Question

show that the points A (2, 1), B(5, 4), C(4, 7), and D(1, 4). are the angular points of the parallelogram ABCD.

Answer 1

Answer:Let A(2,1), B(5,2), C(6,4) and D(3,3) be the vertices of a parallelogram

ABCD. Since, the diagonals of a parallelogram bisect each other.

AC2 = (6-2)2 + (4-1)2 = (4)2+(3)2 =16+9 = 25

BC2 = (6-5)2 + (4-2)2= (1)2+(2)2 = 1+4 = 5

AB2 = (5-2)2+(2-1)2 = (3)2+(1)2 = 9+1 =10

DC2 = (6-3)2+(4-3)2 =( 3)2+(1)2 = 9+1=10

AD2=(3-2)2+(3-1)2 =(1)2+(2)2=1+4= 5

Since BC = AD and DC = AB, ABCD is a parallelogram.

AB2  + BC2 =10+5 = 15

AB2 + BC2 ≠ AC2

∴△ABC is not right angled. Therefore parallelogram ABCD is not a rectangle.