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

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(B) The perpendicular bisector of $AB$ is $2x+3y+1=0$. The slope of this line is $-2/3$. Thus,the slope of $AB$ is $3/2$.The equation of $AB$ is $y-2

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$x-y-3=0$

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(B) The perpendicular bisector of $AB$ is $2x+3y+1=0$. The slope of this line is $-2/3$. Thus,the slope of $AB$ is $3/2$.The equation of $AB$ is $y-2 = \frac{3}{2}(x-3)$,which simplifies to $3x-2y-5=0$.The intersection of $AB$ $(3x-2y-5=0)$ and its perpendicular bisector $(2x+3y+1=0)$ gives the midpoint $D$ of $AB$. Solving these,we get $D=(1,-1)$.Since $D$ is the midpoint of $AB$,$\frac{3+x_B}{2}=1$ and $\frac{2+y_B}{2}=-1$,so $B=(-1,-4)$.The perpendicular bisector of $AC$ is $x+2y-2=0$. The slope of this line is $-1/2$. Thus,the slope of $AC$ is $2$.The equation of $AC$ is $y-2 = 2(x-3)$,which simplifies to $2x-y-4=0$.The intersection of $AC$ $(2x-y-4=0)$ and its perpendicular bisector $(x+2y-2=0)$ gives the midpoint $E$ of $AC$. Solving these,we get $E=(2,0)$.Since $E$ is the midpoint of $AC$,$\frac{3+x_C}{2}=2$ and $\frac{2+y_C}{2}=0$,so $C=(1,-2)$.The equation of side $BC$ passing through $B(-1,-4)$ and $C(1,-2)$ is $y-(-4) = \frac{-2-(-4)}{1-(-1)}(x-(-1))$.$y+4 = \frac{2}{2}(x+1) \implies y+4 = x+1 \implies x-y-3=0$.

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

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