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If \mathrm{x_{1}, x_{2}, x_{3}} as well as \mathrm{y_{1}, y_{2}, y_{3}} are in G.P., with the same common ratio, then the points \left(x_{1}\right.$, $\left.\mathrm{y}_{1}\right),\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$ and $\left(\mathrm{x}_{3}, \mathrm{y}_{3}\right)

Option: 1

 lie on a st. line


Option: 2

lie on an ellipse


Option: 3

 lie on a circle


Option: 4

 Are vertices of a triangle


Answers (1)

best_answer


Since \mathrm{x_{1}, x_{2}, x_{3} \, and\,y_{1}, y_{2}, y_{3} }  are in G.P

\Rightarrow \frac{\mathrm{x}_{2}}{\mathrm{x}_{1}}=\frac{\mathrm{x}_{3}}{\mathrm{x}_{2}}=k_{1} \text { and } \frac{\mathrm{y}_{2}}{\mathrm{y}_{1}}=\frac{\mathrm{y}_{3}}{\mathrm{y}_{2}}=k_{2}

Now,

\text{Area} =\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|=0

Hence the points are collinear
Hence (A) is the correct answer.

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Gunjita

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