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Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
    The equation of the straight line perpendicular to the straight line\[3x+2y=0\]and passing through the point of intersection of the lines\[x+3y-1=0\]and\[x-2y+4=0\]is

    A) \[2x-3y+1=0\]   

    B) \[2x-3y+3=0\]

    C) \[2x-3y+5=0\]

    D)        \[2x-3y+7=0\]

    Correct Answer: D

    Solution :

    The point of intersection of lines\[x+3y-1=0\] and\[x-2y+4=0\]is ( -2 ,1). Let equation of line perpendicular to the given line is\[2x-3y+\lambda =0\]. Since, it passes through (-2 , 1). \[\therefore \]  \[2(-2)-3(1)+\lambda =0\] \[\Rightarrow \]               \[\lambda =7\] \[\therefore \]Required line is\[2x-3y+7=0\]


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