SOLUTION: For what value of k are the graphs of 12y=-3x+8 and 6y=kx-5 perpendicular?

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Question 911963: For what value of k are the graphs of 12y=-3x+8 and 6y=kx-5 perpendicular?
Found 3 solutions by jim_thompson5910, stanbon, MathLover1:
Answer by jim_thompson5910(35256)   (Show Source):
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12y=-3x+8

y=(-3x+8)/12

y = (-3/12)x + 8/12

y = (-1/4)x + 2/3

The slope is -1/4

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6y=kx-5

y=(kx-5)/6

y = (k/6)x - 5/6

The slope is k/6

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The two slopes -1/4 and k/6 must multiply to -1 if they are form perpendicular lines

(-1/4)*(k/6) = -1

-k/24 = -1

-k = -1*24

-k = -24

k = 24

So the answer is k = 24

Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website!
For what value of k are the graphs of 12y=-3x+8 and 6y=kx-5 perpendicular?
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Find the slopes::
y = -3/12 *x + 2/3
slope = -1/4
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y = (k/6)x - (5/6)
slope = k/6
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To be perpendicular, solve
(-1/4)(k/6) = -1
Solve for "k"::
(1/4)(k/6) = 1
k = 24
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Cheers,
Stan H.
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Answer by MathLover1(20850)   (Show Source):
You can put this solution on YOUR website!
For what value of are the graphs of and perpendicular?
the lines are perpendicular if slopes have opposite signs, and their values are reciprocals
or,

...find the slope

.... slope

...find the slope
.... slope

now find that will make

....solve for

...cross multiply

then
=> and =>
so, will be perpendicular to if

let see it on a graph: