Question

Let $x^3-3x^{2}y+x^2+2x{y}^2-x-2y^2+3y-1=0$ represents three sides of triangle $ABC$. If area of triangle $ABC$ is $\Delta$, then value of $\left(111\Delta \right)$ is

Answer 1

$x^3-{3x}^{2}y+x^2+2{xy}^2-x-{2y}^2+3y-1=0$

$\Rightarrow (x-1)(x^2+2x+1+{2y}^2-3xy-3y)=0$

$\therefore$ $x-1=0$ is one side

Now,

$L=x^2+2x+1+{2y}^2-3xy-3y=0$ represents other two sides

Now, $\frac{dL}{dx}=2x+2-3y=0⟶\left(i\right)$

$\frac{dL}{dy}=4y-3x-3=0⟶\left(ii\right)$

$\Rightarrow x=-1,y=0$

$x=1$ intersects $L=0$ at $y=1$ & $y=2$

$\therefore$ area of $ABC$

$=ABD-ACD$

$=\frac{1}{2}\times 2\times 2-\frac{1}{2}\times 2\times 1$

$\Rightarrow \frac{1}{2}\times 2\times 1=1$

$\therefore$ $\Delta =1$

$\therefore$ $111\Delta =111$