The equation
[tex]3x+7y=1[/tex][tex]7y=-3x+1[/tex][tex]y=\frac{-3x}{7}+\frac{1}{7}[/tex]has slope -3/7.
The product of perpendicular lines must be -1, so the slope of the new equation must be:
[tex]m\cdot-\frac{3}{7}=-1[/tex][tex]m=\frac{-1\cdot7}{-3}=\frac{7}{3}[/tex]The line with that slope and that passes throught (1,6) is:
[tex](y-6)=m(x-1)[/tex][tex](y-6)=\frac{7}{3}(x-1)[/tex][tex]y-6=\frac{7}{3}x-\frac{7}{3}[/tex][tex]y=\frac{7}{3}x-\frac{7}{3}+6[/tex][tex]y=\frac{7}{3}x-\frac{7}{3}+\frac{18}{3}[/tex][tex]y=\frac{7}{3}x+\frac{11}{3}[/tex]The following graph shows both equations:
[tex]y=\frac{7}{3}x+\frac{11}{3}[/tex][tex]3\cdot y=(\frac{7}{3}x+\frac{11}{3})\cdot3[/tex][tex]3y=7x+11[/tex][tex]-7x+3y=11[/tex]
