Three particles of masses `1 kg, (3)/(2) kg, and 2 kg` are located the vertices of an equilateral triangle of side `a`. The `x, y` coordinates of the centre of mass are.

Options

A. a. `(5a)/(9) , (2a)/(3sqrt 3)`
B. b. `(2a)/(3sqrt (3)) , (5a)/(9)`
C. c. `(5a)/9 , (2a)/(sqrt 3)`
D. d. `(2a)/(sqrt(3)) , (5a)/(9)`

Explanation

To find the coordinates of the center of mass of three particles located at the vertices of an equilateral triangle, we can follow these steps:

### Step 1: Define the coordinates of the vertices
Let's place the three masses at the vertices of an equilateral triangle with side length \( a \). We can assign the following coordinates to the vertices:
- Mass \( m_1 = 1 \, \text{kg} \) at \( (0, 0) \)
- Mass \( m_2 = \frac{3}{2} \, \text{kg} \) at \( \left(\frac{a}{2}, \frac{\sqrt{3}}{2}a\right) \)
- Mass \( m_3 = 2 \, \text{kg} \) at \( (a, 0) \)

### Step 2: Calculate the total mass
The total mass \( M \) of the system is given by:
\[ M = m_1 + m_2 + m_3 = 1 + \frac{3}{2} + 2 = \frac{7}{2} \, \text{kg} \]

### Step 3: Calculate the x-coordinate of the center of mass
The x-coordinate of the center of mass \( x_{cm} \) is calculated using the formula:
\[ x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M} \]
Substituting the values:
\[ x_{cm} = \frac{(1)(0) + \left(\frac{3}{2}\right)\left(\frac{a}{2}\right) + (2)(a)}{\frac{7}{2}} \]
Calculating each term:
\[ x_{cm} = \frac{0 + \frac{3a}{4} + 2a}{\frac{7}{2}} = \frac{\frac{3a}{4} + \frac{8a}{4}}{\frac{7}{2}} = \frac{\frac{11a}{4}}{\frac{7}{2}} = \frac{11a}{4} \cdot \frac{2}{7} = \frac{22a}{28} = \frac{11a}{14} \]

### Step 4: Calculate the y-coordinate of the center of mass
The y-coordinate of the center of mass \( y_{cm} \) is calculated similarly:
\[ y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{M} \]
Substituting the values:
\[ y_{cm} = \frac{(1)(0) + \left(\frac{3}{2}\right)\left(\frac{\sqrt{3}}{2}a\right) + (2)(0)}{\frac{7}{2}} \]
Calculating each term:
\[ y_{cm} = \frac{0 + \frac{3\sqrt{3}}{4}a + 0}{\frac{7}{2}} = \frac{\frac{3\sqrt{3}}{4}a}{\frac{7}{2}} = \frac{3\sqrt{3}a}{4} \cdot \frac{2}{7} = \frac{6\sqrt{3}a}{28} = \frac{3\sqrt{3}a}{14} \]

### Final Result
Thus, the coordinates of the center of mass are:
\[ \left( \frac{11a}{14}, \frac{3\sqrt{3}a}{14} \right) \]