Question

the diagonals of a parallelogram meet in the point(1,-2,3) and the ends of a side are (0,0,0) and (2,4,-3) find the remaining two vertices of the parallelogram.

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Answer 1

Given:

• In parallelogram ABCD with vertices A(0,0,) and B(2,4,-3). The diagonal AC and BD meet in the point O(1,-2,3)

Find:

• Coordinates of Vertices C and D

Solution:

Diagonals of $\sf parallelo^{gram}$ Bisect each other

So, mid-point of AC = mid-point of BD = Point O

Let, point C = $\sf (x_1 + y_1 + z_1)$ and D = $\sf (x_2 + y_2 + z_2)$

$\sf Then \: \: ( \dfrac{0 + x_1}{2}, \dfrac{0 + y_1}{2}, \dfrac{0 + z_1}{2}) = (1 , - 2,3)$

$\sf \implies \dfrac{ x_1}{2} = 1, \dfrac{ y_1}{2} = - 2, \dfrac{z_1}{2} = 3$

$\qquad\qquad$ ☯ Cross-multiplication☯

$\sf \implies x_1=2, y_1= - 4, z_1= 6$

$\sf \implies (x_1, y_1, z_1) =(2, - 4,6)$

Now,

$\sf \: \: ( \dfrac{2 + x_2}{2}, \dfrac{4+ y_2}{2}, \dfrac{ - 3+ z_2}{2}) = (1 , - 2,3)$

$\sf \implies \dfrac{2 + x_2}{2} = 1, \dfrac{4 + y_1}{2} = - 2, \dfrac{ - 3 + z_1}{2} = 3$

$\qquad\qquad$ ☯Cross-multiplication☯

$\sf \implies 2 + x_2=2, 4 + y_2= - 4, - 3 + z_2= 6$

$\sf \implies x_2=2 - 2, y_2= - 4 - 4, z_2= 6 + 3$

$\sf \implies x_2=0, y_2= - 8, z_2= 9$

$\sf \implies (x_2, y_2, z_2) =(0, -8,9)$

Hence, the other two Vertices are (2,-4,6) and (0,-8,9)