Write an equation for a line passing through the point \( (5,-3) \) that is parallel to the line \( 5 x+7 y=9 \). Then write a second equation for a line passing through the point \( (5,-3) \) that is perpendicular to the line \( 5 x+7 y=9 \).

To solve this problem, we need to find two equations for lines based on the given conditions. ### Step 1: Find the slope of the given line The equation of the line is given in standard form: \[ 5x + 7y = 9 \] To find the slope, we can convert this equation to slope-intercept form \(y = mx + b\), where \(m\) is the slope. 1. Rearranging the equation: \[ 7y = -5x + 9 \] 2. Dividing by 7: \[ y = -\frac{5}{7}x + \frac{9}{7} \] From this, we see that the slope \(m\) of the line is \(-\frac{5}{7}\). ### Step 2: Equation of the parallel line Lines that are parallel have the same slope. Therefore, the slope of the line we want to find is also \(-\frac{5}{7}\). Using the point-slope form of the equation of a line, which is given by: \[ y - y_1 = m(x - x_1) \] where \((x_1, y_1)\) is the point through which the line passes, we can substitute \(m = -\frac{5}{7}\) and the point \((5, -3)\): \[ y - (-3) = -\frac{5}{7}(x - 5) \] Simplifying this: \[ y + 3 = -\frac{5}{7}x + \frac{25}{7} \] Subtracting 3 from both sides: \[ y = -\frac{5}{7}x + \frac{25}{7} - 3 \] Converting 3 to a fraction with a denominator of 7: \[ y = -\frac{5}{7}x + \frac{25}{7} - \frac{21}{7} \] This simplifies to: \[ y = -\frac{5}{7}x + \frac{4}{7} \] ### Step 3: Equation of the perpendicular line Lines that are perpendicular have slopes that are negative reciprocals of each other. The slope of the perpendicular line will be: \[ m = \frac{7}{5} \] Using the point-slope form again with the new slope: \[ y - (-3) = \frac{7}{5}(x - 5) \] Simplifying this: \[ y + 3 = \frac{7}{5}x - \frac{35}{5} \] Subtracting 3 (or \(\frac{15}{5}\)) from both sides: \[ y = \frac{7}{5}x - \frac{35}{5} - \frac{15}{5} \] This simplifies to: \[ y = \frac{7}{5}x - \frac{50}{5} \] Thus, we have: \[ y = \frac{7}{5}x - 10 \] ### Final Equations 1. The equation of the line parallel to \(5x + 7y = 9\) passing through \((5, -3)\) is: \[ y = -\frac{5}{7}x + \frac{4}{7} \] 2. The equation of the line perpendicular to \(5x + 7y = 9\) passing through \((5, -3)\) is: \[ y = \frac{7}{5}x - 10 \]