Question 1011465
to find a line perpendicular to the line {{{8x+y=6}}}, write this equation in a slope-intercept form {{{y=mx+b}}} where {{{m}}} is a {{{slope}}} and {{{b}}} is {{{y-intercept}}}

we will need a slope {{{m}}} because two lines are perpendicular if their slopes are negative reciprocal to each other

{{{y=highlight(-8)x+6}}}=>a slope {{{m=-8}}} 

a line perpendicular to this line will have a slope {{{m[p]=-(1/-8)}}}=> {{{m[p]=(1/8)}}}

so,{{{y=(1/8)x}}} is a line perpendicular to {{{8x+y=6}}}

for each line we need two points to graph them

{{{x}}}|{{{y}}}....for {{{8x+y=6}}}
{{{0}}}|{{{6}}}........{{{8*0+y=6}}}=>{{{y=6}}}
{{{1}}}|{{{-2}}}........{{{8*(1)+y=6}}}=>{{{8+y=6}}}=>{{{y=6-8}}}=>{{{y=-2}}}


{{{x}}}|{{{y}}}....for {{{y=(1/8)x}}}
{{{0}}}|{{{0}}}........{{{y=(1/8)0}}}=>{{{y=0}}}
{{{8}}}|{{{1}}}........{{{y=(1/8)8}}}=>{{{y=1}}}



{{{drawing( 600, 600, -10, 10, -10, 10,
circle(0,6,.13),circle(1,-2,.13),circle(0,0,.14),circle(8,1,.13),
 graph( 600, 600, -10, 10, -10, 10, -8x+6, (1/8)x)) }}}