Question

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

Answer 1

Let the fourth vertex be $$D(a, b)$$. The diagonals $$AC$$ and $$BD$$ bisect each other, so their mid-points are equal.

Mid-point of $$AC$$: $$\left(\frac{-1 + 2}{2}, \frac{0 + 2}{2}\right) = \left(\frac{1}{2}, 1\right)$$

Mid-point of $$BD$$: $$\left(\frac{3 + a}{2}, \frac{1 + b}{2}\right)$$

Hence $$\frac{3 + a}{2} = \frac{1}{2}$$

$$\Rightarrow 3 + a = 1$$

$$\Rightarrow a = -2$$

$$\frac{1 + b}{2} = 1$$

$$\Rightarrow 1 + b = 2$$

$$\Rightarrow b = 1$$

Hence the fourth vertex is $$D(-2, 1)$$.