Question

A, B, C, D,…. are n points in a plane whose coordinates are $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right),.......$ AB is bisected in the point ${{G}_{1}};{{G}_{1}}C$ is divided at ${{G}_{2}}$ in the ratio 1 : 2, ${{G}_{2}}D$ is divided at ${{G}_{3}}$ in the ratio 1 : 3, ${{G}_{3}}E$ at ${{G}_{4}}$ 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: - $\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{n}}}{n}$ and $\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}+....+{{y}_{n}}}{n}$. [This point is called the centre of Mean Position of the n given points].

Answer 1

First consider points A and B and use the mid – point formula given as: - $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$, to calculate the coordinates of ${{G}_{1}}$. Here, $\left( {{x}_{1}},{{y}_{1}} \right)$ are the coordinates of A and $\left( {{x}_{2}},{{y}_{2}} \right)$ are the coordinates of B. Now, consider a line joining the points ${{G}_{1}}$ and C which is divided at ${{G}_{2}}$ in the ratio 1 : 2. Use section formula given as: - $\left( \dfrac{M{{X}_{2}}+N{{X}_{1}}}{M+N},\dfrac{M{{Y}_{2}}+N{{Y}_{1}}}{M+N} \right)$ to calculate the coordinates of ${{G}_{2}}$. Here, M : N is the ratio in which line is divided and $\left( {{X}_{1}},{{Y}_{1}} \right),\left( {{X}_{2}},{{Y}_{2}} \right)$ are the coordinates of ${{G}_{1}},C$ respectively. Use the same section formula to calculate the coordinates of ${{G}_{3}},{{G}_{4}}$ and observe the general pattern to get the coordinates of ${{G}_{n}}$.

Complete step-by-step solution
Here, we have been provided with ‘n’ points: - A, B, C, D ….. n having coordinates $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right),.....$, where $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ are the coordinates of A and B respectively, we get,

$\Rightarrow$ x – coordinate of ${{G}_{1}}=\dfrac{{{x}_{1}}+{{x}_{2}}}{2}$
$\Rightarrow$ y – coordinate of ${{G}_{1}}=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}$
Now, it is given that ${{G}_{1}}C$ is divided at ${{G}_{2}}$ in the ratio 1 : 2.
So, we have,

Using the section formula given as: - $\left( \dfrac{M{{X}_{2}}+N{{X}_{1}}}{M+N},\dfrac{M{{Y}_{2}}+N{{Y}_{1}}}{M+N} \right)$ to calculate the coordinates of ${{G}_{2}}$, where M : N is the ratio in which ${{G}_{1}}C$ is divided and $\left( {{X}_{1}},{{Y}_{1}} \right)$ and $\left( {{X}_{2}},{{Y}_{2}} \right)$ are the coordinates of ${{G}_{1}}$ and C respectively, we get,
$\Rightarrow M:N=1:2$
$\Rightarrow$ Coordinates of C = $\left( {{x}_{3}},{{y}_{3}} \right)=\left( {{X}_{2}},{{Y}_{2}} \right)$
$\Rightarrow$ Coordinates of ${{G}_{1}}=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)=\left( {{X}_{1}},{{Y}_{1}} \right)$
$\Rightarrow$ x – coordinate of ${{G}_{2}}$ = $\dfrac{2\times \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)+1\times {{x}_{3}}}{2+1}$
$\Rightarrow$ x – coordinate of ${{G}_{2}}$ = $\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3}$
$\Rightarrow$ y – coordinate of ${{G}_{2}}$ = $\dfrac{2\times \left( \dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)+1\times {{y}_{3}}}{2+1}$
$\Rightarrow$ y – coordinate of ${{G}_{2}}$ = $\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3}$
Now, we have been given that ${{G}_{2}}D$ is divided at ${{G}_{3}}$ in ratio 1: 3. So, we have,

Using the section formula, we get,
$\Rightarrow M:N=1:3$
$\Rightarrow$ Coordinates of D = $\left( {{x}_{4}},{{y}_{4}} \right)=\left( {{X}_{2}},{{Y}_{2}} \right)$
$\Rightarrow$ Coordinates of ${{G}_{2}}=\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)=\left( {{X}_{1}},{{Y}_{1}} \right)$
$\Rightarrow$ x – coordinate of ${{G}_{3}}=\dfrac{3\times \left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3} \right)+1\times {{x}_{4}}}{3+1}$
$\Rightarrow$ x – coordinate of ${{G}_{3}}=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}}{4}$
Similarly, y – coordinate of ${{G}_{3}}=\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}}{4}$
Therefore, on observing the pattern we can write the coordinates of ${{G}_{4}},{{G}_{5}},....$ easily.
On observing the series in the question we can say that the line joining the points ${{G}_{\left( n-2 \right)}}$ and n will be divided by ${{G}_{n}}$ in the ratio of 1 : (n - 1). Therefore, we have,

Here, M : N = 1 : (n - 1)
$\Rightarrow$ Coordinates of $n=\left( {{x}_{n}},{{y}_{n}} \right)=\left( {{X}_{2}},{{Y}_{2}} \right)$
$\Rightarrow$ Coordinates of ${{G}_{\left( n-2 \right)}}=\left( \dfrac{{{x}_{1}}+{{x}_{2}}+.....+{{x}_{n-1}}}{n-1},\dfrac{{{y}_{1}}+{{y}_{2}}+....+{{y}_{n-1}}}{n-1} \right)=\left( {{X}_{1}},{{Y}_{1}} \right)$
So, applying the section formula, we get,
$\Rightarrow$ x – coordinate of ${{G}_{n}}=\dfrac{\left( n-1 \right)\times \left( \dfrac{{{x}_{1}}+{{x}_{2}}+.....+{{x}_{n-1}}}{n-1} \right)+1\times {{x}_{n}}}{1+\left( n-1 \right)}$
$\Rightarrow$ x – coordinate of ${{G}_{n}}=\dfrac{{{x}_{1}}+{{x}_{2}}+.....+{{x}_{n-1}}+{{x}_{n}}}{1+\left( n-1 \right)}$
Similarly, y – coordinate of ${{G}_{n}}=\dfrac{{{y}_{1}}+{{y}_{2}}+......+{{y}_{n-1}}+{{y}_{n}}}{n}$
Here, ${{G}_{n}}$ will be the final point whose coordinates are: - $\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}.....+{{x}_{n}}}{n}$ and $\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}.....+{{y}_{n}}}{n}$.
Hence, proved

Note: One must note that it is not possible for us to determine the coordinates of each of the points ${{G}_{1}},{{G}_{2}},{{G}_{3}},....$ up to ${{G}_{n}}$ using the formula again and again. That is why we have to identify the general expression for the points. You must remember the question. Remember that the midpoint formula is a special case of section formula where M: N = 1: 1.