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The three vertices of a parallelogram ABCD, taken in order, are A(-1, 0), B(3, 1) and C(2, 2). Find the coordinates of the fourth vertex D.
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Sol. Let the coordinates of fourth vertex D be $(x, y)$
Coordinates of the mid-point of AC = Coordinates of the mid-point of BD
$(\frac{-1+2}{2}, \frac{0+2}{2}) = (\frac{3+x}{2}, \frac{1+y}{2})$
$\therefore \frac{-1+2}{2} = \frac{3+x}{2} \Rightarrow x=-2$ (1/2 Mark)
and $\frac{0+2}{2} = \frac{1+y}{2} \Rightarrow y=1$ (1/2 Mark)
Coordinates of the mid-point of AC = Coordinates of the mid-point of BD
$(\frac{-1+2}{2}, \frac{0+2}{2}) = (\frac{3+x}{2}, \frac{1+y}{2})$
$\therefore \frac{-1+2}{2} = \frac{3+x}{2} \Rightarrow x=-2$ (1/2 Mark)
and $\frac{0+2}{2} = \frac{1+y}{2} \Rightarrow y=1$ (1/2 Mark)