Consider the line $-2x-9y=5$
Find the equation of the line that is parallel to this line and passes through the point $(-8,6)$.
Find the equation of the line that is perpendicular to this line and passes through the point $(-8,6)$.
Equation of parallel line:
Equation of perpendicular line:

Step 1 ：The given line equation is $-2x-9y=5$.

Step 2 ：The slope of a line given by the equation $ax+by=c$ is $-\frac{a}{b}$. So, the slope of the given line is $-\frac{-2}{-9}=\frac{2}{9}$.

Step 3 ：A line parallel to another line will have the same slope. So, the slope of the line parallel to the given line is also $\frac{2}{9}$.

Step 4 ：The equation of a line in slope-intercept form is $y=mx+c$, where $m$ is the slope and $c$ is the y-intercept. We can find the y-intercept by substituting the coordinates of the given point and the slope into this equation.

Step 5 ：The slope of a line perpendicular to another line is the negative reciprocal of the original line's slope. So, the slope of the line perpendicular to the given line is $-\frac{9}{2}$.

Step 6 ：We can find the equation of the line perpendicular to the given line in the same way as we did for the parallel line.

Step 7 ：The equation of the line that is parallel to the line $-2x-9y=5$ and passes through the point $(-8,6)$ is $y=\frac{2}{9}x+\boxed{{\displaystyle 7.78}}$ (rounded to two decimal places).

Step 8 ：The equation of the line that is perpendicular to the line $-2x-9y=5$ and passes through the point $(-8,6)$ is $y=-\frac{9}{2}x-\boxed{{\displaystyle 30}}$.