If one of the diagonals of a square is along the line $x=2y$ and one of its vertices is (3, 0), then its sides through this vertex are given by the equations

Clearly the point (3, 0) does not lie on the diagonal $x=2 y.$

Let $m$ be the slope of a side passing through (3, 0). Then,

its equation is

$y-0=m(x-3).$                                  (i)

Since the angle between a diagonal and a side of a square· is

$\mathrm{\pi }/4.$ Therefore, angle between $x=2y$ and $y-0=m(x-3)$ is

also $\mathrm{\pi }/4.$ Consequently, we have

$\tan \frac{\pi }{4}=\pm \frac{m-1/2}{1+m/2}\Rightarrow m=3,-\frac{1}{3}$

Substituting the values of min (i), we obtain

$y-3x+9=0$ and $3y+x-3=0$ as the required sides.