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(C) Since $ABC$ is an isosceles triangle with base $BC$,vertex $A(x, y)$ must lie on the perpendicular bisector of segment $BC$.The midpoint of $BC$ i

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$\left( \frac{5}{6}, 6 \right)$

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(C) Since $ABC$ is an isosceles triangle with base $BC$,vertex $A(x, y)$ must lie on the perpendicular bisector of segment $BC$.The midpoint of $BC$ is $M = \left( \frac{1 - 2}{2}, \frac{3 + 7}{2} \right) = \left( -\frac{1}{2}, 5 \right)$.The slope of $BC$ is $m_{BC} = \frac{7 - 3}{-2 - 1} = \frac{4}{-3} = -\frac{4}{3}$.The slope of the perpendicular bisector is $m_{\perp} = -\frac{1}{m_{BC}} = \frac{3}{4}$.The equation of the perpendicular bisector is $y - 5 = \frac{3}{4}(x + \frac{1}{2})$.$4y - 20 = 3x + \frac{3}{2}$ $\Rightarrow 8y - 40 = 6x + 3$ $\Rightarrow 6x - 8y + 43 = 0$.Checking option $(C)$: $6(\frac{5}{6}) - 8(6) + 43 = 5 - 48 + 43 = 0$. Since this point satisfies the equation,it is a valid coordinate for vertex $A$.

$ABC$ is an isosceles triangle. If the coordinates of the base are $B(1, 3)$ and $C(-2, 7)$,the coordinates of vertex $A$ can be

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