Question

prove that the points P(-4,3), Q(0,-3), R(6,1) and S(2,7) form a rhombus. Also find whether PQRS is a square​

Answer 1

Step-by-step explanation:

I can't tell the answer

But First U have to find distances between PQ QR RS AND PS

If the distance between all four is same then it's rhombus

Then Find The Distance between PR and QS the Diagonals

And We can Prove that

Since PQ=QR=RS=PS

and Diagonal PR=QS Quadrilateral PQRS is a Square

Answer 2

Answer:

Step-by-step explanation:

P=(-4,3)     Q (0,-3)      R(6,1)      S(2,7)

PQ=$\sqrt{(0+4)^{2}+(-3-3 })^{2}$

=$\sqrt{(4)^{2}+(-6)^{2}$

=$\sqrt{16+36$

PQ=$\sqrt{52\\}$ units

QR=$\sqrt{(6-0)^{2}+(1+3)^{2} }$

=$\sqrt{6^{2}+4^{2} }$

=$\sqrt{36+16$

QR=$\sqrt{52}$ units

RS=$\sqrt{(2-6)^{2}+(7-1)^{2} }$

=$\sqrt{(4)^{2}+(6)^{2}$

=$\sqrt{16+36$

RS=$\sqrt{52\\}$ units

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

=$\sqrt{36+16\\}$

SP=$\sqrt{52}$units

All the sides of the quadrilateral PQRS are equal. Hence PQRS is a Rhombus.

$SP^{2}+ PQ^{2}=SQ^{2}$

=$(\sqrt{52} )^{2} + (\sqrt{52} )^{2}$

=52+52=104

SQ=$\sqrt{104}$ units...........................1

SQ=$\sqrt{(0-2)^{2}+(-3-7)^{2} }$

=$\sqrt{4+100\\}$

SQ=$\sqrt{104}$ units.....................2

Equations  and 2 are equal.

Therefore PQRS is a rhombus with one angle 90°

(Rhombus with one angle 90° is a square)

Therefore PQRS is a square