JEE Main & Advanced Mathematics Straight Line Question Bank Mock Test - Straight Lines

  • question_answer
    If the equation of the locus of a point equidistant from the points (\[{{a}_{1,}}{{b}_{1}}\]) and (\[{{a}_{2,}}{{b}_{2}}\]) is (\[{{a}_{1,-}}{{a}_{2}}\])x+(\[{{b}_{1,-}}{{b}_{2}}\])\[y+c=0\] then the value of c is

    A) \[\frac{1}{2}({{a}_{2}}^{2}+{{b}_{2}}^{2}-{{a}_{1}}^{2}-{{b}_{1}}^{2})\]

    B) \[{{a}_{1}}^{2}+{{a}_{2}}^{2}+{{b}_{1}}^{2}-{{b}_{2}}^{2}\]

    C) \[\frac{1}{2}({{a}_{1}}^{2}+{{a}_{2}}^{2}-{{b}_{1}}^{2}-{{b}_{2}}^{2})\]

    D) \[\sqrt{{{a}_{1}}^{2}+{{b}_{2}}^{2}-{{a}_{2}}^{2}-{{b}_{2}}^{2}}\]

    Correct Answer: A

    Solution :

    [a] The locus is \[{{(h-{{a}_{1}})}^{2}}+{{(k-{{b}_{1}})}^{2}}={{(h-{{a}_{2}})}^{2}}+{{(k-{{b}_{2}})}^{2}}\] \[\Rightarrow ({{a}_{1}}-{{a}_{2}})x+({{b}_{1}}-{{b}_{2}})y+\frac{1}{2}({{a}_{2}}^{2}+{{b}_{2}}^{2}-{{a}_{1}}^{2}-{{b}_{1}}^{2})=0\]\[\Rightarrow c=\frac{1}{2}({{a}_{2}}^{2}+{{b}_{2}}^{2}-{{a}_{1}}^{2}-{{b}^{2}}_{1})\]


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