Question 1169654

Find an equation of the line perpendicular to {{{6y=x+2}}} that passes through the point ({{{3}}},{{{1}}}).

If two lines are perpendicular, the slopes are negative reciprocals.

so, find a slope of {{{6y=x+2}}}

{{{y=x/6+2/6}}}
 
{{{y=(1/6)x+1/3}}}=> slope {{{m=1/6}}}

-> negative reciprocal is {{{-1/m=-1/(1/6)=-6}}}

so, the slope of the perpendicular line is {{{-6}}}

then perpendicular line will be:

{{{y=mx+b}}}........substitute slope ={{{-6}}}

{{{y=-6x+b}}}.......use given point ({{{3}}},{{{1}}}) to calculate {{{b}}}

{{{1=-6*3+b}}}

{{{1=-18+b}}}

{{{b=1+18}}}

{{{b=19}}}

and your equation is: {{{y=-6x+19}}}


{{{drawing ( 600, 600, -10, 10, -10, 20,
circle(3,1,.12),locate(3,1,p(3,1)),
graph( 600, 600, -10, 10, -10, 20, (1/6)x+1/3, -6x+19)) }}}