Final answer:
To find the equation of a line parallel to -8x+5y=7 and passing through the point (4,-5), the equation is y=(8/5)x - 5. To find the equation of a line perpendicular to -8x+5y=7 and passing through the point (4,-5), the equation is y=(-5/8)x - 45/8.
Step-by-step explanation:
To find the equation of a line parallel to -8x+5y=7 and passing through the point (4,-5), we need to find the slope of the given line and use it to construct the equation. Rearranging the equation -8x+5y=7 into slope-intercept form (y=mx+b), we obtain y=(8/5)x + 7/5. The slope of this line is 8/5. A line that is parallel to this line will have the same slope. So the equation of the parallel line passing through the point (4,-5) is y=(8/5)x + b. Plugging in the coordinates of the point, we can solve for b: -5=(8/5)(4) + b, which gives b=-25/5=-5. Therefore, the equation is y=(8/5)x - 5.
To find the equation of a line perpendicular to -8x+5y=7 and passing through the point (4,-5), we again need to find the slope of the given line. The slope of -8x+5y=7 is 8/5. The slope of a line perpendicular to this line will be the negative reciprocal of 8/5, which is -5/8. So the equation of the perpendicular line passing through the point (4,-5) is y=(-5/8)x + b. Plugging in the coordinates of the point, we can solve for b: -5=(-5/8)(4) + b, which gives b=-45/8. Therefore, the equation is y=(-5/8)x - 45/8.