The co-ordinates of the points on the line \(2 x-y=5\) which are the distance of 1 unit from the line \(3 x+4 y=5\) are

Let \(\left(x_1, y_1\right)\) be the required point
\(\therefore \quad 2 x_1-y_1=5\)
Also, \(\left(x_1, y_1\right)\) is at the distance of 1 unit from line \(3 x+4 y=5\)
\(
\begin{array}{ll}
\therefore & 1=\left|\frac{3 x_1+4 y_1-5}{\sqrt{9+16}}\right| \\
\therefore & \pm 5=3 x_1+4 y_1-5 \\
\therefore & 3 x_1+4 y_1-5=5 \quad \text { or } 3 x_1+4 y_1-5=-5 \\
\therefore & 3 x_1+4 y_1=10 \\
& \text { or } \\
& 3 x_1+4 y_1=0
\end{array}
\)
Solving equations (i) and (ii), we get \(x_1=\frac{30}{11}\) and \(y_1=\frac{5}{11}\)
Solving equation (i) and (iii), we get \(x_1=\frac{20}{11}\) and \(y_1=\frac{-15}{11}\)
\(\therefore \quad\left(\frac{30}{11}, \frac{5}{11}\right)\) and \(\left(\frac{20}{11}, \frac{-15}{11}\right)\) are the required points.