Let A(3,5),B(-2,-7) and C(alpha, beta) be thr | vedclass

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(B) Since $\angle ACB = 90^{\circ}$,point $C$ lies on a circle with diameter $AB$.The length of diameter $AB = \sqrt{(3 - (-2))^2 + (5 - (-7))^2} = \s

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$2$

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(B) Since $\angle ACB = 90^{\circ}$,point $C$ lies on a circle with diameter $AB$.The length of diameter $AB = \sqrt{(3 - (-2))^2 + (5 - (-7))^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = 13$.The radius of the circle is $R = \frac{13}{2}$.The area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes h = \frac{1}{2} 	imes 13 	imes h = \frac{82}{3}$.Thus,the height $h = \frac{82 	imes 2}{3 	imes 13} = \frac{164}{39} \approx 4.2$.Since the maximum possible height of the triangle (which is the radius of the circle) is $R = \frac{13}{2} = 6.5$,and $h For each height $h Therefore,there are $2$ such points $C$.

Let $A(3,5)$,$B(-2,-7)$ and $C(\alpha, \beta)$ be three points such that $\angle ACB$ is a right angle and the area of triangle $ABC$ is $\frac{82}{3}$ square units. Then the number of such points $C$ is

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