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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2005

  • question_answer
    A line is perpendicular to the line\[3x+y=3\] and passes through the point\[(-2,2),\]then its intercept on y-axis is

    A)  \[\frac{1}{3}\]

    B)  \[\frac{2}{3}\]

    C)  \[1\]

    D)  \[\frac{8}{3}\]

    Correct Answer: D

    Solution :

     Given, equation of line is\[3x+y=3\] \[\Rightarrow \] \[3x+y-3=0\] Equation of its perpendicular line is \[x-3y+\lambda =0\] and it passes through the point\[(-2,2)\] \[\therefore \] \[-2-3\times 2+\lambda =0\] \[\Rightarrow \] \[\lambda =8\] \[\therefore \]Equation is \[x-3y+8=0\] For the intercept on y-axis, \[x=0\] Then,    \[3y=8\] \[\Rightarrow \] \[y=\frac{8}{3}\]


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