Find the centre of mass of three particles at the vertices of an equilateral triangle. The masses of the particles are $100 \; g$,$150 \; g$,and $200 \; g$ respectively. Each side of the equilateral triangle is $0.5 \; m$ long.

(N/A) Let the vertices of the equilateral triangle be $O(0,0)$,$A(0.5,0)$,and $B(0.25, 0.25\sqrt{3})$.
The masses are $m_1 = 100 \; g$ at $O$,$m_2 = 150 \; g$ at $A$,and $m_3 = 200 \; g$ at $B$.
The coordinates of the centre of mass $(X, Y)$ are given by:
$X = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} = \frac{100(0) + 150(0.5) + 200(0.25)}{100 + 150 + 200} = \frac{75 + 50}{450} = \frac{125}{450} = \frac{5}{18} \; m$
$Y = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} = \frac{100(0) + 150(0) + 200(0.25\sqrt{3})}{450} = \frac{50\sqrt{3}}{450} = \frac{\sqrt{3}}{9} = \frac{1}{3\sqrt{3}} \; m$
Thus,the centre of mass is at $(\frac{5}{18}, \frac{1}{3\sqrt{3}}) \; m$.