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Q.
The area of the triangle formed by the lines $4 \,x^{2}-9\, x y-9\, y^{2}=0$ and $x=2$ is equal
to
Bihar CECEBihar CECE 2010
Solution:
The equations of given lines are $ x=2 $
and $ 4{{x}^{2}}-9\,xy-9{{y}^{2}}=0 $
$ \Rightarrow 4\,x(x-3\,y)+3\,y(x-3\,y)=0 $
$ \Rightarrow $ $ (x-3\,y)(4\,x+3y)=0 $
$ \Rightarrow $ $ (x-3\,y)=0 $ and $ (4\,x+3\,y)=0 $
So, we have the three sides of triangle are
$ x=2,\,x-3\,y=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}\begin{vmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & -8/3 & 2/3 \\ \end{vmatrix} $
$ =\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} $