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

Asked by Ritwika Sharma | 12th Mar, 2013, 08:45: PM

Expert Answer:

Answer : Given : The two opposite vertices of a square are (-1,2) and (3,2).
To Find : the coordinates of the other two vertices.
 

Let ABCD be a square and A (–1, 2) and C (3, 2) be the given vertices. Let the coordinate of vertex B be (x,y).

AB = BC  (as ABCD is a square)

=> AB2 = BC2

=>  [x – (–1)] 2 + (y – 2)= (x – 3)2 + (y – 2)2  (Distance formula)

=>  (x + 1)2 = (x – 3)2

=>  x2 + 2x + 1 = x– 6x + 9

=> 2+ 6x = 9 – 1

=> 8= 8

=> = 1

In triangle ABC, we have

AB2 + BC2 = AC2  (Pythagoras theorem)

=> 2AB2 = AC2             (as AB = BC)

=> 2[(x – (–1))2 + (y – 2)2] = (3 – (–1))2 + (2 – 2)2

=> 2[(x + 1)2 + (y – 2)2] = (4)2 + (0)2

=> 2[(1 + 1)2 + (y – 2)2] = 16                     (as x = 1)

=> 2[ 4 + (y – 2)2] = 16

=> 8 + 2 (y – 2)2 = 16

=> 2 (y – 2)2 = 16 – 8 = 8

=> (y – 2)2 = 4

=> y – 2 = ± 2

=> y – 2 = 2 or y – 2 = –2

=> y = 4 or y = 0

Thus, the other two vertices of the square ABCD are (1, 4) and (1, 0).

Answered by  | 13th Mar, 2013, 12:08: AM

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