If the line x+ay+1=0 is perpendicular to the line 2x-by+1=0, and parallel to the line x-(b-3)y-1=0, then what are a and b?
x + ay + 1 = 0 ---> ay = -x - 1 ---> y = (-1/a)·x - (-1/a) ---> slope = -1/a
2x - by + 1 = 0 ---> -by = -2x - 1 ---> y = (2/b)·x + (1/b) ---> slope = 2/b
Since the above two lines are perpendicular, their slopes are negative reciprocals
---> -1/a = -b/2 ---> a = 2/b
x - (b - 3)·y - 1 = 0 ---> -(b - 3)·y = -x + 1 ---> (b - 3)·y = x - 1 ---> y = ( 1/(b - 3) )·x - 1/(b - 3)
---> slope = 1/(b - 3)
Since this line is parallel to the first line, they have equal slopes: -1/a = 1/(b - 3) ---> 3 - b = a
Combining a = 2/b with 3 - b = a ---> 3 - b = 2/b ---> 3b - b2 = 2 ---> b2 - 3b + 2 = 0
---> (b - 2)(b - 1) = 0 ---> either b = 2 or b = 1
If b = 2 ---> a = 2/b ---> a = 2/2 = 1
If b = 1 ---> a = 2/b ---> a = 2/1 = 2