If the points A(-2, -1), B(1, 0), C(p, 3) and D(1, q) form a parallelogram ABCD, find the values of p and q.

A(-2, -1), B(1, 0), C(p, 3) and D(1, q) are the vertices of a parallelogram ABCD as shown in the figure below:

∴ Diagonal AC and BD bisect each other at O.
O is the mid-point of AC as well as BD. Let the coordinates of O be (x, y).

When O is the mid-point of AC, then by mid-point formula

$\Rightarrow x = \dfrac{p + (-2)}{2} \text{ and y = } \dfrac{3 + (-1)}{2} \\[1em] \Rightarrow x = \dfrac{p - 2}{2} \text{ and y = } \dfrac{2}{2} \\[1em] \therefore x = \dfrac{p - 2}{2} \text{ and } y = 1.$

When O is the mid-point of BD, then by mid-point formula

$\Rightarrow x = \dfrac{1 + 1}{2} \text{ and y = } \dfrac{0 + q}{2} \\[1em] \Rightarrow x = \dfrac{2}{2} \text{ and y = } \dfrac{q}{2} \\[1em] \therefore x = 1 \text{ and } y = \dfrac{q}{2}.$

Now comparing, we get the values of the coordinates of point O in both cases,

$\Rightarrow \dfrac{p - 2}{2} = 1 \text{ and } \dfrac{q}{2} = 1 \\[1em] \Rightarrow p - 2 = 2 \text{ and } q = 2 \\[1em] \Rightarrow p = 4 \text{ and } q = 2.$

Hence, the value of p = 4 and q = 2.