A) \[\sqrt{19}\]
B) \[\frac{1}{2}\sqrt{19}\]
C) \[\frac{1}{2}\sqrt{38}\]
D) \[\frac{1}{2}\sqrt{57}\]
Correct Answer: C
Solution :
\[\because \] \[{{\Delta }_{x}}=\frac{1}{2}\,\left| \begin{matrix} {{y}_{1}} & {{z}_{1}} & 1 \\ {{y}_{2}} & {{z}_{2}} & 1 \\ {{y}_{3}} & {{z}_{3}} & 1 \\ \end{matrix} \right|\] \[=\frac{1}{2}\,\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 0 & 1 & 1 \\ \end{matrix} \right|=\frac{1}{2}\] \[{{\Delta }_{y}}=\frac{1}{2}\,\left| \begin{matrix} {{z}_{1}} & {{x}_{1}} & 1 \\ {{z}_{2}} & {{x}_{2}} & 1 \\ {{z}_{3}} & {{x}_{3}} & 1 \\ \end{matrix} \right|\] \[=\frac{1}{2}\,\left| \begin{matrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & -4 & 1 \\ \end{matrix} \right|=-3\] \[{{\Delta }_{z}}=\frac{1}{2}\,\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|\] \[=\frac{1}{2}\,\left| \begin{matrix} 2 & 1 & 1 \\ 3 & 1 & 1 \\ -4 & 0 & 1 \\ \end{matrix} \right|=-\frac{1}{2}\] \[\therefore \] Area of triangle \[=\sqrt{\Delta _{x}^{2}+\Delta _{y}^{2}+\Delta _{z}^{2}}\] \[=\sqrt{\frac{1}{4}+9+\frac{1}{4}}=\frac{1}{2}\sqrt{38}\]
You need to login to perform this action.
You will be redirected in
3 sec
