Find equations of the lines that pass through the given point and are parallel to and perpendicular to the given line. \( 8 x-2 y=9,(4,1) \) (a) parallel to the given line (b) perpendicular to the given line (

To find the equation of the line parallel to the line given by \( 8x - 2y = 9 \), we first need to determine its slope. By rearranging the equation into slope-intercept form \( y = mx + b \), we get: \[ 2y = 8x - 9 \] \[ y = 4x - \frac{9}{2} \] The slope (m) of the given line is 4. A line that is parallel to this will have the same slope. Thus, using the point (4, 1), we can write the equation of the parallel line as: \[ y - 1 = 4(x - 4) \] Simplifying this gives: \[ y = 4x - 15 \] Now, for the line perpendicular to the given line, the slope will be the negative reciprocal of 4, which is \(-\frac{1}{4}\). Using the point (4, 1), we can write the equation of the perpendicular line as: \[ y - 1 = -\frac{1}{4}(x - 4) \] Simplifying this gives: \[ y = -\frac{1}{4}x + 2 \] So, the equations are: (a) \( y = 4x - 15 \) (parallel) (b) \( y = -\frac{1}{4}x + 2 \) (perpendicular)