Proof

Given points are P(1,-2), Q(5,2), R(3,-1), S(-1,-5).

By distance formula PQ = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

PQ = √[(5-1)^{2}+(2-(-2))^{2}]

= √[(4)^{2}+(4)^{2}]

= √[(16+16)]

= √32

PQ = √32 ………………(i)

By distance formula QR = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

QR = √[(3-5)^{2}+(-1-2)^{2}]

= √[(-2)^{2}+(-3)^{2}]

= √[(4+9)]

= √13

QR = √13 ………………(ii)

By distance formula RS = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

RS = √[(-1-3)^{2}+(-5-(-1))^{2}]

= √[(-4)^{2}+(-4)^{2}]

= √[(16+16)]

= √32

RS = √32 ………………(iii)

By distance formula PS = √[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]

PS = √[(-1-1)^{2}+(-5-(-2))^{2}]

= √[(-2)^{2}+(-3)^{2}]

= √[(4+9)]

= √13

PS = √13 ………………(iv)

Here PQ = RS [From (i) and (iii)]

and QR = PS [From (ii) and (iv)]

Hence PQRS is a parallelogram. [For a parallelogram , opposite sides are equal]

Points P, Q, R, and S are the vertices of a parallelogram.

Hence proved.