Originally Posted by

**mj.alawami** [FONT="Arial"]Find the equation of the bisector of the acute angle whose equation is

$\displaystyle 4x-5y-10 \bold{\color{red}= 0}$ and $\displaystyle x-4y+6\bold{\color{red}= 0} $ **<<<<<< I assume that you mean these equations**

__Attempt__

L1=$\displaystyle \frac{4x'-5y'-10}{-\sqrt{41}}$

L2=$\displaystyle \frac{x'-4y'+6}{-\sqrt{17}}$

Since, the bisect line is in the middle of the lines then we use the formula $\displaystyle \boxed{ -L2=L1 }$

$\displaystyle -[-\frac{x-4y+6}{\sqrt{17}}]=-[\frac{4x-5y-10}{\sqrt{41}}] $

=$\displaystyle \boxed{(\sqrt{41}-4\sqrt{17})x+(-4\sqrt{41}-5\sqrt{17})y+(6\sqrt{41}-10\sqrt{17})=0} $

Is my answer correct?

Thank you.