Question

Show that the point A (-4,-3) B(3,1)C(3,6)D(-4,2) taken in the order form the vertices of a parallelogram are not

Answer 1

Answer:

Heyy friend here are your solutions..

Step-by-step explanation:

You can find using distance formula as follows:-

Hope it helps u..

I guess the question is incomplete..

Answer 2

Answer:

Step-by-step explanation:

Given that ,

the points A( -4 , -3 ) , B( 3, 1 ) , C( 3 , 6 ) and D( -4 , 2 ).

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

     = $\sqrt{7^{2}+4^{2} }$

     = $\sqrt{49 + 16}$

     = $\sqrt{65}$

BC = $\sqrt{(3-3)^{2} + (6-1)^{2} }$

     = $\sqrt{0 + 25}$

     = $5$

CD = $\sqrt{(3 + 4)^{2} + (6-2)^{2} }$

     = $\sqrt{7^{2} + 4^{2} }$

     = $\sqrt{49 + 16 }$

     = $\sqrt{65}$

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

     = $\sqrt{0 + 25 }$

     = $5$

AC = $\sqrt{(3+4)^{2} + (6 + 3 ) ^{2} }$

     = $\sqrt{49 + 81}$

     = $\sqrt{130}$

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

     = $\sqrt{49 + 1}$

     = $\sqrt{50}$

AB = CD = √65

BC = DA = 5

opposite sides are equal .

And diagonals are unequal.

Therefore, it is a parallelogram .

THANKS.

#SPJ2.

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