Question

Find the joint equation of the pair of line which bisect angles between the line given by
$x^2+3xy+{2y}^2=0$

Answer 1

$x^2+3xy+{2y}^2=0$
$\therefore x^2+2xy+xy+{2y}^2=0$
$\therefore x(x+2y)+y(x+2y)=0$
$\therefore (x+2y)(x+y)=0$
$\therefore$ separate equation of the lines represented by $x^2+3xy+2y^2=0$ are $x+2y=0$ and $x+y=0$
Let $P(x,y)$ be any point on one of the angle bisector. Since the points on the angle bisectors are equidistant from both the lines.
the distance of $P(x,y)$ from the line $x+2y=0$
$=$ the distance of $P(x,y)$ from the line $x+y=0$
$\therefore \left|\frac{x+2y}{\sqrt{1+4}}\right|=\left|\frac{x+y}{\sqrt{1+1}}\right|$
$\therefore \frac{(x+2y{)}^2}{5}=\frac{(x+y{)}^2}{2}$
$\therefore 2(x+2y{)}^2=5(x+y{)}^2$
$\therefore 2(x^2+4xy+4y^2)=5(x^2+2xy+y^2)$
$\therefore 2x^2+8xy+8y^2=5x^2+10xy+5y^2$
$\therefore 3x^2+2xy+3y^2=0$
This is the required joint equation of the lines which bisect the angles between the lines represented by $x^2+3xy+2y^2=0$