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:

To find the equation of the line that is parallel to \( -x - 8y = 8 \), we first need to determine its slope. We can rewrite the equation in slope-intercept form (y = mx + b): 1. Rearranging gives us \( -8y = x + 8 \) or \( y = -\frac{1}{8}x - 1 \). The slope of the original line is \( -\frac{1}{8} \). Since parallel lines have the same slope, the equation of the parallel line passing through \( (3, -3) \) can be expressed as: \[ y - (-3) = -\frac{1}{8}(x - 3) \] This simplifies to: \[ y + 3 = -\frac{1}{8}x + \frac{3}{8} \] \[ y = -\frac{1}{8}x - \frac{21}{8} \] Equation of the parallel line: \( y = -\frac{1}{8}x - \frac{21}{8} \). To find the equation of the line that is perpendicular, we take the negative reciprocal of the slope. The negative reciprocal of \( -\frac{1}{8} \) is \( 8 \). Thus, using the point-slope form again: \[ y - (-3) = 8(x - 3) \] This simplifies to: \[ y + 3 = 8x - 24 \] \[ y = 8x - 27 \] Equation of the perpendicular line: \( y = 8x - 27 \).