CBSE Class 10 Answered

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).