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, then the set of values which $k$ can take is given by-

$\because \mathrm{A}(\mathrm{h}, \mathrm{k}), \mathrm{B}(1,1)$ and $\mathrm{c}(2,1)$ are the vertices of a right angled triangle $\mathrm{ABC}$

Now, $A B=\sqrt{(1-h)^{2}+(1-k)^{2}}$

or $B C=\sqrt{(2-1)^{2}+(1-1)^{2}}=1$

or $\mathrm{CA}=\sqrt{(\mathrm{h}-2)^{2}+(\mathrm{k}-1)^{2}}$

Now, pythagorus theorem

$\begin{array}{l} A C^{2}=A B^{2}+B C^{2} \\ 4+h^{2}-4 h+k^{2}+1-2 k \\ =h^{2}+1-2 h+k^{2}+1-2 k+1 \\ 5-4 h=3-2 h \\ \Rightarrow h=1 \end{array}$

Now, given that area of the triangle is 1 ,

Then, area $(\triangle \mathrm{ABC})=\frac{1}{2} \times \mathrm{AB} \times \mathrm{BC}$

$1=\frac{1}{2} \times \sqrt{(1-h)^{2}+(1-k)^{2}} \times 1 \Rightarrow 2=\sqrt{(1-h)^{2}+(1-k)^{2}}$

Putting $h=1$ from equation (1),

we get

$2=\sqrt{(k-1)^{2}}$

Squaring both the sides, we get

$\begin{array}{l} 4=k^{2}+1-2 k \text { or } k^{2}-2 k-3=0 \\ \text { or }(k-3)(k+1)=0 \\ \text { So, } k=-1,3 \end{array}$

Thus the set of values of $\mathrm{k}$ is $\{-1.3\}$