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.