If (2, 3),(4, 5),and (-2, 11) are the vertice | vedclass

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(C) Let the vertices be $A(2, 3)$,$B(4, 5)$,and $C(-2, 11)$.First,check the slopes of the sides:Slope of $AB = \frac{5-3}{4-2} = \frac{2}{2} = 1$.Slop

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$2\sqrt{5}$

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(C) Let the vertices be $A(2, 3)$,$B(4, 5)$,and $C(-2, 11)$.First,check the slopes of the sides:Slope of $AB = \frac{5-3}{4-2} = \frac{2}{2} = 1$.Slope of $BC = \frac{11-5}{-2-4} = \frac{6}{-6} = -1$.Slope of $AC = \frac{11-3}{-2-2} = \frac{8}{-4} = -2$.Since (Slope of $AB$) $	imes$ (Slope of $BC$) $= 1 	imes (-1) = -1$,the triangle is a right-angled triangle with the right angle at vertex $B(4, 5)$.In a right-angled triangle,the circumcenter is the midpoint of the hypotenuse $AC$.Circumcenter $O = \left( \frac{2 + (-2)}{2}, \frac{3 + 11}{2} ight) = \left( \frac{0}{2}, \frac{14}{2} ight) = (0, 7)$.The distance between vertex $B(4, 5)$ and the circumcenter $O(0, 7)$ is given by the distance formula:$d = \sqrt{(4-0)^2 + (5-7)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$.

If $(2, 3)$,$(4, 5)$,and $(-2, 11)$ are the vertices of a triangle,what is the distance between the vertex $(4, 5)$ and the circumcenter?

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