If one side of the square is `2x-y+6=0,` then one of the vertices is `(2,1)` . Find the other sides of the square.

Explanation

Vertex A(2,1) does not lie on the line BC=2x-y+6=0 (1)
Hence, one of the side of the square passing through A is parallel to BC.
Therefore, the equation of AD is
y-1=2(x-2)
`or 2x-y-3=0 " " (2)`
Another side AB of the square is passing through (2,1) and is perpendicular to BC.
Therefore, the equation of AB is
`y-1 = -(1)/(2) (x-2)`
` or x+2y-4=0 " " (3)`
The third side is parallel to AB, which has equation
`x+2y+c=0 " " (4)`
Now, the distance between the lines BC and AD is `9//sqrt(5)`.
Therefore, the distance between the lines (3) and (4) is
`(|C+4|)/(sqrt(5)) = (9)/(sqrt(5))`
`therefore c=5 or -13` Therefore, the equation of the third side is
x+2y+5=0 `(-=C'D')`
`"or " x+2y-13=0 (-=CD)`