Question

Find the equations of the straight lines bisecting the angles between the pairs of straight lines : $12x^2-7xy-12y^2=0;71x^2+94xy-71y^2=0;x^2-y^2=0$

Answer 1

$\text{For a pair of straight lines given by the equation}a{x}^2+2hxy+b{y}^2=0,\text{equation of the angle bisectors is given by}\frac{x^2-y^2}{a-b}=\frac{xy}{h}$

• $\text{Equation of the pair of straight lines is}12x^2-7xy-12y^2=0a=12,b=-12\text{and}h=\frac{-7}{2}\therefore \text{Equation of the angle bisector is}\frac{x^2-y^2}{24}=\frac{-2xy}{7}\text{or}7x^2+48xy-7y^2=0$
  

• $\text{Equation of the pair of straight lines is}71x^2+94xy-71y^2=0a=71,b=-71\text{and}h=47\therefore \text{Equation of the angle bisector is}\frac{x^2-y^2}{142}=\frac{xy}{47}\text{or}47x^2-142xy-47y^2=0$
  

• $\text{Equation of the pair of straight lines is}{x}^2-y^2=0a=1,b=-1\text{and}h=0\therefore \text{Equation of the angle bisector is}\frac{x^2-y^2}{2}=\frac{xy}{0}\text{or}xy=0$