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: Equation of perpendicular line:

To find the equation of the parallel line, we first rewrite the given line \( -x - 8y = 8 \) in slope-intercept form \( y = mx + b \). Rearranging gives us \( 8y = -x - 8 \) or \( y = -\frac{1}{8}x - 1 \). The slope \( m \) here is \( -\frac{1}{8} \). A parallel line will have the same slope, so we use the point-slope form \( y - y_1 = m(x - x_1) \): \( y - (-3) = -\frac{1}{8}(x - 3) \). Simplifying this, we get: \( y + 3 = -\frac{1}{8}x + \frac{3}{8} \) \( y = -\frac{1}{8}x - \frac{21}{8} \). The equation of the parallel line is: \( y = -\frac{1}{8}x - \frac{21}{8} \). To find the equation of the perpendicular line, we take the negative reciprocal of the slope \( -\frac{1}{8} \), which is \( 8 \). Using the point-slope form again: \( y - (-3) = 8(x - 3) \). Expanding and simplifying gives us: \( y + 3 = 8x - 24 \) \( y = 8x - 27 \). So, the equation of the perpendicular line is: \( y = 8x - 27 \). Equation of parallel line: \( y = -\frac{1}{8}x - \frac{21}{8} \) Equation of perpendicular line: \( y = 8x - 27 \)