P is the mid point of A(-1,-1); B(-1,4)

P(x,y)=x₁+x₂/2 , y₁+y₂/2

here x₁=-1, x₂=-1

y₁=-1, y₂=4

P(x,y) = -1+(-1)/2 and -1+4/2

=-1-1/2 and 3/2

=-1 and 1.5

∴P(-1, 1.5)

Q is the mid point of B(-1,4); C(5,4)

Q(x,y)=x₁+x₂/2 , y₁+y₂/2

= -1+5/2 , 4+4/2

=4/2 , 8/2

=2,4

∴Q(2,4)

R is the mid point of C(5,4)D(5,-1)

R(x,y) = 5+5/2 , 4+(-1)/2

=10/2,4-1/2

=5,3/2

=5,1.5

∴R(5,1.5)

S is the mid point of A(-1,-1) ; D(5,-1)

S(x,y)= -1+(-1)/2+5+(-1)/2

=-1-1/2, 5-1/2

=-2/2 , 4/2

=-1,2

∴S(-1,2)

PQRS is a quadrilateral but to prove what type of quad. it is use the distance formula

P(-1, 1.5) ; Q(2,4)

PQ=√[(x₂-x₁)²+(y₂-y₁)²]

PQ = √(2-(-1))²+(4-1.5)²

= √(3)²+(2.5)²

=√9+6.25

=√15.25

∴ PQ = √15.25

Q(2,4); R(5,1.5)

QR = √[(5-2)²+(1.5-4)²]

=√(3)²+(-2.5)²

=√9+6.25

=√15.25

∴QR = √15.25

R(5,1.5) ; S(-1,2)

RS=√[(5+1)²+(2-1.5)²]

= √6²+(0.5)²

=√36+0.25

=√36.25

∴RS=√36.25

S(-1, 2) ; P(-1,1.5)

SP= √[(-1+1)²+(1.5-2)²]

=√(0)²+(0.5)²

=√0+0.25

=√0.25

=0.5

∴SP = 0.5

On comparing all sides joined by the midpoints of the rectangle we find that PQRS is just a quadrilateral not any type of quad.

Thank You,.....

Mark as brainliest if helpful.......

Yours, Jahnavi