Question

Verify the following:
(-1,2, 1),(1,-2,5),(4,-7,8),(2,-3,4) are the vertices of a parallelogram.

Answer 1

$\sf{\underline{\underline{Formula\:to\:be\:used:}}}$

$PQ = \sqrt{(x_{2}-x_{1}) ^{2} + (y_{2} - y_{1}) ^{2} + (z_{2} - z_{1}) ^{2}}$


$\sf{\underline{\underline{Let:}}}$

$( - 1, 2, 1)$ be denoted by A.

$(1, 2, - 5)$ be denoted by B.

$(4, - 7, 8)$ be denoted by C.

$(2, - 3, 4)$ be denoted by D.


$\sf\bold{\underline{\underline{Case\:I}}}$

A = $( - 1, 2, 1)$

$x_{1}=-1$, $y_{1}=2$, $z_{1}=1$

B = $(1, 2, - 5)$

$x_{2}=1$, $y_{2}=2$, $z_{2}=-5$


$\sf{\underline{\underline{Now}}}$:

Substituting these values, in the above formula, we get:

$AB = \sqrt{(x_{2}-x_{1}) ^{2} + (y_{2} - y_{1}) ^{2} + (z_{2} - z_{1}) ^{2}}$

$AB = \sqrt{(1 + 1) ^{2} + ( - 2 - 2) ^{2} + (5 - 1) ^{2} }$

$AB = \sqrt{4 + 16 + 16}$

$AB = \sqrt{36}$

$AB = 6 \: units$

$\sf{\underline{\underline{Therefore}}}$: $\boxed{AB = 6\:units}$


$\sf\bold{\underline{\underline{Case\:II}}}$

B = $(1, 2, - 5)$

$x_{1}=1$, $y_{1}=2$, $z_{1}=-5$

C = $(4, - 7, 8)$

$x_{2}=4$, $y_{2}=-7$, $z_{2}=8$


$\sf{\underline{\underline{Now}}}$:

Substituting these values, in the above formula, we get:

$BC = \sqrt{(x_{2}-x_{1}) ^{2} + (y_{2} - y_{1}) ^{2} + (z_{2} - z_{1}) ^{2}}$

$BC = \sqrt{(4 - 1) ^{2} + ( - 7 + 2) ^{2} + (8 - 5) ^{2} }$

$BC = \sqrt{9 + 25 + 9}$

$BC = \sqrt{43} \: units$

$\sf{\underline{\underline{Therefore}}}$: $\boxed{BC = \sqrt{43 } \: units}$


$\sf\bold{\underline{\underline{Case\:III}}}$

C = $(4, - 7, 8)$

$x_{1}=4$, $y_{1}=-7$, $z_{1}=8$

D = $(2, - 3, 4)$

$x_{2}=2$, $y_{2}=-3$, $z_{2}=4$


$\sf{\underline{\underline{Now}}}$:

Substituting these values, in the above formula, we get:

$CD = \sqrt{(x_{2}-x_{1}) ^{2} + (y_{2} - y_{1}) ^{2} + (z_{2} - z_{1}) ^{2}}$

$CD = \sqrt{(2 - 4) ^{2} + ( - 3 + 7) ^{2} + (4 - 8) ^{2} }$

$CD = \sqrt{4 + 16 + 16}$

$CD = \sqrt{36}$

$CD = 6 \: units$

$\sf{\underline{\underline{Therefore}}}$: $\boxed{CD = 6\:units}$


$\sf\bold{\underline{\underline{Case\:IV}}}$

D = $(2, - 3, 4)$

$x_{1}=2$, $y_{1}=-3$, $z_{1}=4$

A = $( - 1, 2, 1)$

$x_{2}=-1$, $y_{2}=2$, $z_{2}=1$


$\sf{\underline{\underline{Now}}}$:

Substituting these values, in the above formula, we get:

$DA = \sqrt{(x_{2}-x_{1}) ^{2} + (y_{2} - y_{1}) ^{2} + (z_{2} - z_{1}) ^{2}}$

$DA = \sqrt{( - 1 - 2) ^{2} + (2 + 3) ^{2} + (1 - 4) ^{2} }$

$DA = \sqrt{9 + 25 + 9}$

$DA = \sqrt{43} \: units$

$\sf{\underline{\underline{Therefore}}}$: $\boxed{DA = \sqrt{43 } \: units}$


$\sf{\underline{\underline{Now,\:we\:know\:that}}}$:

AB = CD = 6

$\sf{\underline{\underline{And}}}$:

BC = AD = $\sqrt{43}$


$\sf{\underline{\underline{Note}}}$: Opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal.


$\sf{\underline{\underline{Therefore}}}$:

ABCD is a parallelogram.


$\sf{\underline{\underline{Hence}}}$:

The given points are the vertices of a parallelogram.