Let A(h, k),B(1, 1),and C(2, 1) be the vertic | vedclass

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(C) Given vertices are $A(h, k)$,$B(1, 1)$,and $C(2, 1)$.Since $AC$ is the hypotenuse,the right angle is at vertex $B(1, 1)$.Thus,$AB \perp BC$.The sl

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$(-1, 3)$

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(C) Given vertices are $A(h, k)$,$B(1, 1)$,and $C(2, 1)$.Since $AC$ is the hypotenuse,the right angle is at vertex $B(1, 1)$.Thus,$AB \perp BC$.The slope of $BC$ is $m_{BC} = \frac{1-1}{2-1} = 0$. Since $BC$ is horizontal,$AB$ must be vertical.Therefore,the $x$-coordinate of $A$ must be the same as $B$,so $h = 1$.Now,the area of the triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = 1$.Base $BC = \sqrt{(2-1)^2 + (1-1)^2} = 1$.Height $AB = \sqrt{(1-1)^2 + (k-1)^2} = |k-1|$.Area $= \frac{1}{2} 	imes 1 	imes |k-1| = 1$.$|k-1| = 2$.This implies $k-1 = 2$ or $k-1 = -2$.So,$k = 3$ or $k = -1$.The possible values for $k$ are $\{-1, 3\}$.

Let $A(h, k)$,$B(1, 1)$,and $C(2, 1)$ be the vertices of a right-angled triangle with $AC$ as the hypotenuse. If the area of the triangle is $1$,then which of the following can be the set of values for $k$?

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