Question

Find the equation of the bisector of the acute angle between the lines
$3x-4y+7=0$ and $12x+5y-2=0$

Answer 1

The bisectors of given lines are given by
$\frac{3x-4y+7}{\sqrt{25}}=\pm \frac{12x+5y-2}{\sqrt{169}}$
The bisectors are
$21x+77y-101=0$ and $11x-3y+9=0$
Let $\theta$ be the angle between one of the lines say $3x-4y+7=0$ and one of the bisectors say $11x-3y+9=0$. Their slopes are
$m_1=3/4;m_2=11/3$
$\therefore \tan \theta =\frac{\left(3/4\right)-\left(11/3\right)}{1+\left(3/4\right)\left(11/3\right)}=-\frac{35}{45}$
and it is numerically less than $1$
Hence $\theta <{45}^{o}or2\theta <{90}^o$. Therefore this is the bisector of the acute angle between the lines and hence the other bisects the obtuse angle.