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(C) Let the points be $A(1, 2)$ and $B(-2, 0)$.The midpoint of $AB$ is $\left(\frac{1-2}{2}, \frac{2+0}{2}\right) = \left(-\frac{1}{2}, 1\right)$.The

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$6x + 4y = 1$

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(C) Let the points be $A(1, 2)$ and $B(-2, 0)$.The midpoint of $AB$ is $\left(\frac{1-2}{2}, \frac{2+0}{2}\right) = \left(-\frac{1}{2}, 1\right)$.The slope of $AB$ is $m = \frac{0-2}{-2-1} = \frac{-2}{-3} = \frac{2}{3}$.The slope of the perpendicular bisector is $m' = -\frac{1}{m} = -\frac{3}{2}$.The equation of the line passing through $\left(-\frac{1}{2}, 1\right)$ with slope $-\frac{3}{2}$ is:$y - 1 = -\frac{3}{2}(x + \frac{1}{2})$Multiplying by $4$ to clear denominators:$4(y - 1) = -6(x + \frac{1}{2})$$4y - 4 = -6x - 3$$6x + 4y = 1$.

The equation of the perpendicular bisector of the line segment joining the points $(1, 2)$ and $(-2, 0)$ is:

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