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

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#### 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.

∴ Coordiantes of the mid-point of AC = Coordinates of the mid-point of BD

`\Rightarrow ( \frac{-1+2}{2},\ \frac{0+2}{2})=(\frac{3+x}{2},\ \frac{1+y}{2})`

`\Rightarrow ( \frac{1}{2},\ 1)=( \frac{3+x}{2},\frac{y+1}{2})`

`\Rightarrow \frac{3+x}{2}=\frac{1}{2}\text{ and }\frac{y+1}{2}=1`

⇒ x = – 2 and y = 1

Hence, the fourth vertex of the parallelogram is (–2, 1).

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