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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer
    If the area of the triangle with vertices \[(x,\text{ }0),\text{ (}1,1\text{)}\] and \[(0,2)\] is 4 square units, then the value of x is :

    A)  -2                                          

    B)  -4

    C)  - 6                                         

    D)  8

    Correct Answer: C

    Solution :

    Given that \[{{x}_{1}}=x,\,\,{{x}_{2}}=1,\,{{x}_{3}}=0\] \[\therefore \] \[{{y}_{1}}=0,\,{{y}_{2}}=1,\,{{y}_{3}}=2\] \[\therefore \] Area of triangle \[=\frac{1}{2}\,[{{x}_{1}}\,({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})]\] \[=\frac{1}{2}\,[x\,(1-2)+1(2-0)+(0-1)]\] \[=\frac{1}{2}\,[-x\,+2+0]\] \[=\frac{1}{2}\,(2-x)\] Now, \[\frac{1}{2}(2-x)=4\] sq. units (given) \[\Rightarrow \]               \[2-x=8\] \[\Rightarrow \]                               \[x=-6\]


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