The distance (in units) between the circumcen | vedclass

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(C) Let the vertices be $A(1, 2)$,$B(3, -1)$,and $C(4, 0)$.First,calculate the lengths of the sides:$AB = \sqrt{(3-1)^2 + (-1-2)^2} = \sqrt{2^2 + (-3)

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$\frac{11 \sqrt{2}}{30}$

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(C) Let the vertices be $A(1, 2)$,$B(3, -1)$,and $C(4, 0)$.First,calculate the lengths of the sides:$AB = \sqrt{(3-1)^2 + (-1-2)^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4+9} = \sqrt{13}$$AC = \sqrt{(4-1)^2 + (0-2)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9+4} = \sqrt{13}$Since $AB = AC$,the triangle is isosceles.Centroid $G = \left(\frac{1+3+4}{3}, \frac{2-1+0}{3}\right) = \left(\frac{8}{3}, \frac{1}{3}\right)$.For an isosceles triangle,the circumcentre $O$ lies on the altitude from $A$ to $BC$. The slope of $BC$ is $m_{BC} = \frac{0 - (-1)}{4 - 3} = 1$. The altitude from $A$ has slope $-1$ and passes through $(1, 2)$,so its equation is $y - 2 = -1(x - 1) \Rightarrow y = -x + 3$.The circumcentre $O$ is the intersection of the perpendicular bisectors. The perpendicular bisector of $AC$ passes through the midpoint $E\left(\frac{1+4}{2}, \frac{2+0}{2}\right) = \left(\frac{2.5}{1}, 1\right)$ with slope $m = -\frac{1}{m_{AC}} = -\frac{1}{-2/3} = \frac{3}{2}$.Equation of perpendicular bisector of $AC$: $y - 1 = \frac{3}{2}(x - 2.5)$ $\Rightarrow y = 1.5x - 3.75 + 1$ $\Rightarrow y = 1.5x - 2.75$.Solving $y = -x + 3$ and $y = 1.5x - 2.75$: $-x + 3 = 1.5x - 2.75$ $\Rightarrow 2.5x = 5.75$ $\Rightarrow x = 2.3 = \frac{23}{10}$.Then $y = -2.3 + 3 = 0.7 = \frac{7}{10}$. So $O = \left(\frac{23}{10}, \frac{7}{10}\right)$.The distance $OG = \sqrt{(\frac{8}{3} - \frac{23}{10})^2 + (\frac{1}{3} - \frac{7}{10})^2} = \sqrt{(\frac{80-69}{30})^2 + (\frac{10-21}{30})^2} = \sqrt{(\frac{11}{30})^2 + (-\frac{11}{30})^2} = \frac{11}{30} \sqrt{2}$.

The distance (in units) between the circumcentre and the centroid of the triangle formed by the vertices $(1, 2)$,$(3, -1)$ and $(4, 0)$ is

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