If two opposite vertices of a square are (5, 4) and (1, -6), find the coordinates of its remaining two vertices. 

Concept:
Application:

Let ABCD be a square and let A ( 5,4) and c ( 1,-6) be the given angular points.

Let B ( x,y) be the unknown vertex. 

Then AB = BC

⇒ (AB)2(BC)2


⇒  (x5)2 +  (y4)2 =  (x1)2  +  (y+6)2                        
⇒ 8x+20y4=0

2x+5y1=0
x = fraction numerator 1 minus 5 y over denominator 2 end fraction
(1)
  

In right- angled triangle ABC , we have

\(({AB})^{2}\+  \(({BC})^{2}\  = \(({AC})^{2}\         

                             
⇒ (x5)2 + (y4)2 +(x1)2(y+6)2    =  (51)2+(4+6)2

⇒ 2 x2 +2 y2 -12x +4y -38 =0     

                                                                                           
Substituting the value of x from (1) and (2),

we get

2(fraction numerator 1 minus 5 y over denominator 2 end fraction)2+2y26(1-5y)+4y38=0
2 left parenthesis 1 minus 5 y right parenthesis squared plus 8 y squared minus 24 left parenthesis 1 minus 5 y right parenthesis plus 16 y minus 152 equals 02 left parenthesis 1 plus 25 y squared minus 10 y right parenthesis plus 8 y squared minus 24 plus 120 y plus 16 y minus 152 equals 050 y squared plus 2 minus 20 y plus 8 y squared minus 24 plus 120 y plus 16 y minus 152 equals 058 y squared plus 116 y minus 174 equals 02 y squared plus 4 y minus 6 equals 0y squared plus 2 y minus 3 equals 0y squared plus 3 y minus y minus 3 equals 0y left parenthesis y plus 3 right parenthesis minus left parenthesis y plus 3 right parenthesis equals 0left parenthesis y minus 1 right parenthesis left parenthesis y plus 3 right parenthesis equals 0y equals 1 comma negative 3

Putting y =1 and y=3 in (i) we get
x = - 2, 8


Hence, the required vertices of the square are (-2,1) or (8,3)