Question 1151416

If the line represented by the equation 

{{{4x+ky-10=0 }}}
{{{ky=-4x+10 }}}
{{{y=-(4/k)x+10/k }}}=>slope is {{{m=-(4/k)}}}

is perpendicular to the line with the equation 
{{{2x-10y+7=0}}} 
{{{2x+7=10y}}} 
{{{y=(1/5)x+7/10}}}=>slope is {{{m[1]=1/5}}}

perpendicular lines have slopes negative reciprocal to each other

{{{m=-1/m[1]}}}

substitute both

{{{-(4/k)=-1/(1/5)}}}.....solve for {{{k}}}

{{{-(4/k)=-5}}}.......both sides multiply by {{{-1}}}

{{{4/k=5}}}
{{{4=5k}}}
{{{k=4/5}}}

so, the value of {{{k}}} is {{{4/5}}}

{{{y=-(4/(4/5))x+10/(4/5) }}}

{{{y=-5x+25/2 }}} is perpendicular to {{{y=(1/5)x+7/10}}}


{{{ graph( 600, 600, -10, 10, -10, 10, -5x+25/2, (1/5)x+7/10) }}}