Question

Find the bisector of the acute angle between the lines $12x-5y+7=0$ and $3x-4y-1=0$.

Answer 1

Bisector of the acute angle between $12x-5y+7=0$ and $3x-4y-1=0$

$\frac{12x-5y+7}{\sqrt{{12}^2+5^2}}=\frac{3x-4y-1}{\sqrt{3^2+4^2}}\Rightarrow \frac{12x-5y+7}{\sqrt{{13}^2}}=\frac{3x-4y-1}{\sqrt{5^2}}\Rightarrow \frac{12x-5y+7}{13}=\frac{3x-4y-1}{5}\Rightarrow 60x-25y+35=39x-52y-13\Rightarrow 21x+27y+48=0\Rightarrow 21x+27y+48=0\Rightarrow 21x+27y+48=0$