The equation of the bisectors of the angles between the lines joining the origin to the points of…

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
  1. $x^2+4 x y-y^2=0$
  2. $2 x^2+5 x y-y^2=0$
  3. $x^2+6 x y-2 y^2=0$
  4. $2 x^2-4 x y+2 y^2=0$

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

Asked in: AP EAMCET 2019 (20 Apr Shift 2)

Practice more Pair of Lines questions on Aicharya