2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-8

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


You need to login to perform this action.
You will be redirected in 3 sec spinner