Three particles of masses `m`,`m` and `4kg` are kept at a verticals of triangle `ABC`. Coordinates of `A`, `B` and `C` are `(1,2)` , `(3,2)` and `(-2,-2)` respectively such that the centre of mass lies at origin. Find the value of mass `m`.
Hint. `x_(cm)=(sum_(i-1)^(3)m_(i)x_(i))/(sum_(i=1)^(3)m_(i))`, `y_(cm)=(sum_(i=1)^(3)m_(i)y_(i))/(sum_(i=1)^(3)m_(i))`

Explanation

To solve the problem, we need to find the value of mass \( m \) such that the center of mass of the three particles lies at the origin (0,0). The coordinates of the particles are given as follows:

- Particle 1 (mass \( m \)) at point A (1, 2)
- Particle 2 (mass \( m \)) at point B (3, 2)
- Particle 3 (mass 4 kg) at point C (-2, -2)

### Step-by-Step Solution

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