A) \[a({{m}_{2}}-{{m}_{3}})({{m}_{3}}-{{m}_{1}})({{m}_{1}}-{{m}_{2}})\]
B) \[({{m}_{2}}-{{m}_{3}})({{m}_{3}}-{{m}_{1}})({{m}_{1}}-{{m}_{2}})\]
C) \[{{a}^{2}}({{m}_{2}}-{{m}_{3}})({{m}_{3}}-{{m}_{1}})({{m}_{1}}-{{m}_{2}})\]
D) None of these
Correct Answer: C
Solution :
Area \[=\frac{1}{2}\,\,\left| \,\begin{matrix} am_{1}^{2} & 2a{{m}_{1}} & 1 \\ am_{2}^{2} & 2a{{m}_{2}} & 1 \\ am_{3}^{2} & 2a{{m}_{3}} & 1 \\ \end{matrix}\, \right|=\frac{1}{2}{{a}^{2}}\times 2\,\left| \,\begin{matrix} m_{1}^{2} & {{m}_{1}} & 1 \\ m_{2}^{2} & {{m}_{2}} & 1 \\ m_{3}^{2} & {{m}_{3}} & 1 \\ \end{matrix}\, \right|\] \[={{a}^{2}}\,\left| \,\begin{matrix} m_{1}^{2}-m_{2}^{2} & {{m}_{1}}-{{m}_{2}} & 0 \\ m_{2}^{2}-m_{3}^{2} & {{m}_{2}}-{{m}_{3}} & 0 \\ m_{3}^{2} & {{m}_{3}} & 1 \\ \end{matrix}\, \right|\] , by \[\begin{matrix} {{R}_{1}}\to {{R}_{1}}-{{R}_{2}} \\ {{R}_{2}}\to {{R}_{2}}-{{R}_{3}} \\ \end{matrix}\] \[={{a}^{2}}\,(m_{2}^{2}-m_{3}^{2})\,({{m}_{1}}-{{m}_{2}})-({{m}_{2}}-{{m}_{3}})\,(m_{1}^{2}-m_{2}^{2})\] \[={{a}^{2}}({{m}_{1}}-{{m}_{2}})\,({{m}_{2}}-{{m}_{3}})\,({{m}_{3}}-{{m}_{1}})\]. Trick : Let \[a=2,\,\,{{m}_{1}}=0,\,\,{{m}_{2}}=1,\,\,{{m}_{3}}=2,\] then the coordinates are (0, 0), (2, 4), (8, 8). \[\therefore \,\,\Delta =\frac{1}{2}\,\left| \,\begin{matrix} 0 & 0 & 1 \\ 2 & 8 & 1 \\ 4 & 8 & 1 \\ \end{matrix}\, \right|=\frac{1}{2}\,(16-32)=8\,\,sq.\,\,units\].
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