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If \mathrm{x}_{1}, \mathrm{y}_{1} are the roots of \mathrm{x}^{2}+8 \mathrm{x}-20=0, \mathrm{x}_{2}, \mathrm{y}_{2} are the roots of 4 \mathrm{x}^{2}+32 \mathrm{x}-57=0$ and $\mathrm{x}_{3}, \mathrm{y}_{3} are the roots of 9 \mathrm{x}^{2}+72 \mathrm{x}-112=0, then the points, \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right),\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right) \&\left(\mathrm{x}_{3}, \mathrm{y}_{3}\right) 

Option: 1

are collinear


Option: 2

form an equilateral triangle


Option: 3

form a right angled isosceles triangle


Option: 4

are concyclic


Answers (1)

best_answer

\mathrm{Given\; \mathrm{x}_{1}+\mathrm{y}_{1}=-8 \Rightarrow \mathrm{y}_{1}=-8-\mathrm{x}_{1} ; \quad \mathrm{x}_{2}+\mathrm{y}_{2}=-8 \quad \Rightarrow \mathrm{y}_{2}=-8-\mathrm{x}_{2}\, }
and \mathrm{x_{3}+y_{3}= -8\: \Rightarrow y_{3}= -8-x_{3} }

now check
\left|\begin{array}{lll} \mathrm{x}_{1} & -8-\mathrm{x}_{1} & 1 \\ \mathrm{x}_{2} & -8-\mathrm{x}_{2} & 1 \\ \mathrm{x}_{3} & -8-\mathrm{x}_{3} & 1 \end{array}\right| \text{value of this determinant comes out to be zero.}

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