Question 911963
For what value of {{{k}}} are the graphs of {{{12y=-3x+8}}} and {{{6y=kx-5}}} perpendicular?

the lines are perpendicular if slopes  have opposite signs, and their values are reciprocals

or, {{{m[1]=-1/m[2]}}}


{{{12y=-3x+8}}} ...find the slope

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

{{{y= -(1/4)x+2/3 }}} .... slope {{{m[1]= -(3/12) }}}


{{{6y=kx-5}}} ...find the slope

{{{y=(k/6)x-5/6}}}.... slope {{{m[2]= (k/6) }}}


now find {{{k}}} that will make {{{m[1]=-1/m[2]}}}


{{{-(3/12) =-1/ (k/6) }}} ....solve for {{{k}}} 


{{{-(3/12) =-6/k }}} ...cross multiply

{{{-3k =-6*12 }}} 

{{{k =-(6*12)/-3 }}} 

{{{k =(cross(6)2*12)/cross(3) }}} 

{{{k =2*12 }}} 

{{{highlight(k =24) }}}

then 
{{{m[2]= (k/6) }}}=>{{{m[2]= 24/6=4 }}} and {{{y=(24/6)x-5/6}}}=>{{{y= 4x-5/6 }}}

so, {{{y= -(1/4)x+2/3 }}} will be perpendicular to {{{y=(k/6)x-5/6}}} if  {{{highlight(k =24) }}}


let see it on a graph:

{{{ graph( 600, 600, -10, 10, -10, 10, 4x-5/6, -(1/4)x+2/3) }}}