The three vertices of a parallelogram ABCD, taken in order, are A(-1, 0), B(3, 1) and C(2, 2). Find the co-ordinates of the fourth vertex of the parallelogram.

Let coordinates of D be (x, y).

By using mid-point formula,

(x, y) = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

Mid-point of AC :

$\Rightarrow M${AC} = \Big(\dfrac{-1 + 2}{2}, \dfrac{0 + 2}{2}\Big) \\[1em] \Rightarrow M{AC} = \Big(\dfrac{1}{2}, 1\Big).

Midpoint of Diagonal BD :

$\Rightarrow M_{BD} = \Big(\dfrac{3 + x}{2}, \dfrac{1 + y}{2}\Big)$

Since its a parallelogram, the midpoint of diagonal AC must be equal to the midpoint of diagonal BD, as diagonals bisect each other.

Equating the x-coordinates:

$\Rightarrow \dfrac{3 + x}{2} = \dfrac{1}{2}$

⇒ 3 + x = 1

⇒ x = 1 - 3

⇒ x = -2.

Equating the y-coordinates:

$\Rightarrow \dfrac{1 + y}{2} = 1$

⇒ 1 + y = 2

⇒ y = 2 - 1

⇒ y = 1.

Hence, coordinates of D = (-2, 1).