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JEE Main & Advanced Sample Paper JEE Main Sample Paper-11

  • 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)  \[3x+4y25=0\]                 

    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 \[\therefore \] Slope of required line =-1/3 Equation is \[y-4=-\frac{1}{3}(x-3)\]\[x+3y-15=0\]


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