Question 972365
You have 3 lines here:
AB
BC
CA
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You need to find the equations for the
perpendicular bisectors of each of these
lines
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AB
{{{ ( -16 + 1 ) / 2 = -15/2 }}}
{{{ x[AB] = -15/2 }}}
{{{ ( -1 + 6 ) / 2 = 5/2 }}}
{{{ y[AB] = 5/2 }}}
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So the center of the line AB is
at ( -15/2 , 5/2 )
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What's he slope of AB?
{{{ m = ( -1 - 6 ) / ( -16 - 1 ) }}}
{{{ m = 7/17 }}}
The slope perpendicular to this is:
{{{ m = -17/7 }}}
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Now use point-slope formula:
{{{ ( y - 6 ) / ( x - 1 ) = -17/7 }}}
{{{ y - 6 = (-17/7)*( x - 1 ) }}}
{{{ 7y - 42 = -17x + 17 }}}
{{{ 7y = -17x + 59 }}}
{{{ y = (-17/7)*x + 59/7 }}}
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This is the equation of the perpendicular bisector of AB
( if my math is right )
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Now do the exact same steps for BC and CA
The 3 lines you get should meet at the same point.
( the simultaneous solution ) 
This is the point equidistant from the 3 given points
Hope this helps