Question

the two opposite vertices of a square are (-1,2) , (3,2) find the coordinates of the other two vertices

Answer 1

it is right answer. I think

Answer 2

let ABCD is vertices of square in which A (-1,2) and C (3,2) and (x, y) is the co-ordinate of B
we know , properties of square
all side length are equal and between two side form right angle .
now ,
AC^2=BC^2+AB^2

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

16 =2x^2+2y^2-6x+2x-8y+9+1+8
-2=2x^2+2y^2-4x-8y
x^2+y^2-2x-4y+1=0 --------------(1)
again
AB=BC
so,
AB^2=BC^2
(x+1)^2+(y-2)^2=(x-3)^2+(y-2)^2
x/2+2x+1=x^2-6x+9
8x=8
x=1
put x=1 in equation (1)
1+y^2-2-4y+1=0
y^2-4y=0
y=0,4
hence (1,0) and (1,4) two opposite points of square