# Bisector of an acute angle (Help)

• Jun 26th 2009, 09:53 PM
mj.alawami
Bisector of an acute angle (Help)
Find the equation of the line that bisects the acute angles determined by the given lines.

1) $x-6y-12=0$ and $4x-6y-9=0$

Attempt:

$d1=\frac{x-6y-12}{-\sqrt{37}}$ , $\frac{4x-6y-9}{-2\sqrt{13}}$
So d1+d2 =0
$d1=\frac{x-6y-12}{-\sqrt{37}}$ + $\frac{4x-6y-9}{-2\sqrt{13}}$

= $\frac{(-4\sqrt{37} -2\sqrt{13} )x +(6\sqrt{37} +12\sqrt{13} )y+(-9\sqrt{37} -24\sqrt{13} )}{2\sqrt{481}}$

How to make the equation free of square roots?
And is my solution correct ( if not please follow the procedure that i have done which is d1+d2=0 )
Thank you.
• Jun 26th 2009, 10:26 PM
malaygoel
Quote:

$
d1=\frac{x-6y-12}{-\sqrt{37}}
$
, $
\frac{4x-6y-9}{-2\sqrt{13}}
$

why you have found distances from the origin?
• Jun 26th 2009, 10:33 PM
mj.alawami
Quote:

Originally Posted by malaygoel
why you have found distances from the origin?

Oh sorry it should be x prime and y prime.
• Jun 26th 2009, 11:00 PM
malaygoel
Quote:

Originally Posted by mj.alawami
Find the equation of the line that bisects the acute angles determined by the given lines.

1) $x-6y-12=0$ and $4x-6y-9=0$

Attempt:

$d1=\frac{x-6y-12}{-\sqrt{37}}$ , $\frac{4x-6y-9}{-2\sqrt{13}}$

It is
$d1=\frac{|x-6y-12|}{\sqrt{37}}$ , $\frac{|4x-6y-9|}{2\sqrt{13}}$
Quote:

So d1+d2 =0
so there are two equtions:
d1+d2=0
d1=d2
Quote:

$d1=\frac{x-6y-12}{-\sqrt{37}}$ + $\frac{4x-6y-9}{-2\sqrt{13}}$
$0=\frac{x-6y-12}{\sqrt{37}}$ + $\frac{4x-6y-9}{2\sqrt{13}}$
$\frac{x-6y-12}{\sqrt{37}}$ = $\frac{4x-6y-9}{2\sqrt{13}}$

I have not checked calculations...will do in a short time.
• Jun 26th 2009, 11:09 PM
mj.alawami
You have an error in your second equation it should be
d1=-d2