### Video Transcript

A square π΄π΅πΆπ· has side length πΏ. Three masses of 610 grams are placed at π΄, π΅, and π·. Find the coordinates of the center of mass of the system.

In our diagram, we see that masses of 610 grams are placed at three of the vertices
of our square. We are told that the square has side length πΏ. As point π΄ lies at the origin, it has coordinates zero, zero. Point π΅ has coordinates πΏ, zero; point πΆ has coordinates πΏ, πΏ; and point π· has
coordinates zero, πΏ.

We are asked to find the coordinates of the center of mass of the system. This can be written as the ordered pair πΆπ sub π₯ and πΆπ sub π¦, where the center
of mass in the π₯-direction is equal to the sum of the products of the masses
multiplied by the corresponding π₯-coordinates divided by the sum of the masses. The center of mass in the π¦-direction can be calculated in the same way, but this
time using the π¦-coordinates of each mass.

Starting with the π₯-coordinates of points π΄, π΅, and π·, we see that the center of
mass in the π₯-direction is equal to 610 multiplied by zero plus 610 multiplied by
πΏ plus 610 multiplied by zero all divided by 610 plus 610 plus 610. This simplifies to 610 multiplied by πΏ divided by 610 multiplied by three. Dividing the numerator and denominator by 610, this in turn simplifies to πΏ over
three.

We can then repeat this process for the π¦-coordinates at points π΄, π΅, and π·. Once again, our equation simplifies to 610 multiplied by πΏ divided by 610 multiplied
by three. Dividing the numerator and denominator by 610 once again, we see that the center of
mass in the π¦-direction is also equal to πΏ over three.

We can therefore conclude that the coordinates of the center of mass of the system
are πΏ over three, πΏ over three.