Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
    The point on the line\[3x+4y=5\]which is equidistant from (1, 2) and (3, 4) is

    A)  (7, -4)                                  

    B)  (15,-10)

    C)  (1/7, 8/7)      

    D)         (0, 5/4)

    Correct Answer: B

    Solution :

    Let point\[({{x}_{1}},\,\,{{y}_{1}})\]be on the line\[3x+4y=5\].                 \[3{{x}_{1}}+4{{y}_{1}}=5\]                                          ... (i) Also,      \[{{({{x}_{1}}-1)}^{2}}+{{({{y}_{1}}-2)}^{2}}\]                                                 \[={{({{x}_{1}}-3)}^{2}}+{{({{y}_{1}}-4)}^{2}}\] \[\Rightarrow \]               \[x_{1}^{2}+y_{1}^{2}-2{{x}_{1}}-4{{y}_{1}}+5=x_{1}^{2}+y_{1}^{2}-6{{x}_{1}}\]                                                                 \[-8{{y}_{1}}+25\]                 \[4{{x}_{1}}+4{{y}_{1}}=20\]                                       ... (ii) On solving Eqs. (i) and (ii), we get                 \[{{x}_{1}}=15,\,\,{{y}_{1}}=-10\]


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