The equation of the bisectors of the angles between the lines joining the origin to the points of intersection of the curve $x^2+x y+y^2+x+3 y+1=0$ and the line $x+y+2=0$ is

• Correct answer
  A

  $x^2+4 x y-y^2=0$
• B

  $2 x^2+5 x y-y^2=0$
• C

  $x^2+6 x y-2 y^2=0$
• D

  $2 x^2-4 x y+2 y^2=0$

Solution

Homogenise the equation of the given equation of curve
To get the equation of the lines joining the origin to points of intersection of curve (i) and line (ii). $$ \begin{gathered} x^2+y^2+x y+(x+3 y)\left(\frac{x+y}{-2}\right)+\left(\frac{x+y}{-2}\right)^2=0 \\ \Rightarrow x^2+y^2+x y-\frac{1}{2}\left(x^2+4 x y+3 y^2\right) \\ +\frac{1}{4}\left(x^2+2 x y+y^2\right)=0 \end{gathered} $$
Now, equation of the bisectors of the angle between the pair of straight line (iii) is $$ \begin{aligned} \frac{x^2-y^2}{3+1} & =\frac{x y}{-1} \\ \Rightarrow \quad x^2+4 x y-y^2 & =0 \end{aligned} $$ Hence, option (a) is correct.

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