If one of the diagonals of a square is along  he line $x = 2y$ and one of its vertices is (3, 0), then its side through this vertex nearer to the origin is given by the equation

The point (3, 0) does not lie on the diagonal $x = 2y$. Let the equation of a side through the vertex (3, 0) be 

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

Since the angle between a side and a diagonal of a square is $\pi /4$ we have

$\pm \tan \frac{\pi }{4}=\frac{m-1/2}{1+m(1/2)}=\frac{2m-1}{2+m}$

$\Rightarrow m=3,-1/3$

Thus the equation of a side through (3, 0) is

$y=3(x-3)\text{ or }y=\left(-\frac{1}{3}\right)(x-3)$ and the one nearer to the  origin is $3y + x –3 = 0$