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 = $-\frac{5}{18}$ , y = $-\frac{5}{9}$ Thus the lines (i) and (ii) meet at R $\left(-\frac{5}{18}, -\frac{5}{9}\right)$ ∴ The required distance = |QR| = $\sqrt{ \left(-\frac{5}{18} - 1\right)^2 + \left(-\frac{5}{9} - 2\right)^2 }$ = $\sqrt{ \frac{529}{324} + \frac{529}{81} }$ = $\sqrt{ \frac{529}{81} \left( \frac{1}{4} + 1 \right) }$ = $\frac{23}{9} \sqrt{ \frac{5}{4} }$ = $\frac{23 \sqrt{5}}{18}$ units.