The given equation of line is
2x – y = 0 ... (i)
and the point Q(1, 2) lies on this line.
We also have given
4x + 7y + 5 = 0 ... (ii)
On solving (i) and (ii), we get x = −

5

18

, y = −

5

9

Thus the lines (i) and (ii) meet at R (−

5

18

,−

5

9

)
∴ The required distance = |QR|
= √(−

5

18

−1)2+(−

5

9

−2)2
= √

529

324

+

529

81

= √

529

81

(

1

4

+1) =

23

9

√

5

4

=

23√5

18

units.