Vrite an equation of the line passing through the given point and satisfying the give ondition. Give the equation (a) in slope-intercept form and (b) in standard form. \( (-8,-8) \); parallel to \( -x+8 y=48 \) (a) Write the equation of the line in slope-intercept form. \( y=\frac{1}{8 x} x-7 \) (Simplify your answer. Use integers or fractions for any numbers in the equation.) (b) Write the equation of the line in standard form.

To find the equation of the line passing through the point \( (-8, -8) \) and parallel to the line represented by \( -x + 8y = 48 \), we first need to determine the slope of the given line. We can rewrite that equation in slope-intercept form \( y = mx + b \): 1. Rearranging the given line: \(-x + 8y = 48\) \(8y = x + 48\) \(y = \frac{1}{8}x + 6\) So, the slope \( m \) of the given line is \( \frac{1}{8} \). Since parallel lines have the same slope, the slope of our new line is also \( \frac{1}{8} \). (a) Using the point-slope form of the line equation, \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = (-8, -8) \): \[ y + 8 = \frac{1}{8}(x + 8) \] Now, simplify this: \[ y + 8 = \frac{1}{8}x + 1 \] \[ y = \frac{1}{8}x + 1 - 8 \] \[ y = \frac{1}{8}x - 7 \] So, in slope-intercept form, the equation is: \[ y = \frac{1}{8}x - 7 \] (b) Now, to convert this into standard form \( Ax + By = C \): Start with the slope-intercept form: \[ y = \frac{1}{8}x - 7 \] Multiply through by 8 to eliminate the fraction: \[ 8y = x - 56 \] Now, rearranging gives: \[ -x + 8y = -56 \] To further align with standard form (where \( A \) is positive): \[ x - 8y = 56 \] So, in standard form, the equation is: \[ x - 8y = 56 \]