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JEE Main & Advanced Mathematics Straight Line Question Bank Foot of perpendicular, Transformation, Pedal points Image of a point

  • question_answer
    The reflection of the point (4, -13) in the line \[5x+y+6=0\] is                                                             [EAMCET 1994]

    A) \[(-1,-14)\]                                

    B) (3 ,4)

    C) (1, 2)                                         

    D) (- 4, 13)

    Correct Answer: A

    Solution :

               Let \[Q(a,b)\] be the reflection of \[P(4,-13)\] in the line \[5x+y+6=0\].                    Then the mid-point \[R\text{ }\left( \frac{a+4}{2},\frac{b-13}{2} \right)\] lies on \[5x+y+6=0\].                    \ \[5\left( \frac{a+4}{2} \right)+\frac{b-13}{2}+6=0\Rightarrow 5a+b+19=0\]  ......(i)                    Also \[PQ\]is perpendicular to \[5x+y+6=0\].                    Therefore \[\frac{b+13}{a-4}\times \left( -\frac{5}{1} \right)=-1\Rightarrow a-5b-69=0\] .....(ii)                    Solving (i) and (ii), we get\[a=-1,\,\,b=-14\].


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