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(D) The intersection of the perpendicular bisectors of the sides of a triangle is the circumcenter $O$. Solving the equations $x-y+5=0$ and $x+2y=0$,w

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$23x-14y+100=0$

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(D) The intersection of the perpendicular bisectors of the sides of a triangle is the circumcenter $O$. Solving the equations $x-y+5=0$ and $x+2y=0$,we get $x = -10/3$ and $y = 5/3$. Thus,the circumcenter $O$ is $(-10/3, 5/3)$.Since $B$ is the reflection of $A(1,-2)$ across the line $x-y+5=0$,we have $\frac{x_B-1}{1} = \frac{y_B+2}{-1} = -2 \frac{1-(-2)+5}{1^2+(-1)^2} = -2 \frac{8}{2} = -8$. This gives $x_B = -7$ and $y_B = 6$. So $B = (-7, 6)$.Since $C$ is the reflection of $A(1,-2)$ across the line $x+2y=0$,we have $\frac{x_C-1}{1} = \frac{y_C+2}{2} = -2 \frac{1+2(-2)}{1^2+2^2} = -2 \frac{-3}{5} = 6/5$. This gives $x_C = 1+6/5 = 11/5$ and $y_C = -2+12/5 = 2/5$. So $C = (11/5, 2/5)$.The perpendicular bisector of $BC$ passes through the circumcenter $O(-10/3, 5/3)$ and the midpoint $M$ of $BC$. The midpoint $M$ is $(\frac{-7+11/5}{2}, \frac{6+2/5}{2}) = (-12/5, 16/5)$.The slope of the line passing through $O(-10/3, 5/3)$ and $M(-12/5, 16/5)$ is $m = \frac{16/5 - 5/3}{-12/5 - (-10/3)} = \frac{48/15 - 25/15}{-36/15 + 50/15} = \frac{23/15}{14/15} = 23/14$.The equation of the line is $y - 5/3 = 23/14(x + 10/3)$,which simplifies to $14(3y-5) = 69(x+10/3)$ $\Rightarrow 42y - 70 = 69x + 230$ $\Rightarrow 69x - 42y + 300 = 0$ $\Rightarrow 23x - 14y + 100 = 0$.

If the equations of the perpendicular bisectors of the sides $AB$ and $AC$ of a $\triangle ABC$ are $x-y+5=0$ and $x+2y=0$ respectively,and if $A$ is $(1,-2)$,then the equation of the perpendicular bisector of the side $BC$ is

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