1.)
As Parallel: The slopes must be the same, and the yintercepts different. Transform to slopeintercept form, y = mx + b
Equation 1: 9x  6y + 2 = 0
6y = 9x  2
6y/6 = 9x/6  2/6
y =
+
m = 3/2 Equation 2: 3x + ky  6 = 0
ky =  3x + 6
ky/k = 3x/k + 6/k
y = 
+
m = 
Solve for k to make the system parallel:

3k = (3) (2)
3k = 6
3k/3 = 6/3
k = 2 Therefore, Equation 2 is :
3x 2y  6 = 0 2y = 3x + 6
2y/2 = 3x/2 + 6/2
y =
 3
m = 3/2 (same slope as Equation 1)
2.)
As perpendicular, the product of the slopes is 1.
Equation 1: m = 3/2 Equation 2: m = 2/3 (3/2) (2/3) = 1 ⇒ Product of slopes is 1.
Solve for k to make the the two functions perpendicular to each other:
3x + ky  6 = 0
ky = 3x + 6
ky/k = 3x/k + 6/k
y = 
m =
Since the slope of the second equation must be
, then:
3/k = 2/3
2k = (3) (3)
2k = 9
2k/2 = 9/2
k = 9/2 To make the lines perpendicular to each other, Equation 2 must be: 3x + ky  6 = 0
3x +  6 = 0 2/9 (9y/2 = 3x + 6) 2/9
y =
m = 2/3 ANSWER: For parallel equations, k = 2. For perpendicular equations, k = 9/2