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

• A

  $\frac{7}{12}$
• B

  $\frac{5}{12}$
• Correct answer
  C

  $\frac{1}{12}$
• D

  $\frac{1}{6}$

Solution

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

Hence, the correct answer is option: 'C'.