SOLUTION
For two lines to be perpendicular, the product of their slope must be equal to -1.
That is
[tex]\begin{gathered} m1\times m2=-1 \\ \text{Where m1 and m2 are the slopes of the lines } \end{gathered}[/tex]
Equation of a line in slope-intercept form is given as
[tex]\begin{gathered} y=mx+b \\ \text{where m is slope } \end{gathered}[/tex]
Comparing this to
[tex]\begin{gathered} y=3x-2 \\ \text{the slope m1 = 3} \end{gathered}[/tex]
Now, let's find the slope of the other line m2.
[tex]\begin{gathered} m1\times m2=-1 \\ 3\times m2=-1 \\ m2=\frac{-1}{3} \end{gathered}[/tex]
Now, equation of a line in point-slope form is
[tex]y-y_1=m(x-x_1)[/tex]
Relating this to this second slope we have
[tex]y-y_1=m2(x-x_1)[/tex]
The equation becomes
[tex]\begin{gathered} y-y_1=m2(x-x_1) \\ y-y_1=\frac{-1}{3}(x-x_1) \\ \text{from the point }\mleft(6,8\mright) \\ x_1=6,y_1=8 \\ \text{Hence } \\ y-y_1=\frac{-1}{3}(x-x_1) \\ y-8_{}=\frac{-1}{3}(x-6_{}) \\ y-8_{}=\frac{-1}{3}x+\frac{6}{3}_{} \\ y-8=\frac{-1}{3}x+2 \\ y=\frac{-1}{3}x+2+8 \\ y=\frac{-1}{3}x+10 \\ y=-\frac{1}{3}x+10 \end{gathered}[/tex]
Hence, the answer is
[tex]y=-\frac{1}{3}x+10[/tex]
Option B
