A) \[(32,\,\,32)\]
B) \[(-32,\,\,-32)\]
C) \[(3,\,3)\]
D) \[\left( \frac{32}{3},\,\frac{32}{3} \right)\]
Correct Answer: D
Solution :
\[\because \] Area of triangle, \[=\frac{1}{2}\left\| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right\|\] \[\therefore \] \[32=\frac{1}{2}\left\| \begin{matrix} 2a & a & 1 \\ a & 2a & 1 \\ a & a & 1 \\ \end{matrix} \right\|\] \[\Rightarrow \] \[64={{a}^{2}}\left\| \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 1 \\ \end{matrix} \right\|\] \[\Rightarrow \] \[64={{a}^{2}}|2(2-1)-1(1-1)+1(1-2)|\] \[\Rightarrow \] \[64={{a}^{2}}|2-0-1|={{a}^{2}}\] \[\Rightarrow \] \[a=\pm 8\] \[\therefore \] The given points becomes. \[(\pm \,\,16,\,\pm 8),\,(\pm \,\,8,\,\,\pm 16)\] and \[(\pm \,\,8,\,\,\pm 8)\]. \[\therefore \] Centroid \[=\left[ \pm \frac{(16+8+8)}{3},\pm \frac{(8+16+8)}{3} \right]\] \[=\left( \pm \frac{32}{3},\pm \frac{32}{3} \right)\]
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