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earboth · #2 June 26th, 2009, 12:59 PM

Quote:

Originally Posted by mj.alawami

Find the equation of that line that bisects the acute angles determined by the given lines.

$2x+3y-5=0 and x+6y-8=0$

Attempt:
I know that i have to get the distance of d1 and d2 and then apply the equation
$d1+d2=0$

But when i try doing it i get very complex root forms and i guess it is wrong so how to do it ? :'(

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.