A) a - 6 = 2
B) a + b = 2
C) a = 1+ 6
D) a = 1 ? 6
Correct Answer: C
Solution :
Given points will be collinear, if \[\left| \begin{matrix} a & b & 1 \\ b & a & 1 \\ {{a}^{2}} & -{{b}^{2}} & 1 \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[\left| \begin{matrix} a & b & 1 \\ b-a & b & 1 \\ {{a}^{2}}-a & -{{b}^{2}}-b & 0 \\ \end{matrix} \right|=0\] [applying\[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},{{R}_{3}}\to {{R}_{3}}={{R}_{1}}\]] \[\Rightarrow \]\[(a-b)\left| \begin{matrix} a & b & 1 \\ -1 & 1 & 0 \\ {{a}^{2}}-a & -{{b}^{2}}-b & 0 \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[(a-b)({{b}^{2}}+b-{{a}^{2}}+a)=0\] \[\Rightarrow \]\[(a-b)\{(a+b)-({{a}^{2}}-{{b}^{2}})\}=0\] \[\Rightarrow \]\[(a-b)(a+b)(1-a+b)=0\] \[\Rightarrow \]\[a=b\]or\[a+b=0\]or\[a=1+b\]
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