Question 1177470
AB has a midpoint of (5.5, 8)
the slope of AB is -2/7
the equation of the line is y-y1=m(x-x1) m slope (x1, y1) point
this is at y-9=(-2/7)(x-2)
y=(-2/7)x+67/7
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the perpendicular bisector of this line has slope 7/2 and the point is (11/2, 8)
y-8=(7/2)(x-11/2)
y=(7/2)x-45/4
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BC has a midpoint of (5.5, 3.5)
slope is 1
equation is y=x-2
perpendicular bisector has slope -1 and its equation of y-3.5=-1(x-5.5)
y=-x+9
first graph are lines AB and BC
{{{graph(300,300,-10,12,-10,12,-(2/7)x+67/7,x-2)}}}
second is perpendicular bisector of AB
{{{graph(300,300,-10,12,-10,12,(7/2)x-45/4,(-2x/7)+67/7)}}}
third is perpendicular bisector of BC
{{{graph(300,300,-10,12,-10,12,x-2,-x+9)}}}
last is graph of the perpendicular bisectors
{{{graph(300,300,-10,10,-10,10,(7/2)x-45/4,-x+9)}}}

equation of perpendicular bisector of AB is y=(7/2)x-45/4
of BC it is y=-x+9