# ANSWER WITH SOLUTION: Find the value of K so that the lines 3x+ky-6=0 and 9x-6y+2=0 will be parallel and perpendicular

1
by analyticgeometry

• Brainly User
2016-01-29T12:48:57+08:00
1.)  As Parallel:  The slopes must be the same, and the y-intercepts different.
Transform to slope-intercept 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