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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 line \[2x+3y=12\]meets the x-axis at A and y-axis at B. The line through (5, 5) perpendicular to \[AB\]meets the x- axis , y axis and the \[AB\] at C, D and E respectively. If O is the origin of coordinates, then the area of \[OCEB\]is  [IIT 1976]

    A)            \[23\] sq. units                         

    B)            \[\frac{23}{2}sq.\]units

    C)            \[\frac{23}{3}sq.\]units            

    D)            None of these

    Correct Answer: C

    Solution :

               Here O is the point \[(0,\,0)\]. The line \[2x+3y=12\] meets the y-axis at B and so B is the point (0,4). The equation of any line perpendicular to the line \[2x+3y=12\] and passes through (5, 5) is  \[3x-2y=5\]     ......(i)                    The line (i) meets the x-axis at C and so co-ordinates of C are\[\left( \frac{5}{3},\,0 \right).\]Similarly the coordinates of E are (3, 2) by solving the line AB and (i). Thus O(0, 0), \[C\left( \frac{5}{3},0 \right)\], \[E(3,\,2)\]  and B (0, 4). Now the area of figure \[OCEB=\] area of \[\Delta OCE\] + area of \[\Delta OEB=\frac{23}{3}sq.\]units.


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