Chapter 10.3, Problem 3E

To determine

To explain: Centre of gravity of a parallelogram.

Explanation of Solution

Given information

The figure is a parallelogram

Formula used:

Centre of gravity of figures is known as centroid:

  $\begin{array}{l}x=\frac{x_1+x_2+x_3+x_4}{4}\\ y=\frac{y_1+y_2+y_3+y_4}{4}\end{array}$

Proof:

Let PQRSbe a parallelogram with the vertex P, Q, R and S

Let $(x_1,y_1)$ be the coordinates of the point P in a coordinate plane $\text{P = P(x}\text{,}{\text{y}}_{\text{1}}\text{)}$

Let $(x_2,y_2)$ be the coordinates of the point Q, $\text{Q= Q(x}\text{,}{\text{y}}_{\text{2}}\text{)}$

Let $(x_3,y_3)$ be the coordinates of the point R, $\text{R=R(x}\text{,}{\text{y}}_{\text{3}}\text{)}$ and

Let $(x_4,y_4)$ be the coordinates of the point S, $\text{S= S(x}\text{,}{\text{y}}_{\text{4}}\text{)}$

The vector PQ has the projections ${\text{ x}}_{\text{2}}{\text{-x}}_{\text{1}}{\text{, y}}_{\text{2}}{\text{-y}}_{\text{1}}$ in axes x- and y- respectively,

The vector QR has the projections ${\text{ x}}_{\text{3}}{\text{-x}}_{\text{2}}{\text{, y}}_{\text{3}}{\text{-y}}_{\text{2}}$ ,

The vector PS has the projections ${\text{ x}}_{\text{4}}{\text{-x}}_{\text{1}}{\text{, y}}_{\text{4}}{\text{-y}}_{\text{1}}$ , and

The vector SR has the projections ${\text{ x}}_{\text{3}}{\text{-x}}_{\text{4}}{\text{, y}}_{\text{3}}{\text{-y}}_{\text{4}}$

The diagonal PR of this parallelogram represents the sum of vectors:

  $\text{PR = PQ + PS = PQ + QR}$

Let C be the intersection point of the diagonals PR and QS of the parallelogram

PQRS and the diagonals of a parallelogram bisect eachother at the intersection 

The vector PC can be expressed by the two ways:

  $\begin{array}{l}\text{PC = (PQ + PS), }\mathrm{.....}\text{(1)}\\ \text{PC = (PQ + QR)}\text{. }\mathrm{...}\text{(2)}\end{array}$

From the expression (1)

  $\frac{\text{1}}{\text{2}}{\text{((x}}_{\text{2}}\text{-x}\text{)}{\text{ + (x}}_{\text{4}}{\text{-x}}_{\text{1}}\text{))}$

  $\frac{\text{1}}{\text{2}}{\text{((y}}_{\text{2}}\text{-y}\text{)}{\text{ + (y}}_{\text{4}}{\text{-y}}_{\text{1}}\text{))}$

Hence, the x-coordinate of the point C is

  $\begin{array}{l}\text{ C(x)}={\text{x}}_{\text{1}}+\frac{\text{1}}{\text{2}}{\text{((x}}_{\text{2}}\text{-x}\text{)}{\text{ + (x}}_{\text{4}}{\text{-x}}_{\text{1}}\text{))}\\ \frac{\text{x}+x_4}{2}\end{array}$

Similarly, the y-coordinate of the point C is:

  $\begin{array}{l}\text{ C(y)}={\text{y}}_{\text{1}}+\frac{\text{1}}{\text{2}}{\text{((y}}_{\text{2}}\text{-y}\text{)}{\text{ + (y}}_{\text{4}}{\text{-y}}_{\text{1}}\text{))}\\ \frac{\text{y}+y_4}{2}\end{array}$

From the expression (2),

The vector PC has the x- and y- components

  $\frac{\text{1}}{\text{2}}{\text{((x}}_{\text{2}}\text{-x}\text{)}{\text{ + (x}}_{\text{3}}{\text{-x}}_{\text{2}}\text{))}$

  $\frac{\text{1}}{\text{2}}{\text{((y}}_{\text{2}}\text{-y}\text{)}{\text{ + (y}}_{\text{3}}{\text{-y}}_{\text{2}}\text{))}$

Hence, the x-coordinate of the point C is

  $\begin{array}{l}\text{ C(x)}={\text{x}}_{\text{1}}+\frac{\text{1}}{\text{2}}{\text{((x}}_{\text{2}}\text{-x}\text{)}{\text{ + (x}}_{\text{3}}{\text{-x}}_{\text{2}}\text{))}\\ \frac{\text{x}+x_3}{2}\end{array}$

Similarly, the y-coordinate of the point C

  $\begin{array}{l}\text{ C(y)}={\text{y}}_{\text{1}}+\frac{\text{1}}{\text{2}}{\text{((y}}_{\text{2}}\text{-y}\text{)}{\text{ + (y}}_{\text{3}}{\text{-y}}_{\text{2}}\text{))}\\ \frac{\text{y}+y_3}{2}\end{array}$

Since

  $C(x)=\frac{1}{2}(\frac{x_2+x_4}{2}+\frac{x_1+x_3}{2})=\frac{x_1+x_2+x_3+x_4}{4}$

  $C(y)=\frac{1}{2}(\frac{y_2+y_4}{2}+\frac{y_1+y_3}{2})=\frac{y_1+y_2+y_3+y_4}{4}$

This is exactly what has to be proved.