Question

Show that the points A(3, 1), B(0, -2),C(1,1) and D(4, 4) are the vertices of a parallelogram ABCD.

Answer 1

The given points are A (3,1), B(0,-2) and C(1,1), D(4,4). Then

AB $=\sqrt{(0-3{)}^2+(-2-1{)}^2}$

$=\sqrt{(-3{)}^2+(-3{)}^2}$

$=\sqrt{18}$ units

BC $=\sqrt{(1-0{)}^2+(1+2{)}^2}$

$\sqrt{(1{)}^2+(3{)}^2}$

$=\sqrt{10}$ units

CD $=\sqrt{(4-1{)}^2+(4-1{)}^2}$

$=\sqrt{(3{)}^2+(3{)}^2}$

$=\sqrt{18}$ units

AD $=\sqrt{(4-3{)}^2+(4-1{)}^2}$

$=\sqrt{(1{)}^2+(3{)}^2}$

$=\sqrt{10}$ units

Thus, AB = CD $=\sqrt{18}$ units and BC = AD $=\sqrt{10}$ units.

So, quadrilateral ABCD is a parallelogram