Consider a triangle having vertices A(-2, 3), | vedclass

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(C) First,calculate the lengths of the sides of $	riangle ABC$:$AB^2 = (1 - (-2))^2 + (9 - 3)^2 = 3^2 + 6^2 = 9 + 36 = 45 \Rightarrow AB = \sqrt{45}$

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$9$

Vedclass · Answer

(C) First,calculate the lengths of the sides of $	riangle ABC$:$AB^2 = (1 - (-2))^2 + (9 - 3)^2 = 3^2 + 6^2 = 9 + 36 = 45 \Rightarrow AB = \sqrt{45}$$BC^2 = (3 - 1)^2 + (8 - 9)^2 = 2^2 + (-1)^2 = 4 + 1 = 5 \Rightarrow BC = \sqrt{5}$$AC^2 = (3 - (-2))^2 + (8 - 3)^2 = 5^2 + 5^2 = 25 + 25 = 50 \Rightarrow AC = \sqrt{50}$Since $AB^2 + BC^2 = 45 + 5 = 50 = AC^2$,the triangle is a right-angled triangle with $\angle B = 90^{\circ}$.In a right-angled triangle,the circum-center is the midpoint of the hypotenuse $AC$.Circum-center $= \left(\frac{-2 + 3}{2}, \frac{3 + 8}{2}ight) = \left(\frac{1}{2}, \frac{11}{2}ight)$.The midpoint of $BC$ is $\left(\frac{1 + 3}{2}, \frac{9 + 8}{2}ight) = \left(2, \frac{17}{2}ight)$.The line $L$ passes through $\left(\frac{1}{2}, \frac{11}{2}ight)$ and $\left(2, \frac{17}{2}ight)$.The slope $m = \frac{\frac{17}{2} - \frac{11}{2}}{2 - \frac{1}{2}} = \frac{3}{\frac{3}{2}} = 2$.The equation of line $L$ is $y - \frac{11}{2} = 2(x - \frac{1}{2})$ $\Rightarrow y = 2x - 1 + \frac{11}{2}$ $\Rightarrow y = 2x + \frac{9}{2}$.Since the line intersects the $y$-axis at $\left(0, \frac{\alpha}{2}ight)$,we have $\frac{\alpha}{2} = \frac{9}{2}$,which gives $\alpha = 9$.

Consider a triangle having vertices $A(-2, 3)$,$B(1, 9)$,and $C(3, 8)$. If a line $L$ passing through the circum-center of triangle $ABC$ bisects line $BC$ and intersects the $y$-axis at point $\left(0, \frac{\alpha}{2}\right)$,then the value of the real number $\alpha$ is $.....$

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