Question

vєriƒy ρ(-1,2,1) q(1,-2,5) r(4.-7.8) ɑท∂ s (2.-3.4) ɑrє τнє vєrτicєs σƒ ɑ ρɑrɑℓℓєℓσgrɑм​

Answer 1

Given that,

P = (-1,2,1)

Q = (1,-2,5)

R = (4,-7,8)

S = (2,-3,4)

Assume PQ, RS, QR and PS are the sides of the parallelogram.

Now using distance formula,

$\dashrightarrow\sf PQ = \sqrt{(x_2 -x_1)^2 + (y_2-y_1)^2 + (z_2 - z_1)^2}$

$\dashrightarrow\sf PQ = \sqrt{(1 - ( - 1))^2 + ( - 2 - 2)^2 + (5 - 1)^2}$

$\dashrightarrow\sf PQ = \sqrt{4 + 16 + 16}$

$\dashrightarrow\sf PQ = \sqrt{36}$

$\dashrightarrow\boxed{\sf PQ =6}$

Similarly,

$\dashrightarrow\sf QR = \sqrt{(x_2 -x_1)^2 + (y_2-y_1)^2 + (z_2 - z_1)^2}$

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

$\dashrightarrow\sf QR = \sqrt{9 + 25 + 9}$

$\dashrightarrow\boxed {\sf QR = \sqrt{43}}$

$\:$

$\dashrightarrow\sf RS = \sqrt{(x_2 -x_1)^2 + (y_2-y_1)^2 + (z_2 - z_1)^2}$

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

$\dashrightarrow\sf RS = \sqrt{4 + 16 + 16}$

$\dashrightarrow\sf RS = \sqrt{36}$

$\dashrightarrow\boxed{\sf RS = 6}$

$\:$

$\dashrightarrow\sf PS = \sqrt{(x_2 -x_1)^2 + (y_2-y_1)^2 + (z_2 - z_1)^2}$

$\dashrightarrow\sf PS = \sqrt{(2 - ( - 1))^2 + ( - 3 - 2)^2 + (4 - 1)^2}$

$\dashrightarrow\sf PS = \sqrt{9 + 25 + 9}$

$\dashrightarrow\boxed{ \sf PS = \sqrt{43}}$

Here, PQ = RS and QR = PS Opposite side are equal so, the given points are vertices of the parallelogram.

Answer 2

Answer:

• Given points are vertices of a parallelogram.

Step-by-step explanation:

Given:

• Point p = (-1, 2, 1)
• Point q = (1, -2, 5)
• Point r = (4, -7, 8)
• Point s = (2, -3, 4)

To Prove:

• Following points are vertices of parallelogram.

Now, we know about distance formula.

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

Now, put the Values in the formula,

$\implies{\tt{PQ = \sqrt{(1+1)^{2}+(-2-2)^{2}+(5-1)^{2}}}}$

$\implies{\tt{PQ = \sqrt{(2)^{2}+(-4)^{2}+(4)^{2}}}}$

$\implies{\tt{PQ = \sqrt{4+16+16}}}$

$\implies{\tt{PQ = \sqrt{36}}}$

$\implies{\boxed{\tt{PQ =6\;units}}}$

Again by distance formula,

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

Now, put the Values in the formula,

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

$\implies{\tt{QR = \sqrt{(3)^{2}+(-5)^{2}+(3)^{2}}}}$

$\implies{\tt{QR = \sqrt{9+25+9}}}$

$\implies{\boxed{\tt{QR = \sqrt{43}\;units}}}$

Again by distance formula,

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

Now, put the Values in the formula,

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

$\implies{\tt{RS = \sqrt{(-2)^{2}+(4)^{2}+(-4)^{2}}}}$

$\implies{\tt{RS = \sqrt{4+16+16}}}$

$\implies{\tt{RS = \sqrt{36}}}$

$\implies{\boxed{\tt{RS =6\;units}}}$

Again by distance formula,

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

Now, put the Values in the formula,

$\implies{\tt{SP = \sqrt{(-1-2)^{2}+(2+3)^{2}+(1-4)^{2}}}}$

$\implies{\tt{SP = \sqrt{(-3)^{2}+(5)^{2}+(-3)^{2}}}}$

$\implies{\tt{SP = \sqrt{9+25+9}}}$

$\implies{\boxed{\tt{SP = \sqrt{43}\;units}}}$

Now, we see that

$\implies{\tt{PQ = RS = 6\;units}}$

$\implies{\tt{QR = SP = \sqrt{43}\;units}}$

We, know that opposite sides of parallelogram are equal. So PQRS is a parallelogram.

∴ Given points are vertices of a parallelogram.