Question

CAN ANYONE SOLVE THE QUESTION ATTACHED ?

Plz help me

BeBrainly guys ^_^

Answer 1

Here's the answer

P ( 2, -2 )
Q ( 7 , 3 )
R ( 11 , -1 )
S ( 6 , -6 )

Using distance formula

$\sf{PQ = \sqrt{(7 - 2) {}^{2} + (3 - ( - 2) ){}^{2}}}$

$\sf{PQ = \sqrt{5 {}^{2} - 5 {}^{2} }}$

$\sf{PQ = \sqrt{25 + 25}}$

$\sf{PQ = \sqrt{50}}$

$\sf{PQ = 5 \sqrt{2}}$
Let this be eq. ( 1 )

_______________________________
Now,

$\sf{QR = \sqrt{(11 - 7) {}^{2} + ( - 1 - 3) {}^{2}} }$

$\sf{QR = \sqrt{4 {}^{2} + ( - 4) {}^{2} }}$

$\sf{QR = \sqrt{ 16 + 16}}$

$\sf{QR = \sqrt{32}}$

$\sf{QR = 4 \sqrt{2}}$
Let this be eq. (2)
____________________________

Now,
$\sf{RS = \sqrt{(6 - 11) {}^{2} + ( - 6 - (- 1)) {}^{2} }}$

RS = √25 + √25

RS = √50

RS = 5√2

Let this be eq. (3)
_________________________

Now,

PS = $\sf { \sqrt{(6-2)^2 + (-6 - (-2)^2)}}$

PS = $\sf { \sqrt{16 + 16}}$

PS = $\sf { \sqrt{32}}$

PS = 4√2
Let this be eq.(4)

_________________________
In Quadrilateral PQRS ,

PQ = RS.. From eq.1 and eq.3

QR = PS..From eq.,2 and eq.4

A quadrilateral is a parallelogram,if both the pairs of it's opposite sides are congruent.

Therefore

PQRS is a parallelogram.

Thus, P ( 2, -2 ), Q ( 7 , 3 ) , R ( 11 , -1 ) , S ( 6 , -6 ) are the vertices of a parallelogram.

Thanks!

Answer 2

HEY MATE HERE IS THE ANSWER IN THE ATTACHMENT :-


$<marquee>$❣❣HOPE IT HELPS U ❣❣

$<marquee>$❣❣PLEASE MARK MY ANSWER AS BRAINLILIST ❣❣

$<marquee>$❣❣THANKS ❣❣

☺☺☺

$\boxed{BE BRAINLY}$