Consider the line \( -x-8 y=8 \). Find the equation of the line that is parallel to this line and passes through the point \( (3,-3) \). Find the equation of the line that is perpendicular to this line and passes through the point \( (3,-3) \). Equation of parallel line:

First, we need to express the given line \( -x - 8y = 8 \) in slope-intercept form \( y = mx + b \). Rearranging, we get \( 8y = -x - 8 \) or \( y = -\frac{1}{8}x - 1 \). The slope of this line is \( -\frac{1}{8} \). A parallel line will have the same slope. Using the point-slope form, the equation of the parallel line through the point \( (3, -3) \) is: \[ y + 3 = -\frac{1}{8}(x - 3). \] Simplifying this gives: \[ y = -\frac{1}{8}x - \frac{21}{8}. \] Equation of parallel line: \( y = -\frac{1}{8}x - \frac{21}{8} \). Now, for the perpendicular line, the slope will be the negative reciprocal of \( -\frac{1}{8} \), which is \( 8 \). Again using the point-slope form for the point \( (3, -3) \): \[ y + 3 = 8(x - 3). \] This simplifies to: \[ y = 8x - 27. \] Equation of perpendicular line: \( y = 8x - 27 \).