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(D) The slope of line segment $AB$ is $m_{AB} = \frac{-5-3}{6-(-2)} = \frac{-8}{8} = -1$. Since the perpendicular bisector is perpendicular to $AB$,it

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$x-y=3$

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(D) The slope of line segment $AB$ is $m_{AB} = \frac{-5-3}{6-(-2)} = \frac{-8}{8} = -1$. Since the perpendicular bisector is perpendicular to $AB$,its slope $m$ must satisfy $m 	imes m_{AB} = -1$,so $m = 1$. The midpoint of $AB$ is $M = \left( \frac{-2+6}{2}, \frac{3-5}{2} ight) = (2, -1)$. The equation of the line with slope $m=1$ passing through $(2, -1)$ is given by $y - y_1 = m(x - x_1)$. Substituting the values: $y - (-1) = 1(x - 2)$,which simplifies to $y + 1 = x - 2$,or $x - y = 3$.

The equation of the perpendicular bisector of the line segment joining $A(-2, 3)$ and $B(6, -5)$ is

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