In a triangle ABC,2x+3y+1=0 and x+2y-12=0 are | vedclass

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(C) The slope of the perpendicular bisector of $AB$ is $m_1 = -\frac{2}{3}$. Thus,the slope of $AB$ is $m_{AB} = -\frac{1}{m_1} = \frac{3}{2}$.Since $

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$\frac{5}{3}$

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(C) The slope of the perpendicular bisector of $AB$ is $m_1 = -\frac{2}{3}$. Thus,the slope of $AB$ is $m_{AB} = -\frac{1}{m_1} = \frac{3}{2}$.Since $AB$ passes through $A(3,2)$,its equation is $y-2 = \frac{3}{2}(x-3) \Rightarrow 3x-2y-5=0$.The intersection of $AB$ and its perpendicular bisector $2x+3y+1=0$ gives the midpoint $E$ of $AB$. Solving $3x-2y-5=0$ and $2x+3y+1=0$,we get $E(1,-1)$.Let $B$ be $(x_1, y_1)$. Since $E$ is the midpoint of $AB$,$\frac{x_1+3}{2} = 1 \Rightarrow x_1 = -1$ and $\frac{y_1+2}{2} = -1 \Rightarrow y_1 = -4$. So,$B(-1,-4)$.The slope of the perpendicular bisector of $AC$ is $m_2 = -\frac{1}{2}$. Thus,the slope of $AC$ is $m_{AC} = -\frac{1}{m_2} = 2$.Since $AC$ passes through $A(3,2)$,its equation is $y-2 = 2(x-3) \Rightarrow 2x-y-4=0$.The intersection of $AC$ and its perpendicular bisector $x+2y-12=0$ gives the midpoint $D$ of $AC$. Solving $2x-y-4=0$ and $x+2y-12=0$,we get $D(4,4)$.Let $C$ be $(x_2, y_2)$. Since $D$ is the midpoint of $AC$,$\frac{x_2+3}{2} = 4 \Rightarrow x_2 = 5$ and $\frac{y_2+2}{2} = 4 \Rightarrow y_2 = 6$. So,$C(5,6)$.The slope of $BC$ is $\frac{6-(-4)}{5-(-1)} = \frac{10}{6} = \frac{5}{3}$.

In a $\triangle ABC$,$2x+3y+1=0$ and $x+2y-12=0$ are the perpendicular bisectors of its sides $AB$ and $AC$ respectively. If $A$ is $(3,2)$,then the slope of the side $BC$ is

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