Question

Prove that the points P(0,-1), Q(-2,3) R(6,7) and S(8,3) are the vertices of a rectangle PQRS.

Answer 1

Let A (0 , -1) , B (6 , 7) , C ( -2 ,3) and D (8 , 3) be the given points. Then,

AB=√(6−0)2+(7+1)2=√36+64=10

CD=√(8+2)2+(3−3)2=√100+0=10

AD=√(8−0)2+(3+1)2=√64+16=4√5

BC=√(6+2)2+(7−3)2=√64+16=4√5

∴ AB = CD and AD = BC

So, ABCD is a parallelogram

Now , AC=√(−2−0)2+(3−(−1)2=√4+16=2√5

and , BD=√(8−6)2+(3−7)2=√4+16=2√5

Clearly, opposite sides are equal and diagonals are equal
.
Hence, ABCD is a rectangle.

Now, Area of rectangle ABCD=AB×BC

AB×BC=10×4√5 sq.units

⇒AB×BC=40√5 sq.units

Answer 2

U can prove it by showing SIDES PQ AND RS equal using distance formula