Answer:
Given, area of triangle = 5 sq units ...(i) Also, vertices of triangle are (k, 0), (1, 1) and (0, 3). Then. area of triangle = absolute value of
\[\left[ \because \text{area of triangle=absolute value}\,\,\text{of}\,\frac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right| \right]\]\[=\left| \frac{1}{2}[k(1-3)-0\,+1(3)] \right|=\left| \frac{1}{2}(-\,2k+3) \right|\] ?(ii) Now, from Eqs. (i) and (ii), we have \[\left| \frac{1}{2}(-\,2k+3) \right|=5\] \[\Rightarrow \] \[\frac{1}{2}(-\,2k+3)=\pm \,5\] \[\Rightarrow \] \[-\,2k+3=\pm \,10\] \[\Rightarrow \] \[-\,2k=\pm \,10-3\] \[\Rightarrow \] \[-\,2k=7,\,\,-\,13\] \[\therefore \] \[k=-\frac{7}{2},\,\,\frac{13}{2}\]
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