question_answer

If area of a triangle with vertices (k, 0), (1, 1) and (0, 3) is 5 sq units, find the value(s) of k.

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 \[\left| \frac{1}{2}(-\,2k+3) \right|=5\] \[\Rightarrow \]   \[\frac{1}{2}(-\,2k+3)=\pm \,5\] \[\Rightarrow \] \[-\,2k+3=\pm \,10\] \[\Rightarrow \]   \[-\,2k=\pm \,10-3\] \[\Rightarrow \] \[-\,2k=7,\,\,-\,13\] \[\therefore \]      \[k=-\frac{7}{2},\,\,\frac{13}{2}\]