• # 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)         $\text{2}\,\text{sq}\,\text{units}$

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 $=\frac{1}{2}\left| \begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & -8/3 & 2/3 \\ \end{matrix} \right|$                 $=\frac{1}{2}\left[ 2.\frac{2}{3}+2.\frac{8}{3} \right]$                 $=\frac{1}{2}.\frac{20}{3}=\frac{10}{3}\,\text{sq}\,\text{units}$