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**earboth** 1. Calculate the intersection point. I've got $\displaystyle P\left(\frac23\ ,\ \frac{11}9\right)$

2. Now transform the given equations such that the direction vector (or in your case the normal vector to the direction) is a unit vector:

$\displaystyle 2x+3y-5=0~\implies~\frac2{\sqrt{13}}x+\frac3{\sqrt{13}}y-\frac5{\sqrt{13}}=0$

$\displaystyle x+6y-8=0~\implies~\frac1{\sqrt{37}}x+\frac6{\sqrt{37}}y-\frac8{\sqrt{37}}=0$

3. The angle bisector passes through P and have the direction:

$\displaystyle

\left(\frac2{\sqrt{13}} + \frac1{\sqrt{37}} \right)x + \left(\frac3{\sqrt{13}} + \frac6{\sqrt{37}} \right)y - C =0$

Plug in the coordinates of C to complete the equation of the line.