Question

show that A (-1,0) B (3,1) C (2,2) and D(-2,1) are the vertices of a parallelogram ABCD?

Answer 1

since opposite side are equal and diagonal s are not equal therefore it is a parallelogram

Answer 2

$\huge \red {Hello \: flFriend}$

$\bf{Given}$

Let the coordinates of

A = (X₁, Y₁)

B = (X₂, Y₂)

C = (X₃, Y₃)

D = (X₄ , Y₄)

$\green{By \: distance \: formula : }$

$AB \: = \sqrt{(X2 - X1) + (Y2 - Y1)}$

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

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

$AB \: = \sqrt{(4) {}^{2} + 1}$

$AB = \sqrt{16 + 1} = \sqrt{17}$

Now,

$BC = \sqrt{(X3 - X2) {}^{2} + (Y3 - Y2) {}^{2} }$

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

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

$BC = \sqrt{1 + 1} = \sqrt{2}$

Now,

$CD = \sqrt{(X4 - X3) {}^{2} + (Y4 - Y3) {}^{2} }$

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

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

$CD = \sqrt{16 + 1} = \sqrt{17}$

Now,



$AD = \sqrt{(X4 - X1) {}^{2} + (Y4 - Y1) {}^{2} }$

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

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

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

$AD = \sqrt{1 + 1} = \sqrt{2}$

Here, AB = CD

and

BC = AD

Therefore, by the property of parallelogram the given vertices are of parallelogram .

$\huge \boxed { \boxed{thans}}$