The position vector of the center of mass r → cm of an asymmetric uniform bar of negligible area of…

The position vector of center of mass ${\overrightarrow{\mathrm{r}}}_{\mathrm{cm}}$ is given as 

${\overrightarrow{\mathrm{r}}}_{\mathrm{cm}}={\mathrm{X}}_{\mathrm{cm}}\hat{\mathrm{x}}+{\mathrm{Y}}_{\mathrm{cm}} \hat{\mathrm{y}}$

Where, ${\mathrm{X}}_{\mathrm{cm}}=$$\mathrm{x}-$coordinate of center of mass  and ${\mathrm{Y}}_{\mathrm{cm}}=\mathrm{y}-$coordinate of center of mass

${\overrightarrow{\mathrm{r}}}_{\mathrm{cm}}=\left(\frac{{\mathrm{m}}_1{\mathrm{x}}_1+{\mathrm{m}}_2{\mathrm{x}}_2+{\mathrm{m}}_3{\mathrm{x}}_3}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3}\right) \hat{\mathrm{x}} +\left(\frac{{\mathrm{m}}_1{\mathrm{y}}_1+{\mathrm{m}}_2{\mathrm{y}}_2+{\mathrm{m}}_3{\mathrm{y}}_3}{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3}\right) \hat{\mathrm{y}}$

${\overrightarrow{\mathrm{r}}}_{\mathrm{cm}}=\left[\frac{\left(2\mathrm{m}\right)\times \mathrm{L}+\left(\mathrm{m}\right)\times 2\mathrm{L}+\mathrm{m}\left(\frac{5\mathrm{L}}{2}\right)}{2\mathrm{m}+\mathrm{m}+\mathrm{m}}\right]\hat{\mathrm{x}}+\left[\frac{\left(2\mathrm{m}\right)\times \mathrm{L}+\left(\mathrm{m}\right)\times {\displaystyle \frac{\mathrm{L}}{2}}+\mathrm{m}\times \left(0\right)}{2\mathrm{m}+\mathrm{m}+\mathrm{m}}\right]\hat{\mathrm{y}}$

${\overrightarrow{\mathrm{r}}}_{\mathrm{cm}}=\frac{13\mathrm{L}}{8}\hat{\mathrm{x}}+\frac{5\mathrm{L}}{8}\hat{\mathrm{y}}$