Question

Find the equation of the obtuse angle bisector of lines $4x-3y+10=0$ and $8y-6x-5=0$.

Answer 1

First, we make the constant terms positive in the given two equations.

Making constant terms positive, the two-equation becomes

$4x-3y+10=0$ and $6x-8y+5=0$

Now, $a_1{a}_2+b_1{b}_2=4\times 6+(-3)\times (-8)=24+24=48$, which is positive.

Hence, the obtuse angle bisector is

$\frac{4x-3y+10}{\sqrt{4^2+(-3{)}^2}}=\frac{6x-8y+5}{\sqrt{6^2+(-8{)}^2}}$

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

$10x+10y+150=0$

$x+y+15=0$, which is the required obtuse angle bisector.