ABC is an isosceles triangle. If the coordina | vedclass

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(A) Let the vertices of the base be $B(1, 3)$ and $C(-2, 7)$.The midpoint of $BC$ is $M = \left( \frac{1-2}{2}, \frac{3+7}{2} \right) = \left( -\frac{

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$\left( -\frac{1}{2}, 5 \right)$

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(A) Let the vertices of the base be $B(1, 3)$ and $C(-2, 7)$.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 $m = \frac{7-3}{-2-1} = \frac{4}{-3} = -\frac{4}{3}$.The perpendicular bisector of $BC$ passes through $M\left( -\frac{1}{2}, 5 \right)$ with slope $m' = -\frac{1}{m} = \frac{3}{4}$.The equation of the perpendicular bisector is $y - 5 = \frac{3}{4}(x + \frac{1}{2})$.Vertex $A$ must lie on this perpendicular bisector.Checking the options:For option $A$: $\left( -\frac{1}{2}, 5 \right)$ is the midpoint $M$ itself,which cannot be vertex $A$ as it would result in a degenerate triangle.For option $B$: $x = -\frac{1}{8}, y = 5$. Substituting into the equation: $5 - 5 = \frac{3}{4}(-\frac{1}{8} + \frac{1}{2}) \Rightarrow 0 = \frac{3}{4}(\frac{3}{8}) = \frac{9}{32}$,which is false.Checking the slope condition for $A$ to be equidistant from $B$ and $C$,$A$ must lie on the perpendicular bisector. Re-evaluating the options provided,the point $\left( -\frac{1}{2}, 5 \right)$ is the midpoint. If the question implies $A$ lies on the perpendicular bisector,and given the options,option $A$ is the midpoint,which is usually excluded. However,if we test the coordinates,$\left( -\frac{1}{2}, 5 \right)$ satisfies the perpendicular bisector equation.

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

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