Question 1127763


find the equation of the perpendicular bisect of {{{AB}}} given that {{{A }}} is ({{{2}}},{{{7}}}) and {{{B}}} is ({{{6}}},{{{-1}}}). 

first find equation of a line passing through points 
{{{A }}} ({{{2}}},{{{7}}}) and {{{B}}} ({{{6}}},{{{-1}}})

{{{y=mx+b}}} where {{{m}}} is a slope and {{{b}}} is y-intercept

{{{m=(y[2]-y[1])/(x[2]-x[1])}}}

{{{m=(-1-7)/(6-2)}}}
{{{m=-8/4}}}
{{{m=-2}}}

{{{y=-2x+b}}} ....plug in ({{{2}}},{{{7}}})

{{{7=-2*2+b}}}

{{{7=-4+b}}}

{{{7+4=b}}}

{{{b=11}}}

so, {{{y=-2x+11}}}

recall that the perpendicular line will have a slope negative reciprocal to a slope of this line

so, {{{-1/m=-1/-2=1/2}}}

 {{{y=(1/2)x+b}}}

the perpendicular bisect of {{{AB}}}, we can find coordinates o midpoint of {{{AB}}}
{{{A }}} ({{{2}}},{{{7}}}) and {{{B}}} ({{{6}}},{{{-1}}})

({{{(2+6)/2}}},{{{(7-1)/2}}})
=>({{{4}}},{{{3}}})

-> the perpendicular bisect is passing through ({{{4}}},{{{3}}}), use it to find  y-intercept {{{b}}}

 {{{3=(1/2)4+b}}}
 {{{3=2+b}}}
 {{{3-2=b}}}
{{{b=1}}}

and equation is:  {{{y=(1/2)x+1}}}


{{{drawing ( 600, 600, -10, 10, -10, 10,
circle(2,7,.12),circle(6,-1,.12),
locate(2,7,A(2,7)),locate(6,-1,B(6,-1)),
green(line(2,7,6,-1)),
graph( 600, 600, -10, 10, -10, 10, -2x+11, (1/2)x+1)) }}}