CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    If\[{{x}_{1}},{{x}_{2}},{{x}_{3}}\]as well as\[{{y}_{1}},{{y}_{2}},{{y}_{3}}\]are in GP with the same common ratio, then the points\[({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}})\]and\[({{x}_{3}},{{y}_{3}})\]:

    A)  lie on a parabola

    B)  lie on an ellipse

    C)  lie on a circle

    D)  are the vertices of a triangle

    E)  lie on a straight line

    Correct Answer: E

    Solution :

    Let \[{{x}_{1}}=x,{{x}_{2}}=xr,{{x}_{3}}=x{{r}^{2}}\] and \[{{y}_{1}}=y,{{y}_{2}}=yr,{{y}_{3}}=y{{r}^{2}}\] \[\because \]\[{{x}_{1}},{{x}_{2}},{{x}_{3}}\]and\[{{y}_{1}},{{y}_{2}},{{y}_{3}}\]are in GP. \[\because \]\[\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{{{y}_{3}}-{{y}_{2}}}{{{x}_{3}}-{{x}_{2}}}=\frac{{{y}_{1}}-{{y}_{3}}}{{{x}_{1}}-{{x}_{3}}}\] \[\therefore \]The points\[({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}})\]and\[({{x}_{3}},{{y}_{3}})\]lies in a straight line.


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