Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2). Find the coordinates of the fourth vertex.

Asked by Pragya Singh | 11 months ago |  126

##### Solution :-

We are given the three vertices of a parallelogram ABCD are A(3,-1,2) , B(1,2,-4) and C(-1,1,2) .

Let the coordinates of the fourth vertex of the parallelogram ABCD be D(x,y,z) .

According to the property of parallelogram, the diagonals of the parallelogram bisect each other. In this parallelogram ABCD, AC and BD at point O.
So, Mid-point of AC = Mid-point of BD

$$( \dfrac{3-1}{2}, \dfrac{-1+1}{2}, \dfrac{2+2}{2})$$

$$( \dfrac{x+1}{2}, \dfrac{y+1}{2}, \dfrac{z-4}{2})$$

(1,0,2) = $$( \dfrac{x+1}{2}, \dfrac{y+1}{2}, \dfrac{z-4}{2})$$

$$\dfrac{x+1}{2}=1, \dfrac{y+2}{2}=0, \dfrac{z-4}{2}=2$$

We get, x = 1, y = 2 and z = 8

Therefore, the coordinates of the fourth vertex of the parallelogram ABCD are D(1,-2,8) .

Answered by Abhisek | 11 months ago

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