Geo2.39

From Exampleproblems

Prove that the perpendicular bisectors of the sides of a triangle are concurrent.

Let $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)\,$ be the vertices of the triangle ABC.

If D,E,F are the midpoints of the sides $BC,CA,AB\,$ respectively,then

$D=\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right),E=\left(\frac{x_3+x_1}{2},\frac{y_3+y_1}{2}\right),F=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\,$

Slope of BC is $y_2-\frac{y_3}{x_2-x_3}\,$

Therefore,slope of the perpendicular bisector to the side is $-\frac{x_2-x_3}{y_2-y_3}\,$

Equation to the perpendicular bisector thro'D is

$y-\frac{y_2+y_3}{2}=-\frac{x_2-x_3}{y_2-y_3}(x-\frac{x_2+x_3}{2}\,$

Simplifying, we have

$L_1\equiv 2x(x_2-x_3)+2y(y_2-y_3)-(x_2^{2}-x_3^{2}-(y_2^{2}-y_3^{2})=0\,$

Similarly the other perpendicular bisectors are

$L_2\equiv 2x(x_3-x_1)+2y(y_3-y_1)-(x_3^{2}-x_1^{2}-(y_3^{2}-y_1^{2})=0\,$

$L_3\equiv 2x(x_1-x_2)+2y(y_1-y_2)-(x_1^{2}-x_2^{2}-(y_1^{2}-y_2^{2})=0\,$

Now $1.L_1+1.L_2+1.L_3=0\,$ implies $L_1=0,L_2,L_3=0\,$ are concurrent.

Hence the required is proved.

Main Page:Geometry:Straight Lines-I

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