The area (in square unit) of the triangle formed by $x+y+1=0$ and the pair of straight lines $x^2-3 x y+2 y^2=0$ is

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Aicharya · Accepted Answer

$\begin{aligned} 	ext { Given, } & & x^2-2 x y-x y+2 y^2 & =0 \ \Rightarrow & & (x-2 y)(x-y) & =0\end{aligned}$ On solving Eqs. (i) and (ii), we get $A\left(-\frac{2}{3},-\frac{1}{3}ight), B\left(-\frac{1}{2},-\frac{1}{2}ight), C(0,0)$ $	herefore$ Area of $	riangle A B C=\frac{1}{2}\left|\begin{array}{ccc}-\frac{2}{3} & -\frac{1}{3} & 1 \ -\frac{1}{2} & -\frac{1}{2} & 1 \ 0 & 0 & 1\end{array}ight|$ $=\frac{1}{2}\left[\frac{1}{3}-\frac{1}{6}ight]=\frac{1}{2}\left[\frac{1}{6}ight]=\frac{1}{12}$

The area (in square unit) of the triangle formed by $x+y+1=0$ and the pair of straight lines $x^2-3 x y+2…

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