JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer
    If the \[x-\] intercept of some line L is double as that of the line, \[3x+4y=12\] and the \[y-\]intercept of L is half as that of the same line. Then the slope of L is :     jEE Main  Online Paper (Held On 22 April 2013)

    A)  -3                                          

    B)  -3/8

    C)  -3/2                                      

    D)  -3/16

    Correct Answer: D

    Solution :

     Given line \[3x+4y=12\] can be rewritten as \[\frac{3x}{12}+\frac{4y}{12}=1\Rightarrow \frac{x}{4}+\frac{y}{3}=1\] \[\Rightarrow \]\[x-\]intercept = 4 and \[y-\]intercept = 3 Let the required line be \[L:\frac{x}{a}+\frac{y}{b}=1\]where \[a=x-\]mtercept and \[b=y-\]intercept According to the question \[a=4\times 2=8\]and \[b=3/2\] \[\therefore \]  Required line is \[\frac{x}{8}+\frac{2y}{3}=1\] \[\Rightarrow \]               \[3x+16y=24\] \[\Rightarrow \]               \[y=\frac{-3}{16}x+\frac{24}{16}\] Hence, required slope \[=\frac{-3}{16}.\]

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