_{4}$ in the ratio $1 : 3; G_{4} E$ at $G_{5}$ in the ratio 1 : 4, and so on until all the points are exhausted. Show that the coordinates of the final point so obtained are
$\frac{x_{1} + x_{2} + x_{3} + . . . . . + x_{n}}{n}$ and $\frac{y_{1} + y_{2} + y_{3} + . . . + y_{n}}{n}$
[This point is called the Centre of Mean Position of the n given points.]

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Solution

$G_{1} (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) G_{2} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{(\frac{x_{1} + x_{2}}{2}) 2 + 1 (x_{3})}{2 + 1}, \frac{(\frac{y_{1} + y_{2}}{2}) 2 + 1 (y_{3})}{2 + 1} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ \Rightarrow G_{2} (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}) G_{3} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{(\frac{x_{1} + x_{2} + x_{3}}{3}) 3 + 1 (x_{4})}{3 + 1}, \frac{(\frac{y_{1} + y_{2} + y_{3}}{3}) 3 + 1 (y_{4})}{3 + 1} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ \Rightarrow G_{3} (\frac{x_{1} + x_{2} + x_{3} + x_{4}}{4}, \frac{y_{1} + y_{2} + y_{3} + y_{4}}{4}) G_{n} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{(\frac{x_{1} + x_{2} + x_{3} + x_{4} . . . . . . . . . . x_{n}}{n - 1}) (n - 1) + 1 (x_{n})}{n - 1 + 1}, \frac{(\frac{y_{1} + y_{2} + y_{3} + y_{4} + . . . . . . . . . . . . y_{n}}{n - 1}) (n - 1) + 1 (y_{n})}{n - 1 + 1} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ \Rightarrow G_{n} (\frac{x_{1} + x_{2} + x_{3} + x_{4} . . . . . . . . . . x_{n}}{n}, \frac{y_{1} + y_{2} + y_{3} + y_{4} + . . . . . . . . . . . . y_{n}}{n})$

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Q 1

If x₁, x₂, x₃ as well as y₁, y₂, y₃ are in G.P. with the same common ratio, then the points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃)

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