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JEE Main & Advanced Mathematics Straight Line Question Bank Slope of line, Equation of line in different forms

  • question_answer
    Equation of the perpendicular bisector of the line segment joining the points (7, 4) and (-1, -2), is       [AMU 1979]

    A) \[4x-3y=15\]                           

    B) \[3x+4y=15\]

    C) \[4x+3y=15\]                         

    D) None of these

    Correct Answer: C

    Solution :

               Midpoint \[\equiv (3,\,1)\]                    Slope of perpendicular \[=\frac{-1}{\frac{-2-4}{-1-4}}=\frac{-4}{3}\]                    Therefore the required equation is                    \[4x+3y=4(3)+3(1)=15\].


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