First find the slope of the line through $(2,11)$ and $(8,-1)$:

Using point-slope form with the point $(8,-1)$ gives

so

Because the repair station is where the path crosses the $x$-axis, its $y$-coordinate is $0$. Substitute $y=0$:

which is equivalent to

Since $\frac{-12}{6}=-2$, the equation is

A point $(x,y)$ is on the line if its coordinates satisfy the equation $y=4x-3$. For $(1,1)$, substituting $x=1$ gives $y=4(1)-3=1$, which matches the $y$-value. So $(1,1)$ is on the line.

A translation left $2$ units means replace $x$ with $x+2$ in the original equation:

Then translate up $3$ units by adding $3$ to the output:

So the translated line has equation $y=4x+4$.

Because $f$ is linear, its slope can be found from the table. From $x=-1$ to $x=2$, the change in $x$ is $3$ and the change in $f(x)$ is $1-7=-6$, so the slope is $\frac{-6}{3}=-2$. Since $g$ is parallel to $f$, $g$ also has slope $-2$. Using the point $(4,10)$ on $g$, write $g(x)=-2x+b$. Then $10=-2(4)+b$, so $10=-8+b$ and $b=18$. Therefore, $g(0)=18$.

The line passes through $(0,6)$, so its $y$-intercept is $6$. That means the equation must have the form $y=mx+6$. Next, use the two given points to find the slope:

So the equation is