ABCD is a parallelogram. If the coordinates of A, B and D are (10, -6), (2, -6) and (4, -2) respectively, find the coordinates of C.

Let the coordinates of C be (x, y) and other three vertices of the given parallelogram are A(10, -6), B(2, -6) and D(4, -2).

Since, ABCD is a parallelogram, its diagonals bisect each other.

Let AC and BD intersect each other at O.

So, O is the mid-point of BD, so coordinates of O are,

$\Rightarrow \Big(\dfrac{2 + 4}{2}, \dfrac{-6 + (-2)}{2}\Big) \\[1em] = \Big(\dfrac{6}{2}, \dfrac{-8}{2}\Big) \\[1em] = (3, -4).$

The mid-point of AC is

$\Rightarrow \Big(\dfrac{10 + x}{2}, \dfrac{-6 + y}{2}\Big)$

Since O is the mid-point of AC so comparing,

$\Rightarrow \dfrac{10 + x}{2} = 3 \text{ and } \dfrac{-6 + y}{2} = -4 \\[1em] \Rightarrow 10 + x = 6 \text{ and } -6 + y = -8 \\[1em] \Rightarrow x = 6 - 10 \text{ and } y = -8 + 6 \\[1em] \Rightarrow x = -4 \text{ and } y = -2.$

Hence, the coordinates of P are (-4, -2).