ABCD is a parallelogram whose vertices A and B have coordinates (2, -3) and (-1, -1) respectively. If the diagonals of the parallelogram meet at the point M(1, -4), find the coordinates of C and D. Hence, find the perimeter of the parallelogram.

Coordinates of A are (2, -3) and B (-1, -1).

Since, M is the point where diagonals meet, hence it is midpoint of AC and BD.

Let coordinates of C and D be (x₁, y₁) and (x₂, y₂).

When M(1, -4) is the midpoint of AC then by midpoint formula,

$\Rightarrow 1 = \dfrac{2 + x$1}{2} \text{ and } -4 = \dfrac{-3 + y1}{2} \\[1em] \Rightarrow 2 + x1 = 2 \text{ and } -3 + y1 = -8 \\[1em] \Rightarrow x1 = 2 - 2 \text{ and } y1 = -8 + 3 \\[1em] \Rightarrow x1 = 0 \text{ and } y1 = -5.

∴ Coordinates of C are (0, -5)

When M(1, -4) is the midpoint of BD then by midpoint formula,

$\Rightarrow 1 = \dfrac{-1 + x$2}{2} \text{ and } -4 = \dfrac{-1 + y2}{2} \\[1em] \Rightarrow -1 + x2 = 2 \text{ and } -1 + y2 = -8 \\[1em] \Rightarrow x2 = 2 + 1 \text{ and } y2 = -8 + 1 \\[1em] \Rightarrow x2 = 3 \text{ and } y2 = -7.

∴ Coordinates of D are (3, -7).

By distance formula, the length of AB is,

$= \sqrt{[2 - (-1)]^2 + [(-3) - (-1)]^2} \\[1em] = \sqrt{(2 + 1)^2 + (-3 + 1)^2} \\[1em] = \sqrt{3^2 + (-2)^2} \\[1em] = \sqrt{9 + 4} \\[1em] = \sqrt{13}.$

By distance formula, the length of BC is,

$= \sqrt{[0 - (-1)]^2 + [(-5) - (-1)]^2} \\[1em] = \sqrt{(1)^2 + (-5 + 1)^2} \\[1em] = \sqrt{1 + (-4)^2} \\[1em] = \sqrt{1 + 16} \\[1em] = \sqrt{17}.$

Perimeter of parallelogram ABCD = 2(AB + BC) = $2(\sqrt{13} + \sqrt{17}).$

Hence, the coordinates of C and D are (0, -5) and (3, -7) respectively. The perimeter of parallelogram ABCD is $2(\sqrt{13} + \sqrt{17})$ units.