x-y=0 and frac{x}{2}+frac{y}{2}=1 are respect | vedclass

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(A) Given that $x-y=0$ is the perpendicular bisector of side $AB$. Since the slope of the bisector is $1$,the slope of $AB$ is $-1$. The equation of $

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$x-2y+1=0$

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(A) Given that $x-y=0$ is the perpendicular bisector of side $AB$. Since the slope of the bisector is $1$,the slope of $AB$ is $-1$. The equation of $AB$ is $y-3 = -1(x-2)$,which simplifies to $x+y-5=0$. Solving $x-y=0$ and $x+y-5=0$ gives the midpoint $D$ of $AB$ as $(\frac{5}{2}, \frac{5}{2})$. Let $B = (x_1, y_1)$. Using the midpoint formula,$\frac{x_1+2}{2} = \frac{5}{2}$ and $\frac{y_1+3}{2} = \frac{5}{2}$,so $B = (3, 2)$. Given that $\frac{x}{2}+\frac{y}{2}=1$ (or $x+y-2=0$) is the perpendicular bisector of $AC$. Since the slope of the bisector is $-1$,the slope of $AC$ is $1$. The equation of $AC$ is $y-3 = 1(x-2)$,which simplifies to $x-y+1=0$. Solving $x+y-2=0$ and $x-y+1=0$ gives the midpoint $E$ of $AC$ as $(\frac{1}{2}, \frac{3}{2})$. Let $C = (x_2, y_2)$. Using the midpoint formula,$\frac{x_2+2}{2} = \frac{1}{2}$ and $\frac{y_2+3}{2} = \frac{3}{2}$,so $C = (-1, 0)$. The equation of side $BC$ passing through $(3, 2)$ and $(-1, 0)$ is $\frac{y-0}{2-0} = \frac{x-(-1)}{3-(-1)}$. $\frac{y}{2} = \frac{x+1}{4}$ $\Rightarrow 2y = x+1$ $\Rightarrow x-2y+1=0$.

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

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