The equation of the angle bisectors of the lines represented by the equation 2x^2+xy-3y^2=…

Question

The equation of the angle bisectors of the lines represented by the equation $2x^2+xy-3y^2=0$ is:

Accepted Answer

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

Here $a=2,\ h=\dfrac12,\ b=-3$:
$$\frac{x^2-y^2}{2-(-3)} = \frac{xy}{1/2} \implies \frac{x^2-y^2}5 = 2xy \implies x^2-y^2=10xy$$

$x^2-y^2-10xy=0$

Solutions · 1