A) \[-1\]
B) \[1\]
C) \[1/2\]
D) \[a(a+1)\,(a+2)\,(a+3)\]
Correct Answer: B
Solution :
Area of triangle whose vertices \[\{a(a+1),\,(a+1)\},\,\,\{(a+1)\,(a+2),(a+2)\}\]and \[\{(a+2)+(a+3),(a+3)\}\] are: \[\Delta =\frac{1}{2}\,\,\left| \begin{matrix} a(a+1) & (a+1) & 1 \\ (a+1),\,(1+2) & (a+2) & 1 \\ (a+2)\,(a+3) & (a+3) & 1 \\ \end{matrix} \right|\] Use operation, \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\] and \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\] \[\Delta =\frac{1}{2}\left| \begin{matrix} a(a+1) & (a+1) & 1 \\ 2(a+1) & 1 & 0 \\ (a+2)(a+3)-a(a+1) & 2 & 0 \\ \end{matrix} \right|\] \[\Delta =\frac{1}{2}\,\{4\,(a+1)-(a+2)\,(a+3)+a(a+1)\}\] \[\Delta =\frac{1}{2}\,\{4a+4-{{a}^{2}}-5a-6+{{a}^{2}}+a\}\] \[\Delta =\frac{1}{2}\,\{-2\}\] \[\Rightarrow \] \[\Delta =-1\] Since, the value of area of triangle is numeric. So, we neglect \[-ve\] sign. \[\Rightarrow \] Area of triangle \[=1\]
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