The two opposite vertices of a square are (-1 , 2) and (3,2) then find the other two vertices.

Asked by  | 26th Feb, 2012, 07:40: PM

Expert Answer:

Let ABCD be a square and let A(-1,2) and C(3,2) be the given two vertices. Also, let (x,y) be the coordinates of point B.
Since AB=BC

­­(x+1)2+(y-2)2= ­­(3-x)2+(2-y)2

Or x2+2x+1+y2-4y+4=9+x2-6x+4-4y+y2

2x +1- 4y + 4 = -6x + 9 - 4y + 4

8x=8

x=1                                                          

Now, in triangle ABC using Pythagoras Theorem

AC2=AB2+BC2

  ­­(3+1)2+(2-2)2 = ­­(x+1)2+(y-2)2 +(x-3)2+(y-2)2

16 = 2x2 + 2y2 + 2x + 1 - 4y + 4 - 6x + 9 - 4y + 4

Using x=1 we get,

16 = 2 + 2y2 + 2 + 1 - 4y + 4 – 6 + 9 - 4y + 4

2y2 - 8y = 0

y2 - 4y = 0

y(y - 4) = 0 or y = 0,4                                   

So, the other two vertices are (1,0) and (1,4).

Answered by  | 26th Feb, 2012, 10:02: PM

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