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(C) Let the vertices be $A(-2,-1)$,$B(6,-1)$,and $C(2,5)$.First,find the perpendicular bisector of $AB$. Since $A$ and $B$ have the same $y$-coordinat

Vedclass · Accepted Answer

$3x^2-8x+4=0$

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(C) Let the vertices be $A(-2,-1)$,$B(6,-1)$,and $C(2,5)$.First,find the perpendicular bisector of $AB$. Since $A$ and $B$ have the same $y$-coordinate,the line $AB$ is horizontal. The midpoint is $(\frac{-2+6}{2}, -1) = (2,-1)$. The perpendicular bisector is the vertical line $x=2$ ... $(i)$.Next,find the perpendicular bisector of $BC$. The midpoint of $BC$ is $(\frac{6+2}{2}, \frac{-1+5}{2}) = (4,2)$. The slope of $BC$ is $m = \frac{5-(-1)}{2-6} = \frac{6}{-4} = -\frac{3}{2}$. The slope of the perpendicular bisector is $\frac{2}{3}$. The equation is $y-2 = \frac{2}{3}(x-4)$ $\Rightarrow 3y-6 = 2x-8$ $\Rightarrow 2x-3y=2$ ... (ii).Substitute $x=2$ from $(i)$ into (ii): $2(2)-3y=2$ $\Rightarrow 4-3y=2$ $\Rightarrow 3y=2$ $\Rightarrow y=\frac{2}{3}$.The circumcentre is $(2, \frac{2}{3})$.The quadratic equation with roots $2$ and $\frac{2}{3}$ is $x^2 - (2+\frac{2}{3})x + 2(\frac{2}{3}) = 0$.$x^2 - \frac{8}{3}x + \frac{4}{3} = 0 \Rightarrow 3x^2-8x+4=0$.Thus,option $(c)$ is correct.

The quadratic equation whose roots are the coordinates of the circumcentre of the triangle formed by the points $(-2,-1), (6,-1),$ and $(2,5)$ is

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