The origin is a corner of a square and two of its sides are $y+2x=0$ and $y+2x=3$. Find the equation of other sides.
My Attempt: Let $OA$ and $CB$ be the sides of the square $OABC$ with equations $y+2x=0$ and $y+2x=3$. The equation of line $OC$ perpendicular to $OA$ is $$2x-y+k=0$$ Above equation passes through $O(0,0)$. Then $$2.0-0+k=0$$ $$k=0$$ So, $2x-y=0$ is the required equation.
On
Since one of the sides lies on $y + 2 x = 0$ which passes though one vertex (the origin), then the other side that passes through this vertex, is the perpendicular line which is
$ -2 y + x = 0$
We also know the third side lies on $y+2x = 3$ , which means the side length which is the distance between the given vertex (the origin) and this line is
$s = \dfrac{3}{\sqrt{5}} $
The fourth side is parallel to $-2 y + x = 0$ and thus lies on the line
$ -2 y + x = c $ where
$ \dfrac{|c|}{ \sqrt{5} } = \dfrac{3}{\sqrt{5}} $
Thus $c $ can be $3$ or $-3$.
Hence the fourth side can lie on $ -2 y + x = 3 $ or $ - 2y + x = -3 $
You've got the equation of line $OC$ wrong. It should be $$2y - x + k = 0$$ This is using the fact that the product of slopes of perpendicular lines is $-1$.