find the equation of a line perpendicular to y -9= 2/3(x-3) that passes through the point (6,1) in slope intercept form

The Solution.
The given equation is
[tex]y-9=\frac{2}{3}(x-3)[/tex]To determine the slope of the line, we shall express the equation in the y =mx + c , where m is the slope.
[tex]\begin{gathered} y-9=\frac{2}{3}(x-3) \\ \text{Clearing the bracket, we get} \\ y-9=\frac{2}{3}x-2 \\ \text{Collecting the like terms, we get} \\ y=\frac{2}{3}x-2+9 \\ \\ y=\frac{2}{3}x+7 \end{gathered}[/tex]The slope of the given equation is 2/3.
But to obtain the slope a line that is perpendicular to the given line, we use the formula below:
[tex]m_2=-\frac{1}{m_1}[/tex][tex]undefined[/tex]