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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    . The position of reflection of the point (4, 1) about the line\[y=x-1\]is:

    A)  (1, 2)                                    

    B)  (3, 4)

    C)  \[(-\text{ }1,0)\]          

    D)        \[(-2,-1)\]

    E)  (2, 3)

    Correct Answer: E

    Solution :

    Let\[Q(x,\text{ }y)\]be the image of the point\[P(4,\text{ 1})\]to the line\[y-x+1=0\]. Then, PQ is\[\bot \]to \[y-x+1=0\] \[\therefore \]  \[\frac{y-1}{x-4}\times 1=-1\] \[\Rightarrow \]               \[y-1=-x+4\] \[\Rightarrow \]               \[y+1=4+1=5\]                              ...(i) And also midpoint of PQ i. e.,\[\left( \frac{4+x}{2},\frac{y+1}{2} \right)\] lies on \[y-x+1=0\] \[\therefore \]  \[\frac{y+1}{2}-\frac{(4+x)}{2}+1=0\] \[\Rightarrow \]               \[y+1-4-x+2=0\] \[\Rightarrow \]               \[y-x-1=0\] \[\Rightarrow \]               \[y-x=1\]                                         ...(ii) On solving Eqs. (i) and (ii), we get \[y=3\] \[x=2\] \[\therefore \]Required point is (2, 3).


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