SOLUTION: Find the value on K for which the graphs of Kx-2y+13=0 and 9x-6y+11=0 are perpendicular

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Question 523777: Find the value on K for which the graphs of Kx-2y+13=0 and 9x-6y+11=0 are perpendicular
Answer by Maths68(1474) (Show Source):
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Kx-2y+13=0
-2y=-kx-13=0
-2y/-2=(-kx-13)/-2
y=(k/2)x+13/2............(1)
Slope = k/2

9x-6y+11=0
-6y=-9x-11
-6y/-6=(-9x-11)/-6
y=(9/6)x+11/6
y=(3/2)x+11/6
y=3x/2+11/6.............(3)
slope = 3/2

Since lines are perpendicular, multiplication of their slopes will result in (-1)
3/2 * k/2 = -1
3k/4=-1
3k=-4
k=-4/3
Value of K will be -4/3

Check
======
y=(k/2)x+13/2............(1)
Put the value of k in above equation
y=(-4/3)(1/2)x+13/2
y=(-2/3)x+13/2

.........(4)
.............(3)
Graph
Red Line y=-2x/3+13/2
Green Line y=3x/2+11/6