Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days
8104911739
For Business Enquiry

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number

022-62211530

Mon to Sat - 11 AM to 8 PM

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 March 2013, 8:45 PM
Answered by 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 Expert 13th March 2013, 12:08 AM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Chat with us on WhatsApp