JEE Main & Advanced JEE Main Solved Paper-2014

  • question_answer
    Let a, b, c and d be non−zero numbers. If the point of intersection of the lines \[4ax+2ay+c=0\]and \[5bx+2by+d=\] lies in the fourth quadrant and is equidistant from the two axes then : JEE Main  Solved  Paper-2014

    A)  2bc − 3ad = 0    

    B) 2bc + 3ad = 0

    C) 3bc − 2ad = 0

    D) 3bc + 2ad = 0

    Correct Answer: C

    Solution :

    Solving the given lines we get the point of intersection P\[P\left( \frac{bc-ad}{ab},\frac{4ad-5bc}{2ab} \right)\] As point is equidistant from the axes. \[\Rightarrow \]\[\frac{bc-ad}{ab}=\left| \frac{4ad-5bc}{2ab} \right|\]which gives 3bc − 2ad = 0

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