Step by step explanation..
Don't copy❌
Answers
Answer:
PQRS is a rhombus
Step-by-step explanation:
Given vertices are A(-1,-1), B(-1,4), C(5,4), D(5,-1)
(i) Coordinates of P are:
= (-1 - 1/2, -1 + 4/2)
= (-1, 3/2)
(ii) Coordinates of Q are:
= (-1 + 5/2, 4 + 4/2)
= (2,4).
(iii) Coordinates of R are:
= (5 + 5/2, 4 - 1/2)
= (5,3/2)
(iv) Coordinates of S are:
= (-1 + 5/2, -1 - 1/2)
= 2, -1
Now,
PQ = √(2 + 1)² + (4 - 3/2)²
= √3² + (5/2)²
= √9 + 25/4
= √61/4
QR = √(5 - 2)² + (3/2 - 4)²
= √3² + (5/2)²
= √9 + 25/4
= √61/4
RS = √(2 - 5)² + (-1 - 3/2)²
= √(-3)² + (-5/2)²
= √9 + 25/4
= √61/4
SP = √(-1 - 2)² + (3/2 + 1)²
= √(-3)² + (5/2)²
= √9 + 25/4
= √61/4
∴ PQ = QR = RS = SP ⇒ Sides are equal.
Now,
PR = √(5 + 1)² + (3/2 - 3/2)²
= 6
SQ = √(2 - 2)² + (4 + 1)²
= 5
∴ PR ≠ SQ ⇒ Diagonals are not equal.
Therefore, PQRS is a rhombus.
Hope it helps!
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.