Question

The area of the triangle formed by the lines $4x^2-9xy-9y^2=0$ and $x=2$ is equal to

• A. $\frac{20}{-3}$ sq units
• B. $3$ sq units
• C. $\frac{10}{3}$ sq units
• D. $2$ sq units

Answer 1

Answer: (C)

The equations of given lines are $x=2$
and $4x^2-9xy-9y^2=0$
$\Rightarrow 4x(x-3y)+3y(x-3y)=0$
$\Rightarrow$ $(x-3y)(4x+3y)=0$
$\Rightarrow$ $(x-3y)=0$ 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
$=\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ 0& 2& 2\\ 0& -8/3& 2/3\end{array}\right|$
$=\frac{1}{2}\left[2.\frac{2}{3}+2.\frac{8}{3}\right]$
$=\frac{1}{2}.\frac{20}{3}=\frac{10}{3}\text{sq}\text{units}$