Question

The three vertices of a parallelogram taken in order are $(-1,0),(3,1),(2,2)$ respectively. Find the coordinates of the fourth vertex.

Answer 1

Compute the coordinates of the sides of the parallelogram.

Let $A(-1,0),B(3,1),C(2,2)\mathrm{and}D(x,y)$ be the vertices of a parallelogram $ABCD$ taken in order.

Since the diagonals of a parallelogram bisect each other.
So, coordinates of the mid point of $AC=$ Coordinates of the mid point of $BD$.

$$\begin{gathered} \Rightarrow \left(\frac{-1+2}{2},\frac{0+2}{2}\right)=\left(\frac{3+x}{2},\frac{y+1}{2}\right) \\ \Rightarrow \left(\frac{1}{2},\frac{2}{2}\right)=\left(\frac{3+x}{2},\frac{y+1}{2}\right) \\ \Rightarrow \frac{3+x}{2}=\frac{1}{2} \\ \therefore x=-2 \end{gathered}$$
Also,

$$\begin{gathered} \frac{y+1}{2}=1 \\ \Rightarrow y+1=2 \\ \therefore y=1 \\ \begin{array}{}\end{array} \end{gathered}$$
Thus, the fourth vertex of the parallelogram is $\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{)}$.