Question

if there are four points P(2,-1) ,Q(3,4) ,R(-2,3) and S(-3,-2) in a plane ,then prove that PQRS is not a square but a rhombus .

Answer 1

See answer in the picture

Answer 2

Answer:

The proof is explained strep-wise below :

Step-by-step explanation:

First finding the lengths of all the sides :

$PQ=\sqrt{(2-3)^2+(-1-4)^2}\\\\\implies PQ=\sqrt{26}\\\\QR=\sqrt{(3+2)^2+(4-3)^2}\\\\\implies QR=\sqrt{26}\\\\RS=\sqrt{(-2+3)^2+(3+2)^2}\\\\\implies RS=\sqrt{26}\\\\PS=\sqrt{(2+3)^2+(-1+2)^2}\\\\\implies PS=\sqrt{26}$

Now, length of all the sides are equal. So PQRS may be square or rhombus

So, finding the length of diagonals :

$PR=\sqrt{(2+2)^2+(-1-3)^2}\\\\\implies PR=\sqrt{32}\\\\QS=\sqrt{(3+3)^2+(4+2)^2}\\\\\implies QS =\sqrt{72}$

The length of diagonals are not equal so PQRS cannot be a square.

Therefore, PQRS is a rhombus not a square.

Hence Proved.