Question 1195341

Points A and B are the coordinates of the x and y intercepts of the line 

{{{2x-3y=6}}}

x intercept: 
{{{2x-3*0=6}}}
{{{2x=6}}}
{{{x=3}}}=> A=({{{3}}},{{{0}}})

y intercept:
{{{2*0-3y=6}}}
{{{-3y=6}}}
 {{{y=6/-3}}}
{{{y=-2}}} => B=({{{0}}},{{{-2}}})

slope of given line (also a segment AB) is {{{m=(-2-0)/(0-3)=-2/-3=2/3}}}


  the equation of a perpendicular bisector to the line segment □(→┬AB ) will have a slope negative reciprocal to the slope of given line

so, slope is {{{m=-1/(2/3)=-3/2}}}

intersection of the line segment {{{AB}}} and a perpendicular bisector will be midpoint of the line segment {{{AB}}} which is

M({{{(3+0)/2}}},{{{(0-2)/2}}})=({{{3/2}}},{{{-1}}})

 the equation of a perpendicular bisector will be

{{{y-y[1]=m(x-x[1])}}}....substitute {{{m=-3/2}}} and M=({{{3/2}}},{{{-1}}})

{{{y-(-1)=-(3/2)(x-3/2)}}}

{{{y+1=-(3/2)x-(3/2)(-3/2)}}}

{{{y=-(3/2)x+9/4-1}}}

{{{y=-(3/2)x+5/4}}}-> the equation of a perpendicular bisector 



{{{ drawing( 600, 600, -10, 10, -10, 10, circle(0,-2,.12), locate(0,-2,B(0,-2)),
circle(3,0,.12), locate(3,0.5,A(3,0)), green(line(3,0,0,-2)),circle(3/2,-1,.12), locate(3/2,-1,M(3/2,-1)),
graph( 600, 600, -10, 10, -10, 10,(2x)/3-2, -(3/2)x+5/4)) }}}