# Show that the Points A(1, 2, 3), B(–1, –2, –1), C(2, 3, 2) and D(4, 7, 6) Are the Vertices of a Parallelogram Abcd, but Not a Rectangle. - Mathematics

Show that the points A(1, 2, 3), B(–1, –2, –1), C(2, 3, 2) and D(4, 7, 6) are the vertices of a parallelogram ABCD, but not a rectangle.

#### Solution

To show that ABCD is a parallelogram, we need to show that its two opposite sides are equal.

$AB = \sqrt{\left( - 1 - 1 \right)^2 + \left( - 2 - 2 \right)^2 + \left( - 1 - 3 \right)^2}$
$= \sqrt{4 + 16 + 16}$
$= \sqrt{36}$
$= 6$
$BC = \sqrt{\left( 2 + 1 \right)^2 + \left( 3 + 2 \right)^2 + \left( 2 + 1 \right)^2}$
$= \sqrt{9 + 25 + 9}$
$= \sqrt{43}$
$CD = \sqrt{\left( 4 - 2 \right)^2 + \left( 7 - 3 \right)^2 + \left( 6 - 2 \right)^2}$
$= \sqrt{4 + 16 + 16}$
$= \sqrt{36}$
$= 6$
$DA = \sqrt{\left( 1 - 4 \right)^2 + \left( 2 - 7 \right)^2 + \left( 3 - 6 \right)^2}$
$= \sqrt{9 + 25 + 9}$
$= \sqrt{43}$
$AB = CD and BC = DA$
$\text{ Since, opposite pairs of sides are equal } .$
$\therefore \text{ ABCD is a parallelogram }$
$AC = \sqrt{\left( 2 - 1 \right)^2 + \left( 3 - 2 \right)^2 + \left( 2 - 3 \right)^2}$
$= \sqrt{1 + 1 + 1}$
$= \sqrt{3}$
$BD = \sqrt{\left( 4 + 1 \right)^2 + \left( 7 + 2 \right)^2 + \left( 6 + 1 \right)^2}$
$= \sqrt{25 + 81 + 49}$
$= \sqrt{155}$
$\text{ Since }, AC \neq BD$
Thus, ABCD is not a rectangle.

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 28 Introduction to three dimensional coordinate geometry
Exercise 28.2 | Q 23 | Page 10