Question

Prove that the point A(1, 3, 0), B(–5, 5, 2), C(–9, –1, 2) and D(–3, –3, 0) taken in order are the vertices of a parallelogram. Also, show that ABCD is not a rectangle.

Answer 1

Let A(1,3,0),B( \[-\]5,5,2), C(\[-\]9,\[-\]1,2) and D(\[-\]3,\[-\]3,0) be the coordinates of quadrilateral \[\square ABCD\]

\[AB = \sqrt{\left( - 5 - 1 \right)^2 + \left( 5 - 3 \right)^2 + \left( 2 - 0 \right)^2}\]
\[ = \sqrt{36 + 4 + 4}\]
\[ = \sqrt{44}\]
\[BC = \sqrt{\left( - 9 + 5 \right)^2 + \left( - 1 - 5 \right)^2 + \left( 2 - 2 \right)^2}\]
\[ = \sqrt{16 + 36 + 0}\]
\[ = \sqrt{52} \]
\[CD = \sqrt{\left( - 3 + 9 \right)^2 + \left( - 3 + 1 \right)^2 + \left( 0 - 2 \right)^2}\]
\[ = \sqrt{36 + 4 + 4}\]
\[ = \sqrt{44}\]
\[DA = \sqrt{\left( 1 + 3 \right)^2 + \left( 3 + 3 \right)^2 + \left( 0 - 0 \right)^2}\]
\[ = \sqrt{16 + 36 + 0}\]
\[ = \sqrt{52}\]
Here, we see that AB = CD& BC = DA

 Since, opposite pair of sides are equal .
Therefore, 

\[\square ABCD\] is a parallelogram .

\[AC = \sqrt{\left( - 9 - 1 \right)^2 + \left( - 1 - 3 \right)^2 + \left( 2 - 0 \right)^2}\]
\[ = \sqrt{\left( - 10 \right)^2 + \left( - 4 \right)^2 + \left( 2 \right)^2}\]
\[ = \sqrt{100 + 16 + 4}\]
\[ = \sqrt{120}\],m

\[BD = \sqrt{\left( - 3 + 5 \right)^2 + \left( - 3 - 5 \right)^2 + \left( 0 - 2 \right)^2}\]
\[ = \sqrt{\left( 2 \right)^2 + \left( - 8 \right)^2 + \left( - 2 \right)^2}\]
\[ = \sqrt{4 + 64 + 4}\]
\[ = \sqrt{72}\]

\[\therefore AC \neq BD\]
∴ ABCD is not a rectangle.