Question

Find the equation of the bisector of the angles between the straight lines $4x-3y+4=0$ and $6x+8y-9=0$.

Answer 1

The equations of the bisectors of the angles between $4x-3y+4=0$ and $6x+8y-9=0$ are :

$\frac{4x-3y+4}{\sqrt{4^2+(-3{)}^2}}=\pm \frac{6x+8y-9}{\sqrt{6^2+8^2}}$

$\frac{4x-3y+4}{5}=\pm \frac{6x+8y-9}{10}$

$40x-30y+40=\pm (30x+40y-45)$

Taking positive sign, we get,

$40x-30y+40=30x+40y-45$

$2x-14y+17=0$

Taking negative sign, we get,

$40x-30y+40=-(30x+40y-45)$

$70x+10y-5=0$

Therefore, the required equations are $2x-14y+17=0$ and $70x+10y-5=0$.