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(A) The midpoint $M$ of the line segment $AB$ is given by $(\frac{1+5}{2}, \frac{1+7}{2}) = (3, 4)$.The slope of the line segment $AB$ is $m_{AB} = \f

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$2x + 3y = 18$

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(A) The midpoint $M$ of the line segment $AB$ is given by $(\frac{1+5}{2}, \frac{1+7}{2}) = (3, 4)$.The slope of the line segment $AB$ is $m_{AB} = \frac{7-1}{5-1} = \frac{6}{4} = \frac{3}{2}$.The slope of the perpendicular bisector is $m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{3/2} = -\frac{2}{3}$.The equation of the line passing through $(3, 4)$ with slope $-\frac{2}{3}$ is given by $y - 4 = -\frac{2}{3}(x - 3)$.Multiplying by $3$,we get $3y - 12 = -2x + 6$,which simplifies to $2x + 3y = 18$.

If the coordinates of $A$ and $B$ are $(1, 1)$ and $(5, 7)$,then the equation of the perpendicular bisector of the line segment $AB$ is

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