Question 1091924
At A,{{{y=0}}},
{{{3x+2(0)=24}}}
{{{x=8}}}
A:(8,0)
At B, {{{x=0}}}
{{{3(0)+2y=24}}}
{{{y=12}}}
B:(0,12)
The slope of AB is,
{{{m=(12-0)/(0-8)=-12/8=-3/2}}}
The perpendicular bisector to AB would have a slope,
{{{m*(-3/2)=-1}}}
{{{m=2/3}}}
and it goes through (0,-1),
{{{y-(-1)=(2/3)(x-0)}}}
{{{y+1=(2/3)x}}}
So now is (0,-1) point C? 
Here's a picture of the perpendicular bisector to AB through (0,-1) but now unless (0,-1) is point C, I'm confused as to where C would be.
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If that's the case,
{{{A=(1/2)(12-(-1))(8)}}}
{{{A=52}}}
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*[illustration dx2.JPG].
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If not, please provide additional information.