JEE Main & Advanced Mathematics Pair of Straight Lines Question Bank Critical Thinking

  • question_answer The square of distance between the point of intersection of the lines represented by the equation \[a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\] and origin, is

    A)            \[\frac{c(a+b)-{{f}^{2}}-{{g}^{2}}}{ab-{{h}^{2}}}\]   

    B)            \[\frac{c(a-b)+{{f}^{2}}+{{g}^{2}}}{\sqrt{ab-{{h}^{2}}}}\]

    C)            \[\frac{c(a+b)-{{f}^{2}}-{{g}^{2}}}{ab+{{h}^{2}}}\]  

    D)            None of these

    Correct Answer: A

    Solution :

               Let the lines represented by given equation be            \[y={{m}_{1}}x+{{c}_{1}}\]and\[y={{m}_{2}}x+{{c}_{2}}\].            Then \[a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\]            \[=b(y-{{m}_{1}}x-{{c}_{1}})(y-{{m}_{2}}x-{{c}_{2}})=0\]            Comparing the coefficients of \[{{x}^{2}},\ xy,\ x,\ y\] and constant term, we get  \[{{m}_{1}}{{m}_{2}}=\frac{a}{b},\ {{m}_{1}}+{{m}_{2}}=\frac{-2h}{b},\ {{m}_{1}}{{c}_{2}}+{{m}_{2}}{{c}_{1}}=\frac{2g}{b},\]            \[{{c}_{1}}+{{c}_{2}}=-\frac{2f}{b}\]and \[{{c}_{1}}{{c}_{2}}=\frac{c}{b}\]            Also the point of intersection of \[y={{m}_{1}}x+{{c}_{1}}\]and \[y={{m}_{2}}x+{{c}_{2}}\] is \[\left( \frac{{{c}_{2}}-{{c}_{1}}}{{{m}_{1}}-{{m}_{2}}},\frac{{{m}_{1}}{{c}_{2}}-{{m}_{2}}{{c}_{1}}}{{{m}_{1}}-{{m}_{2}}} \right)\]            Therefore, the square of distance of this point from origin is \[{{\left( \frac{{{c}_{2}}-{{c}_{1}}}{{{m}_{1}}-{{m}_{2}}} \right)}^{2}}+\frac{{{({{m}_{1}}{{c}_{2}}-{{m}_{2}}{{c}_{1}})}^{2}}}{{{({{m}_{1}}-{{m}_{2}})}^{2}}}\]            \[=\frac{[{{({{c}_{1}}+{{c}_{2}})}^{2}}-4{{c}_{1}}{{c}_{2}}]+[{{({{m}_{1}}{{c}_{2}}+{{m}_{2}}{{c}_{1}})}^{2}}-4{{m}_{1}}{{m}_{2}}{{c}_{1}}{{c}_{2}}]}{{{({{m}_{1}}+{{m}_{2}})}^{2}}-4{{m}_{1}}{{m}_{2}}}\]            Now putting the value defined above, we get the required distance i.e., \[\frac{-c(a+b)+{{f}^{2}}+{{g}^{2}}}{{{h}^{2}}-ab}\].


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