JEE Main & Advanced AIEEE Solved Paper-2010

  • question_answer
      The line L given by \[\frac{x}{5}+\frac{y}{b}=1\]passes through the point (13, 32). The line K is parallel to L and has the equation \[\frac{x}{c}+\frac{y}{3}=1\].Then the distance between L and K is -       AIEEE  Solved  Paper-2010

    A) \[\frac{23}{\sqrt{15}}\] 

    B)        \[\sqrt{17}\]      

    C)        \[\frac{17}{\sqrt{15}}\] 

    D)        \[\frac{23}{\sqrt{17}}\]

    Correct Answer: D

    Solution :

    \[\frac{x}{5}+\frac{y}{b}=1\]       Passes through (13, 32) \[\frac{13}{5}+\frac{32}{b}=1\] \[\Rightarrow \] \[13b+160=5b\] \[\Rightarrow \]\[b=-20\] so line is \[20x+5y=100\]               (1) second line \[\frac{x}{c}+\frac{y}{3}=1\] \[3x+cy=3c\]                      (2) (1) and (2) are parallel \[\frac{3}{-20}=\frac{c}{5}\] \[c=\frac{-3}{4}\] Line \[3x-\frac{3}{4}y=-\frac{9}{4}\] \[12x3y=9\] \[-20x+5y=-9\times \left( -\frac{5}{3} \right)\] \[-20x+5y=15\]                 ......(2) Distance between (1) and (2) \[=\frac{|-100-15|}{\sqrt{400+25}}=\frac{115}{\sqrt{425}}=\frac{115}{5\sqrt{17}}=\frac{23}{\sqrt{17}}\]


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