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Find the circumcenter of the triangle formed by the points (2,1),(1,-2),(-2,3)\,

Let the points be A,B,C,D\, respectively.

Midpoint of BC=D=\left({\frac  {-1}{2}},{\frac  {1}{2}}\right)\,

Midpoint of AB=F=\left({\frac  {3}{2}},{\frac  {-1}{2}}\right)\,

Slope of BC={\frac  {-2-3}{1+2}}=-{\frac  {5}{3}}\,

Let S be the circum center.Therefore,the slope of SD is{\frac  {3}{5}}\,

Equation of the perpendicular bisector SD of BC is y-{\frac  {1}{2}}={\frac  {3}{5}}(x+{\frac  {1}{2}}),6x-10y+8=0\,

Slope of AB=3\,. Slope ofSF={\frac  {-1}{3}}\,

Equation of the perpendicular bisector SF of AB is y+{\frac  {1}{2}}={\frac  {-1}{3}}(x-{\frac  {3}{2}}),2x+6y=0,x+3y=0\,

Solving these two equations,we get the circumcenter as \left(-{\frac  {6}{7}},{\frac  {2}{7}}\right)\,

Main Page:Geometry:Straight Lines-I