We have to find the equation of the line "L" that is perpendicular to y=3x and passes through the point (1,3)
First let's find the slope of the perpendicular line, the slope of the line with equation y = 3 x is 3. If we multiply the slopes of two perpendicular lines, we get -1.
[tex]3x\cdot-\frac{1}{3}=-1[/tex]Therefore the slope of our perpendicular line "L" is -1/3
Second, we will make sure that the slope passes through the point (1,3)
Now use the intercept-slope form to find the equation
[tex]y-y_1=m(x_{}-x_1)[/tex]In this case, y1 = 3 and x1 = 1 and remember that m = -1/3, we substitute these values into the equation and operate
[tex]\begin{gathered} y-3=-\frac{1}{3}(x-1) \\ y-3=-\frac{1}{3}x+\frac{1}{3} \\ \end{gathered}[/tex]Add 3 units to both sides
[tex]\begin{gathered} y-3+3=-\frac{1}{3}x+\frac{1}{3}+3 \\ y=-\frac{1}{3}x+\frac{10}{3} \end{gathered}[/tex]The last result is our line "L" which is perpendicular to y=3x and passes through the point (1,3).
This is the answer
