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JEE Main & Advanced Sample Paper JEE Main Sample Paper-20

  • question_answer
    Equation of straight line \[ax+by+c=0\] where\[3a+4b+c=0\], which is at maximum distance from \[(1,\,\,-2)\], is

    A) \[3x+y-17=0\]   

    B) \[4x+3y-24=0\]

    C) \[3x+4y-25=0\]

    D) \[x+3y-15=0\]

    Correct Answer: D

    Solution :

     It passes through a fixed point\[(3,\,\,4)\] Slope of line joining \[(3,\,\,4)\] and \[(1,\,\,-2)\] is\[-6/-2=3\] 3 \[\therefore \]Slope of required line\[=-1/3\] Equation is\[y-4=-\frac{1}{3}(x-3)\]                 \[x+3y-15=0\]


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