Recall that two lines are perpendicular if the product of their slopes is equal to -1.
Taking the given equation to its slope-intercept form we get:
[tex]\begin{gathered} \frac{3y}{3}=\frac{x}{3}+\frac{4}{3}, \\ y=\frac{x}{3}+\frac{4}{3}\text{.} \end{gathered}[/tex]
Therefore the slope of the given line is 1/3, therefore, the slope of a perpendicular line to the given line is:
[tex]-\frac{1}{\frac{1}{3}}=-3.[/tex]
Now we will use the following slope-point formula for the equation of a line:
[tex]y-y_1=s(x-x_1)\text{.}[/tex]
Substituting s=-3 and (x₁,y₁)=(-6,8) we get:
[tex]y-8=-3(x-(-6))\text{.}[/tex]
Simplifying the above equation we get:
[tex]\begin{gathered} y-8=-3(x+6), \\ y-8=-3x-18. \end{gathered}[/tex]
Adding 8 to the above equation we get:
[tex]\begin{gathered} y-8+8=-3x-18+8, \\ y=-3x-10. \end{gathered}[/tex]
Answer:
[tex]y=-3x-10.[/tex]
