Show that points P(1,-2), Q(5,2), R(3,-1), S(-1,-5) are the vertices of a parallelogram.

Proof
Given points are P(1,-2), Q(5,2), R(3,-1), S(-1,-5).
By distance formula PQ = √[(x₂-x₁)²+(y₂-y₁)²]
PQ = √[(5-1)²+(2-(-2))²]
= √[(4)²+(4)²]
= √[(16+16)]
= √32
PQ = √32 ………………(i)
By distance formula QR = √[(x₂-x₁)²+(y₂-y₁)²]
QR = √[(3-5)²+(-1-2)²]
= √[(-2)²+(-3)²]
= √[(4+9)]
= √13
QR = √13 ………………(ii)
By distance formula RS = √[(x₂-x₁)²+(y₂-y₁)²]
RS = √[(-1-3)²+(-5-(-1))²]
= √[(-4)²+(-4)²]
= √[(16+16)]
= √32
RS = √32 ………………(iii)
By distance formula PS = √[(x₂-x₁)²+(y₂-y₁)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.