2. A parallelogram has 3 of its vertices at $(1,2),(3,8)$, and $(4,1)$. Compute the sum of the possible $x$-coordinates for the 4 th vertex.
Solution:
8
There are 3 possible locations for the 4 th vertex. Let $(a, b)$ be its coordinates. If it is opposite to vertex $(1,2)$, then since the midpoints of the diagonals of a parallelogram coincide, we get $\left(\frac{a+1}{2}, \frac{b+2}{2}\right)=\left(\frac{3+4}{2}, \frac{8+1}{2}\right)$. Thus $(a, b)=(6,7)$. By similar reasoning for the other possible choices of opposite vertex, the other possible positions for the fourth vertex are $(0,9)$ and $(2,-5)$, and all of these choices do give parallelograms. So the answer is $6+0+2=8$.