Two point masses $m$ and $M$ are separated by a distance $L$. The distance of the centre of mass of the system from $m$ is

Four particles of masses $m_1 = 2m$,$m_2 = 4m$,$m_3 = m$,and $m_4$ are placed at the four corners of a square. What should be the value of $m_4$ so that the centre of mass of all the four particles is exactly at the centre of the square?

The centre of mass of two particles lies:

$A$ rod of length $3 \; m$ has a linear mass density that is directly proportional to the distance $x$ from one of its ends. Find the position of its centre of gravity from that end. (in $; m$)

Four particles of masses $m, 2m, 3m$ and $4m$ are arranged at the corners of a parallelogram with each side equal to $a$ and one of the angles between two adjacent sides is $60^o$. The parallelogram lies in the $x-y$ plane with mass $m$ at the origin and $4m$ on the $x$-axis. The centre of mass of the arrangement will be located at

Three equal masses are placed at $(0,0)$,$(a,0)$,and $\left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right)$. Find the coordinates of the center of mass.