Answer
$$\left( {\overline x ,\overline y } \right) = \left( {2,\frac{{48}}{{25}}} \right)$$
Work Step by Step
\[\begin{gathered}
{\text{}} \hfill \\
\boxed{\begin{array}{*{20}{c}}
{{m_i}}&{12}&6&{4.5}&{15} \\
{\left( {{x_i},{y_i}} \right)}&{\left( {2,3} \right)}&{\left( { - 1,5} \right)}&{\left( {6,8} \right)}&{\left( {2, - 2} \right)}
\end{array}} \hfill \\
\end{gathered} \]
$$\eqalign{
& m{\text{ is the total mass of the system}} \cr
& m = {m_1} + {m_2} + {m_3} + {m_4} \cr
& m = 12 + 6 + 4.5 + 15 \cr
& m = 37.5 \cr
& \cr
& {\text{*The moment about the }}y{\text{ - axis is }} \cr
& {M_y} = {m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3} + {m_4}{x_4} \cr
& {M_y} = \left( {12} \right)\left( 2 \right) + \left( 6 \right)\left( { - 1} \right) + \left( {4.5} \right)\left( 6 \right) + \left( {15} \right)\left( 2 \right) \cr
& {M_y} = 75 \cr
& \cr
& {\text{*The moment about the }}x{\text{ - axis is }} \cr
& {M_x} = {m_1}{y_1} + {m_2}{y_2} + {m_3}{y_3} \cr
& {M_x} = \left( {12} \right)\left( 3 \right) + \left( 6 \right)\left( 5 \right) + \left( {4.5} \right)\left( 8 \right) + \left( {15} \right)\left( { - 2} \right) \cr
& {M_x} = 72 \cr
& \cr
& {\text{The center of mass }}\left( {\overline x ,\overline y } \right)({\text{or center of gravity}}){\text{ is}} \cr
& \overline x = \frac{{{M_y}}}{m} = \frac{{75}}{{37.5}} = 2 \cr
& \overline y = \frac{{{M_x}}}{m} = \frac{{72}}{{37.5}} = \frac{{48}}{{25}} \cr
& \left( {\overline x ,\overline y } \right) = \left( {2,\frac{{48}}{{25}}} \right) \cr} $$
