Question

The point of intersection of the lines represented by the equation $2x^2+3y^2+7xy+8x+14y+8=0$ is

• A. $\left(0,2\right)$
• B. $\left(1,2\right)$
• C. $\left(-2,0\right)$
• D. $\left(-2,1\right)$

Answer 1

The correct option is C

$\left(-2,0\right)$

Explanation for the correct option:

Find the point of intersection of the given lines

Given: $2x^2+3y^2+7xy+8x+14y+8=0$

Rewrite the given equation as follows:

$$\begin{gathered} 2x^2+3y^2+7xy+8x+14y+8=0 \\ \Rightarrow 2x^2+3y^2+6yx+4x+yx+2y+4x+12y+8=0 \\ \Rightarrow 2x\left(x+3y+2\right)+y\left(x+3y+2\right)+4\left(x+3y+2\right)=0 \\ \Rightarrow \left(2x+y+4\right)\left(x+3y+2\right)=0 \end{gathered}$$

Let line $L_1$ be $2x+y+4=0\left(1\right)$

And line $L_2$ be $x+3y+2=0\left(2\right)$

Multiplying equation $\left(2\right)$ by $2$ and Subtracting equation $\left(1\right)$ from $\left(2\right)$, we get,

$$\begin{gathered} \Rightarrow 2x+6y+4-\left(2x+y+4\right)=0 \\ \Rightarrow 2x+6y+4-2x-y-4=0 \\ \Rightarrow 5y=0 \\ \Rightarrow y=0 \end{gathered}$$

Putting $y=0$ in equation $\left(1\right)$,

$$\begin{gathered} \Rightarrow 2x+0+4=0 \\ \Rightarrow 2x=-4 \\ \Rightarrow x=-2 \end{gathered}$$

Now, $x=-2$ and $y=0$.

So, the point of intersection is $\left(-2,0\right)$.

Hence, option (C) is the correct answer.