Answer to Question #139729 in Analytic Geometry for Aravind

Question #139729
The coordinates of the vertices of a triangle are A( x1, y1), B( x2, y2), and C(x3,y3) .Then what's the co ordinates of the circumcentre.
1
Expert's answer
2020-10-26T20:37:00-0400

A(x1,y1), B(x2,y2), C(x3,y3)A(x_1,y_1), \ B(x_2,y_2),\ C(x_3,y_3)

To calculate the coordinate of the circumcenter, we have to solve any 2 bisector equations of any of the three lines AB, BC, or AC, then find out the intersection points. i.e.



The midpoint of side AB = x1+x22,y1+y22=(x12,y12)\lgroup \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \rgroup= (x_{12}, y_{12})


And the slope of AB = y2y1x2x1=m12\dfrac{y_2-y_1}{x_2-x_1} = m_{12}

\therefore The slope of the perpendicular bisector of side AB = 1m12=m21\frac{-1}{m_{12}} = m_{21}


The equation of the perpendicular bisector of AB is therefore,

yy12=m21(xx12) (i)y-y_{12} = m_{21}(x-x_{12}) \ ---(i)


Similarly,

For side AC

The midpoint of side AC =

x1+x32,y1+y32=(x13,y13)\lgroup \dfrac{x_1 + x_3}{2}, \dfrac{y_1 + y_3}{2} \rgroup= (x_{13}, y_{13})


And its slope = y3y1x3x1=m13\dfrac{y_3-y_1}{x_3-x_1} = m_{13}

Making the slope of its perpendicular bisector = 1m13=m31\frac{-1}{m_{13}} = m_{31}


The equation of the perpendicular bisector of length AC is therefore,

yy13=m31(xx13) (ii)y-y_{13} = m_{31}(x-x_{13}) \ ---(ii)


On solving equations (i) and (ii), the values of x and y is the coordinate of the circumcenter


OR, using formula method;


Circumcenter(X,Y)=x1sin2A+x2sin2B+x3sin2Csin2A+sin2B+sin2C,y1sin2A+y2sin2B+y3sin2Csin2A+sin2B+sin2CCircumcenter(X, Y) = \lgroup \dfrac{x_1 sin 2A + x_2 sin 2B + x_3 sin 2C}{sin 2A + sin 2B + sin 2C},\dfrac{y_1 sin 2A + y_2 sin 2B + y_3 sin 2C}{sin 2A + sin 2B + sin 2C}\rgroup

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