Question

Show that the points A(2, 1), B(5, 2), C(6, 4) and D(3, 3) are the angular points of a parallelogram. Is this figure a rectangle ?

Answer 1

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.$A{C}^2=(6-2{)}^2+(4-1{)}^2$
$=(4{)}^2+(3{)}^2$
$=16+9=25$
$B{C}^2=(6-5{)}^2+(4-2^{)}2$
$=\left(1\right)2+\left(2\right)2$
$=1+4=5$
$A{B}^2=(5-2{)}^2+(2-1{)}^2$
$=(3{)}^2+(1{)}^2=9+1=10$
$D{C}^2=(6-3{)}^2+(4-3{)}^2$
$=(3{)}^2+(1{)}^2$
$=9+1=10$
$A{D}^2=(3-2{)}^2+(3-1{)}^2$
$=(1{)}^2+(2{)}^2$
$=1+4=5$
Since $BC=ADandDC=AB,$ ABCD is a parallelogram.
$A{B}^2+B{C}^2=10+5=15$
$A{B}^2+B{C}^2\ne A{C}^2$
$\therefore \Delta ABC$ is not right angled. Therefore parallelogram ABCD is not a rectangle.