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