# The Points (5, –4, 2), (4, –3, 1), (7, 6, 4) and (8, –7, 5) Are the Vertices of - Mathematics

MCQ

The points (5, –4, 2), (4, –3, 1), (7, 6, 4) and (8, –7, 5) are the vertices of

#### Options

• a rectangle

•  a square

• a parallelogram

•  none of these

#### Solution

None of these
Suppose:
A(5, –4, 2)
B(4, –3, 1)
C(7, 6, 4)
D(8, –7, 5)

$AB = \sqrt{\left( 4 - 5 \right)^2 + \left( - 3 + 4 \right)^2 + \left( 1 - 2 \right)^2}$
$= \sqrt{\left( - 1 \right)^2 + \left( 1 \right)^2 + \left( - 1 \right)^2}$
$= \sqrt{1 + 1 + 1} = \sqrt{3}$
$BC = \sqrt{\left( 7 - 4 \right)^2 + \left( 6 + 3 \right)^2 + \left( 4 - 1 \right)^2}$
$= \sqrt{\left( 3 \right)^2 + \left( 9 \right)^2 + \left( 3 \right)^2}$
$= \sqrt{9 + 81 + 9} = \sqrt{99} = 3\sqrt{11}$
$CD = \sqrt{\left( 8 - 7 \right)^2 + \left( - 7 - 6 \right)^2 + \left( 5 - 4 \right)^2}$
$= \sqrt{\left( 1 \right)^2 + \left( - 13 \right)^2 + \left( 1 \right)^2}$
$= \sqrt{1 + 169 + 1} = \sqrt{171}$
$DA = \sqrt{\left( 8 - 5 \right)^2 + \left( - 7 + 4 \right)^2 + \left( 5 - 2 \right)^2}$
$= \sqrt{\left( 3 \right)^2 + \left( - 3 \right)^2 + \left( 3 \right)^2}$
$= \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3}$

We see that none of the sides are equal.

Concept: Three Dimessional Space
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 28 Introduction to three dimensional coordinate geometry
Q 5 | Page 23