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

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

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$\{-1, 3\}$

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(A) Given vertices are $A(h, k)$,$B(1, 1)$,and $C(2, 1)$.Since $AC$ is the hypotenuse,the right angle is at $B$. Thus,$AB \perp BC$.The slope of $AB$ is $m_1 = \frac{k-1}{h-1}$ and the slope of $BC$ is $m_2 = \frac{1-1}{2-1} = 0$.Since $AB \perp BC$,the line $AB$ must be vertical,so $h = 1$.Now,the area of the triangle is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = 1$.Here,base $BC = \sqrt{(2-1)^2 + (1-1)^2} = 1$.Height $AB = \sqrt{(1-1)^2 + (k-1)^2} = |k-1|$.So,$\frac{1}{2} 	imes 1 	imes |k-1| = 1$.$|k-1| = 2$.$k-1 = 2$ or $k-1 = -2$.$k = 3$ or $k = -1$.Therefore,the set of values for $k$ is $\{-1, 3\}$.

Let $A(h, k)$,$B(1, 1)$,and $C(2, 1)$ be the vertices of a right-angled triangle with $AC$ as its hypotenuse. If the area of the triangle is $1$ square unit,then the set of values which $k$ can take is given by

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