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Q.
The three vertices of a parallelogram taken in order are (-1, 0), (3, 1) and (2, 2) respectively. Find the coordinate of the fourth vertex.
Introduction to Three Dimensional Geometry
Solution:
Let A(-1, 0) , B(3, 1) , C(2, 2) and D(x, y) be the vertices of a parallelogram ABCD taken in order. Since, the diagonals of a parallelogram bisect each other.
$\therefore $ Coordinates of the mid point of AC
= Coordinates of the mid-point of BD
$\Rightarrow \left(\frac{1-+2}{2}, \frac{0+2}{2}\right) = \left(\frac{3+x}{2}, \frac{1+y}{2}\right)$
$ \Rightarrow \left(\frac{1}{2} , 1 \right) = \left(\frac{3+x}{2} , \frac{y+1}{2}\right)$
$ \Rightarrow \frac{3+x}{2} = \frac{1}{2}$ and $ \frac{y+1}{2} = 1 $
$ \Rightarrow $ x = -2 and y = 1.
Hence the fourth vertex of the parallelogram is (-2, 1)