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J & K CET Engineering J and K - CET Engineering Solved Paper-2010

  • question_answer
    The equation of a straight line which is parallel to \[x-2y+8=0\]and passes through \[(0,4)\] is

    A)  \[x-2y+8=0~~~\]

    B)  \[x-2y+7=0\]

    C)  \[x-2y+4=0\]

    D)  \[2x+2y-13=0\]

    Correct Answer: A

    Solution :

    Equation of straight line parallel to \[x-2y+8=0\] is \[x-2y+\lambda =0\] But, it passes through the point \[(0,\,\,4)\]. \[\therefore \] \[0-8+\lambda =0\] \[\Rightarrow \] \[\lambda =8\] Hence, required line is \[x-2y+8=0\]


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