Sides of a rhombus are parallel to the lines x + y – 1 = 0 and 7x – y – 5 = 0. It is given that diagonals of the rhombus intersect at (1, 3) and one vertex ‘A’ of the rhombus lies on the line  y = 2x. Then the coordinates of the vertex A are 

It is clear that diagonals of the rhombus will be parallel to the bisectors of the given lines and will pass through (1, 3). Equations of bisectors of the given lines are 

$\frac{x+y-1}{\sqrt{2}} =\pm \left(\frac{7x-y-5}{5\sqrt{2}}\right)$  Or  2x – 6y =0, 6x + 2y = 5

Therefore, the equations of diagonals are x – 3y + 8 = 0 and 3x + y – 6 = 0. Thus the required vertex will be the point where these lines meet the line 
y = 2x. Solving these lines we get possible coordinates as $\left(\frac{8}{5},\frac{16}{5}\right)$ and $\left(\frac{6}{5},\frac{12}{5}\right)$.