Q90 / 160 · AP EAMCET 2004 · Mathematics

Area of the triangle formed by the lines $3 x^2-4 x y+y^2=0,2 x-y=6$ is

  • A
    16 sq units
  • B
    25 sq units
  • Correct answer
    C
    36 sq units
  • D
    49 sq units

Solution

Given lines are $$ \begin{aligned} & 3 x^2-4 x y+y^2=0 \\ & \Rightarrow 3 x^2-3 x y-x y+y^2=0 \\ &(3 x-y)(x-y)=0 \\ & \Rightarrow 3 x-y=0, x-y=0 \\ & \text { and } 2 x-y=6 \end{aligned} $$ The point of intersection of these lines are $(0,0),(-6,-18)$ and $(6,6)$. $$ \begin{aligned} \therefore \text { Area of triangle } & =\frac{1}{2}\left|\begin{array}{lll} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array}\right| \\ & =\frac{1}{2}\left|\begin{array}{ccc} 0 & 0 & 1 \\ -6 & -18 & 1 \\ 6 & 6 & 1 \end{array}\right| \\ & =\frac{1}{2}(36+108)=\frac{1}{2} \\ & =36 \text { sq unit } \end{aligned} $$

Hence, the correct answer is option: 'C'.

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