One diagonal of a square is along the line $8\mathrm{x}-15\mathrm{y}=0$ and one of its vertex is $\left(1,2\right).$ Then the equations of the sides of the square passing through this vertex are 

The slope of BD is 8/15 and the angle made by BD with DC and BC is ${45}^0.$ So, let the slope of DC be m, then.

$\tan {45}^0=\pm \frac{\mathrm{m}-\frac{8}{15}}{1+\frac{8}{15}\mathrm{m}}$

or $\left(15+8\mathrm{m}\right)=\pm \left(15\mathrm{m}-8\right)$

or $\mathrm{m}=\frac{23}{7}$ and $-\frac{7}{23}$

Hence, the equations of DC  and BC are $\mathrm{y}-2=\frac{23}{7}\left(\mathrm{x}-1\right)$

or $23\mathrm{x}-7\mathrm{y}-9=0$

and $\mathrm{y}-2=\frac{7}{23}\left(\mathrm{x}-1\right)$

or $7\mathrm{x}+23\mathrm{y}-53=0$