If $A(6,1)$, $B(8,2)$, $C(9,4)$ and $D(p,3)$ are the vertices of a parallelogram taken in order, what is the value of $p$?

Principle: diagonals of a parallelogram bisect each other. 1. Midpoint of $AC$ is $\left(\dfrac{6+9}{2},\dfrac{1+4}{2}\right)=\left(\dfrac{15}{2},\dfrac{5}{2}\right)$. 2. Midpoint of $BD$ is $\left(\dfrac{8+p}{2},\dfrac{2+3}{2}\right)=\left(\dfrac{8+p}{2},\dfrac{5}{2}\right)$. 3. Since these midpoints are equal, $\dfrac{15}{2}=\dfrac{8+p}{2}$, so $15=8+p$ and hence $p=7$. Option D is tempting if one adds the last coordinates incorrectly, while A and B do not make the diagonal midpoints match.