BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer The area of the triangle formed by the lines  \[4{{x}^{2}}-9xy-9{{y}^{2}}=0\] and \[x=2\]is equal to

    A) \[\frac{20}{3}\,\text{sq}\,\text{units}\]

    B)         \[3\,\text{sq}\,\text{units}\]    

    C)         \[\frac{10}{3}\,\text{sq}\,\text{units}\]               

    D)         \[\text{2}\,\text{sq}\,\text{units}\]

    Correct Answer: C

    Solution :

    The equations of given lines are \[x=2\]                 and        \[4{{x}^{2}}-9xy-9{{y}^{2}}=0\]                 \[\Rightarrow \]\[4x(x-3y)+3y(x-3y)=0\]                 \[\Rightarrow \]\[(x-3y)(4x+3y)=0\]                 \[\Rightarrow \]\[(x-3y)=0\]and \[(4x+3y)=0\] So, we have the three sides of triangle are \[x=2,\,x-3y=0\]and \[4x+3y=0\] Solving these equations taking any two at a time, we get the vertices of the triangle \[(0,0),(2,-8/3)\]and\[(2,2/3).\] Now, area of the triangle \[=\frac{1}{2}\left| \begin{matrix}    1 & 1 & 1  \\    0 & 2 & 2  \\    0 & -8/3 & 2/3  \\ \end{matrix} \right|\]                 \[=\frac{1}{2}\left[ 2.\frac{2}{3}+2.\frac{8}{3} \right]\]                 \[=\frac{1}{2}.\frac{20}{3}=\frac{10}{3}\,\text{sq}\,\text{units}\]

adversite


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