Question

Draw a quadrilateral ABCD in a cartesian plane, whose vertices are(-5,5)(5'5)(5,-5) and (-5,-5). Name the quadrilateral ABCD

Answer 1

ANSWER =

Given = a quadrilateral ABCD whose vertices are [-5,5] , [5,5] , [5,-5] , [-5,-5]

Lets find the distance b\w AB By using the distance formula that is ,

AB = √[X2-X1]²+[Y2-Y1]²

 =√[5+5]²+[5-5]²= √10²+ 0 = √100= 10 units

Similarly lets find the distance b\w BC by using the distance formula ,

BC = √[X2-X1]²+[Y2-Y1]²

      √[5-5]²+[-5-5]² = √0+[-10]²= √0 +100 = √100= 10 units

Now we have come to know that the distance b\w AB and BC is the same that is 10 units

Now , Lets find the distance b\w CD

CD = √[X2-X1]²+[Y2-Y1]²

   √[-5-5]²+ [-5+5]²=  √[-10]²+ 0= √100+0=√100= 10units

Similarly find the distance b\w DA

DA= √[X2-X1]²+[Y2-Y1]²

    = √[-5+5]²+[5+5]²=√0 + 10²= √100 = 10 units

Now we have come to know that the ditance b\w AB , BC , CD and DA is the same

Hence , AB = BC =CD = DA

SO , it is a square because we know that in a square all the 4 sides are equal , in this question also we have come to know that it is a square as the distance b\w every point is the sam ethat id 10 units or 10cm .

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