Question

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

• A. $\frac{7}{12}$
• B. $\frac{5}{12}$
• C. $\frac{1}{12}$
• D. $\frac{1}{6}$

Answer 1

The correct option is C

$\frac{1}{12}$

Explanation for correct option:

Given that a triangle is formed by $x+y+1=0$ and pair of straight lines $x^2-3xy+2y^2=0$

Determining the points,

Now we have,

$$\begin{gathered} \Rightarrow x^2-3xy+2y^2=0 \\ \Rightarrow x^2-2xy-xy+2y^2=0 \\ \Rightarrow x\left(x-2y\right)-y\left(x-2y\right)=0 \\ \Rightarrow \left(x-y\right)\left(x-2y\right)=0 \end{gathered}$$

Therefore, the equation of lines is,

$$\begin{gathered} \Rightarrow x-y=0 \\ \Rightarrow x=y\cdots \left(1\right) \end{gathered}$$

And,

$$\begin{gathered} \Rightarrow x-2y=0 \\ \Rightarrow x=2y\cdots \left(2\right) \end{gathered}$$

And, also, $x+y+1=0\cdots \left(3\right)$

Using $\left(1\right)$ in the equation $\left(3\right)$ we get,

$$\begin{gathered} \Rightarrow y+y+1=0 \\ \Rightarrow 2y=-1 \\ \Rightarrow y=-\frac{1}{2} \end{gathered}$$

Put this in equation $\left(1\right)$ we get,

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

Now using $\left(1\right)$ and $\left(2\right)$ we get,

$x=0,y=0$

Now again using $\left(2\right)$ and $\left(3\right)$, we get,

$$\begin{gathered} \Rightarrow 2y+y+1=0 \\ \Rightarrow 3y=-1 \\ \Rightarrow y=-\frac{1}{3} \end{gathered}$$

Therefore,

$$\begin{gathered} \Rightarrow x=2y \\ \Rightarrow x=2\left(-\frac{1}{3}\right) \\ \Rightarrow x=-\frac{2}{3} \end{gathered}$$

So, the points of the triangle are $\left(0,0\right),\left(\frac{-2}{3},\frac{-1}{3}\right)$ and $\left(\frac{-1}{2},\frac{-1}{2}\right)$.

Now, the area of a triangle is given by,

$$\begin{gathered} \mathrm{Area}=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ \frac{-2}{3}& \frac{-1}{3}& 1\\ \frac{-1}{2}& \frac{-1}{2}& 1\end{array}\right| \\ =\frac{1}{2}\left[1\left(\frac{2}{6}-\frac{1}{6}\right)\right] \\ =\frac{1}{2}\left(\frac{1}{6}\right) \\ =\frac{1}{12}\mathrm{square}\mathrm{units} \end{gathered}$$

Therefore, the area of a triangle formed by given curves is equal to $\frac{1}{12}\mathrm{square}\mathrm{units}$.

Hence, the correct answer is option (C).