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I've been given a parallelogram with two of its vertices, $A$ and $B$, being equal to (-5, -8, 3) and (4, 7, -5) respectively, and a centroid $S$ at (-10, 4, 6).

How do I go around finding remaining coordinates of points $D$ and $C$?

  • 0
    Use http://gradestack.com/CBSE-Class-9th-Complete/Quadrilterals/Theorem-4-In-a/14902-2953-4006-study-wtw and http://gradestack.com/CBSE-Class-9th-Complete/Quadrilterals/Theorem-3-In-a/14902-2953-4004-study-wtw – 2017-01-22

2 Answers 2

1

Let $C = (x_1, y_1, z_1)$

Then mid point of AC is $(-10, 4, 6)$

$\left(\frac{-5 + x_1}{2}, \frac{-8 + y_1}{2}, \frac{3 + z_1}{2}\right) = (-10, 4, 6)$

$\frac{-5 + x_1}{2} = -10, \frac{-8 + y_1}{2} = 4, \frac{3 + z_1}{2} = 6$

$x_1 = -15, y_1 = 16, z_1 = 9$

Compare and find values. Similarly for BD.

0

The centroid is the midpoint between each pair of oposite vertices of the parallelogram. So if one vertex is at $(2,3,0)$ and the opposite vertex is at $(0,1,2)$, the centroid would be halfway between at $(1,2,1)$. If instead you are given $(0,1,2)$ for a vertex and $(1,2,1)$ for the centroid, the opposite vertex must be at $(2,3,0)$ to get the midpoint in the right place.

Now what do you get with your given coordinates?