Let, ABCD be a square And

A( 3,4 ) AND C ( 1,-1 ) are the given angular points.

Let N ( x,y ) be the unknown vertex.

Then after that, AB = BC

AB² = BC²

( x-3 )² + ( y-4 )² = ( x-1 )² + ( y+1 )²

4x + 10y - 23 = 0

x = 23-10y/4  --- ( 1 )

Now, In Δ ABC,

AB² + BC² = AC²

(x-3)² + (y-4)² + (x-1)² + (y+1)² = (3-1)² + (4+1)²  --- (ii)

x²+y²-4x-3y-1 = 0

Substitute (i) into (ii) ,

(23-10y/4 )² + y² - (23-10y) -3y-1 = 0

4y² - 12y + 5 = 0

( 2y-1 ) ( 2y-5 )= 0

y = 1/2 or 5/2

Put, y = 1/2 and y = 5/2 in ( i) equation!

Than we get the value of x :-

x = 9/2  and x = -1/2.

Hence, The required vertices of square = ( 9/2 , 1/2 ) and ( -1/2, 5/2 ).

Thank you.