CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer
    \[ax+by-{{a}^{2}}=0,\]where a, b are non-zero, is the equation to the straight line perpendicular to a line \[l\]and passing through the point where (crosses the\[x-\]axis. Then equation to the line\[l\]is:

    A)  \[\frac{x}{b}-\frac{y}{a}=1\]                      

    B)  \[\frac{x}{a}-\frac{y}{b}=1\]

    C)  \[\frac{x}{b}+\frac{y}{a}=ab\]  

    D)         \[\frac{x}{a}-\frac{y}{b}=ab\]

    E)  \[\frac{x}{a}+\frac{y}{b}=ab\]

    Correct Answer: B

    Solution :

    Equation of line perpendicular to \[ax+by-{{a}^{2}}=0\]is\[bx-ay+\lambda =0\]and line \[ax+by-{{a}^{2}}=0\]is passes through\[\left( -\frac{\lambda }{b},0 \right)\], then \[\lambda =-ab\] \[\therefore \]  \[bx-ay=ab\]      \[\Rightarrow \]               \[\frac{x}{a}-\frac{y}{b}=1\]


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