Question

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

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Answer 1

Step-by-step explanation:

HelloflFriend 

\bf{Given}Given 

Let the coordinates of 

A = (X₁, Y₁)

B = (X₂, Y₂)

C = (X₃, Y₃)

D = (X₄ , Y₄)

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

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

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

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

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

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

Now,

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

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

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

BC = \sqrt{1 + 1} = \sqrt{2}BC=1+1=2 

Now,

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

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

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

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

Now,

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

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

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

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

AD = \sqrt{1 + 1} = \sqrt{2}AD=1+1=2 

Here, AB = CD 

and 

BC = AD 

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

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

Answer 2

Here is your answer user.