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mj.alawami · #3 June 26th, 2009, 01:16 PM

Quote:

Originally Posted by earboth

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

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:

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

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

3. The angle bisector passes through P and have the direction:
$\left(\frac2{\sqrt{13}} + \frac1{\sqrt{37}} \right)x + \left(\frac3{\sqrt{13}} + \frac6{\sqrt{37}} \right)y - C =0$

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

I am getting different answers :

$\frac{13+2\sqrt{481}}{13} x +\frac{78+2\sqrt{481}}{13}y-\frac{104-5\sqrt{481}}{13}=0$