Question

ABCD is a parallelogram with vertices $A(x_1,y_1),B\left(x_2,y_2\right),C(x_3,y_3)$ . Find the coordinates of the fourth vertex D in terms of $x_1,x_2,x_3,y_1,y_2\mathrm{and}{y}_3$.

Answer 1

Suppose the coordinates of D be (x, y).
Since diagonals of a parallelogram bisect each other.


Therefore the midpoint of AC is the midpoint of BD, i.e $\left(\frac{x_1+x_3}{2},\frac{{\displaystyle y_1+y_3}}{{\displaystyle 2}}\right)and\left(\frac{{\displaystyle x_2+x}}{{\displaystyle 2}},\frac{{\displaystyle y_2+y}}{{\displaystyle 2}}\right)$ respectively.
$$\begin{gathered} \Rightarrow x_1+x_3=x_2+xand{y}_1+y_3=y_2+y \\ \Rightarrow x=x_1+x_3-x_{2}andy=y_1+y_3-y_2 \\ ThuscoordinatesofDare\left(x_1+x_3-x_2,y_1+y_3-y_2\right) \end{gathered}$$