Question 1207326


what is the equation of the perpendicular bisector of ag, where 

a ({{{12}}},{{{-1}}}) and g({{{3}}},{{{9}}})?


the equation of the ag will have a slope

{{{m=(9-(-1))/(3-12)=-10/9}}}

{{{y-y[1]= m(x-x[1])}}}....plug in slope and point g({{{3}}},{{{9}}})

{{{y-9= -(10/9)(x-3)}}}

{{{y-9= -(10/9)x-(-(10/9)3)}}}

{{{y-9= -(10/9)x+10/3}}}

{{{y= -(10/9)x+10/3+9}}}

{{{y= -(10/9)x+37/3}}}


the equation of the perpendicular bisector will have a slope negative reciprocal to {{{-10/9 }}}and it is

{{{m=-1/(-10/9)=9/10}}}

bisector will pass through midpoint of ag which is ({{{(12+3)/2}}}, {{{(-1+9)/2}}})=({{{15/2}}},{{{4}}})

use slope - point form

{{{y-y[1]= m(x-x[1])}}}....plug in slope and point ({{{15/2}}},{{{4}}})

{{{y-4=(9/10)(x-15/2)}}}

{{{y-4=(9/10)x-27/4}}}

{{{y=(9/10)x-27/4+4}}}

{{{y=(9/10)x-11/4 }}}=> answer


{{{ drawing( 600, 600, -10, 15, -10, 10, 
circle(12,-1,.12), circle(3,9,.12),
locate(12,-1,a), locate(3,9,g),

graph( 600, 600, -10, 15, -10, 10, -(10/9)x+37/3, (9/10)x-11/4 )) }}}