By vector method show that the quadrilateral with vertices $A (1, 2, - 1), B (8, - 3, - 4), C (5, - 4, 1), D (- 2, 1, 4)$ is a parallelogram.

Open in App

Solution

As we know that diagonals of parallelogram bisect each other.
In the figure, $O$ is midpoint of diagonals $A C$ and $A D$ then

$(\frac{1 + 5}{2}, \frac{2 - 4}{2}, \frac{- 1 + 1}{2}) = (\frac{8 - 2}{2}, \frac{- 3 + 1}{2}, \frac{- 4 + 4}{2})$ [Finding coordinates of $O$ ]

$(3, - 1, 0) = (3, - 1, 0) .$

Hence, $A B C D$ is a parallelogram.

Suggest Corrections

Similar questions

A quadrilateral has vertices (4, 1), (1, 7), (-6, 0) and (-1, -9). Show that the mid-points of the sides of this quadrilateral form a parallelogram.

A (- 2, 5), B (4, - 1), C (9, 1), D (3, 7)

Show that A(1, 2), B(4, 3), C(6, 6) and D(3, 5) are the vertices parallelogram. Show that ABCD is not a rectangle.

If A(3,-4),B(4,5),C(5,-5),D(4,-10) are the vertices of a quadrilateral prove that ABCD is Rhombus?

Join BYJU'S Learning Program