Question 1185570

 an equation of the line perpendicular to {{{5y=x+3 }}}that passes through ({{{8}}},{{{6}}})

 {{{5y=x+3 }}} ....solve for {{{y}}}

{{{y=(1/5)x+3/5 }}} -> a slope is {{{1/5}}}

the line perpendicular to given line will have a slope negative reciprocal

{{{m=-1/(1/5)=-5}}}


using point slope formula we have

{{{y-y[1]=m(x-x[1])}}}.........substitute {{{m=-5}}} and ({{{8}}},{{{6}}})

{{{y-6=-5(x-8)}}}

{{{y=-5x+40 +6}}}

{{{y=-5x+46}}}-> the slope intercept form of this line


{{{drawing( 600, 600, -10, 50, -10, 50, circle(8,6,.12), locate(8,6,p(8,6)),
 graph( 600, 600, -10, 50, -10, 50, (1/5)x+3/5 , -5x+46)) }}}