The points on the straight line 3x – 4y + 1 = 0 which are at distance of 5 units from the point (3, 2) are

$\begin{array}{l}Slopeof3x-4y+1=0is\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\\ \mathrm{Tan}\theta =\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,r=5\\ \mathrm{Cos}\theta =\frac{4}{5},\sin \theta =\frac{3}{5}\\ \therefore Q(x_1\pm r\cos \theta ,y_1\pm r\sin \theta )\\ =(3\pm 5\left(\frac{4}{5}\right),2\pm 5\left(\frac{3}{5}\right))\\ =(3\pm 4,2\pm 3)\\ =(7,5)or(-1,-1)\end{array}$