A) lie on a straight
B) line lie on an ellipse
C) lie on a circle
D) are vertices of a triangle
Correct Answer: A
Solution :
If \[{{x}_{1}},{{x}_{2}},{{x}_{3}}\] and \[{{y}_{1}},{{y}_{2}},{{y}_{3}}\] are in GP, then let \[{{x}_{2}}=r\,{{x}_{1}}\], \[{{x}_{3}}={{r}^{2}}{{x}_{1}}\] and \[{{y}_{2}}=r\,{{y}_{1}}\], \[{{y}_{3}}={{r}^{2}}\,{{y}_{1}}\] with common ratio /", then the points are \[({{x}_{1}},{{y}_{1}}),\,(r{{x}_{1}},r{{y}_{1}})\] and \[({{r}^{2}}{{x}_{1}},{{r}^{2}}{{y}_{1}})\]. Now, \[\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|=\left| \begin{matrix} {{x}_{1}} & y{{ & }_{1}} & 1 \\ r\,{{x}_{1}} & r\,{{y}_{1}} & 1 \\ {{r}^{2}}{{x}_{1}} & {{r}^{2}}{{y}_{1}} & 1 \\ \end{matrix} \right|\] Taking \[{{x}_{1}}\] common from \[{{C}_{1}}\] and \[{{y}_{1}}\] from \[{{C}_{2}}\] \[={{x}_{1}}{{y}_{1}}\left| \begin{matrix} 1 & 1 & 1 \\ r & r & 1 \\ {{r}^{2}} & {{r}^{2}} & 1 \\ \end{matrix} \right|\] \[={{x}_{1}}{{y}_{1}}=(0)=0\] (since, two columns are identical) Thus, these points lie on a straight line. Alternate Solution Let \[{{x}_{1}}=a\Rightarrow {{x}_{2}}=ar\] and \[{{x}_{3}}=a{{r}^{2}}\] and \[y=b\Rightarrow {{y}_{2}}=br\] and \[{{y}_{3}}=b{{r}^{2}}\] Let the points are A(a, b), B(ar, br) and \[C(a{{r}^{2}},b{{r}^{2}})\]. Now, slope of \[AB=\frac{b(r-1)}{a(r-1)}=\frac{b}{a}\] and slope of \[BC=\frac{b({{r}^{2}}-r)}{a({{r}^{2}}-r)}=\frac{b}{a}\] \[\because \] Slope of AB = Slope of BC \[\Rightarrow \] \[AB||BC\] But B is a common point. \[\therefore \] A, B and C are collinear points. i.e., the points \[({{x}_{1}},{{y}_{1}}),\,({{x}_{2}},{{y}_{2}})\] and \[({{x}_{3}},{{y}_{3}})\] lie on a straight line.
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