You have the following equation:
[tex]6y=x-1[/tex]if you divide by 6 both sides, you obtain:
[tex]y=\frac{1}{6}x-\frac{1}{6}[/tex]the slope of the previous line is m = 1/6.
Take into accout that the relation between the slopes of two perpendicular lines is given by:
[tex]m^{\prime}=-\frac{1}{m}[/tex]For the required perpendicular line you have:
[tex]m^{\prime}=-\frac{1}{\frac{1}{6}}=-6[/tex]Now, use the following general equation for a line:
[tex]y-y_o=m^{\prime}(x-x_o)[/tex]where (xo,yo) is any point on the line.
Replace (xo,yo) = (3,5) and m' = -6 into the previous formula and solve for y:
[tex]\begin{gathered} y-5=-6(x-3) \\ y-5=-6x+18 \\ y=-6x+18+5 \\ y=-6x+23 \end{gathered}[/tex]Hence, the equation of the perpendicular line is, in intercept slope form:
y = -6x + 23
