How do you find a like that goes through a point and is perpendicular to another line (algebra) HELP!?

Question

I need to know how to solve this:

Write the equation of the line through (5,9) perpendicular to 3x+8y=75.

Answer 1

First determine the slope of the line.

3x+8y = 75

8y = -3x+75

y = (-3/8)x+75/8

The slope of the line is -3/8. Any perpendicular to this line will have slope +8/3.

(5,9) is a point on the perpendicular. The equation of the perpendicular, in point-slope form:

y-9 = (8/3)(x-5)

If you want it in slope-intercept form, you need to determine the y-intercept b using point (5,9):

b = y-mx

= 9-(8/3)5

= 9 - 40/3

= -13/3

The equation of the perpendicular, in slope-intercept form:

y = (8/3)x-13/3