A) lie on a straight line
B) lie on an ellipse
C) lie on a circle
D) are vertices of a triangle
Correct Answer: A
Solution :
Let\[\frac{{{x}_{2}}}{{{x}_{1}}}=\frac{{{x}_{3}}}{{{x}_{2}}}=r\]and\[\frac{{{y}_{2}}}{{{y}_{1}}}=\frac{{{y}_{3}}}{{{y}_{2}}}=r\] \[\Rightarrow \]\[{{x}_{2}}={{x}_{1}}r,{{x}_{3}}={{x}_{1}}{{r}^{2}}\]and\[{{y}_{2}}={{y}_{1}}r,{{y}_{3}}={{y}_{1}}{{r}^{2}}\] Let\[\Delta =\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|=\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}}r & {{y}_{2}}r & 1 \\ {{x}_{1}}{{r}^{2}} & {{y}_{1}}{{r}^{2}} & 1 \\ \end{matrix} \right|\] Applying \[{{R}_{2}}\to {{R}_{2}}-r{{R}_{1}}\]and \[{{R}_{3}}\to {{R}_{3}}-r{{R}_{2}}\] \[\Delta =\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ 0 & 0 & 1-r \\ 0 & 0 & 1-r \\ \end{matrix} \right|=0\] (\[\because \]Rows \[{{R}_{2}}\]and\[{{R}_{3}}\]are identical) Hence, \[({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}}),({{x}_{3}},{{y}_{3}})\]lie on a straight line.You need to login to perform this action.
You will be redirected in
3 sec