Given Line is: 2x - y = 0

=> y = 2x

So slope of the line  = 2

=> tanθ = 2/1

Now hypotenuse² = base² + perpendicular²

                          = 1² + 2²

                          = 1+4

                          = 5

=> hypotenuse² = √5

Sinθ = 2/√5 and cosθ = 1/√5

The equation of a line passing thrugh (1,2) and making angle θ is

      (x-1)/cosθ = (y-2)/sinθ = r

=> (x-1)/1/√5 = (y-2)/2/√5 = r

Now (x-1)/1/√5 = r

=> x-1 = r/√5

=> x = 1+ r/√5

and

       (y-2)/2/√5 = r

=> y-2 = r*2/√5

=> y = 2 + 2r/√5

So point is (1+ r/√5, 2 + 2r/√5)

If this point lies on the line 4x + 7y + 5 = 0

=> 4*(1+ r/√5) + 7*(2 + 2r/√5) + 5 = 0

=> 4+ 4r/√5 + 14 +14r/√5 = 0

=> 18+ 18r/√5 = 0

=> 1+ r/√5 = 0

=> r/√5 = -1

=> r = -√5

So the distance is |r| = |-√5| = √5 unit