1. ## Analytic geometry help?

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 ? :'(

2. 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.

3. 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 :

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

4. Originally Posted by mj.alawami
I am getting different answers :

$\frac{13+2\sqrt{481}}{13} x +\frac{78+2\sqrt{481}}{13}y-\frac{104-5\sqrt{481}}{13}=0$
I have attached a screenshot. Obviously we both have made a mistake somewhere.

If (and when) I find my mistake I'll be back.

5. Back again.

I've attached my complete calculations. The graph shows the equations #1, #2 and #11

6. Originally Posted by earboth
Back again.

I've attached my complete calculations. The graph shows the equations #1, #2 and #11

$\boxed {(-\sqrt{13}-2\sqrt{37})x+(-6\sqrt{13}-3\sqrt{37})y+(8\sqrt{15}+5\sqrt{37}}$

7. ## equation of angle bisector

posted by mj.alawami and earboth

This is an interesting problem which may still be looking for an answer

I get the following

y=-.3955x +1.486

Method used employs trig to convert slopes to elevation angles.The angle for the bisector is the angle of lower line plus half the angle between the lines 21.58 degrees slope -.3955. line equation from point slope formula using the point (2/3,11/9)

bj

8. Originally Posted by bjhopper
posted by mj.alawami and earboth

This is an interesting problem which may still be looking for an answer ..... No

I get the following

y=-.3955x +1.486

...
The result from my previous post was:

$\left(\dfrac{2\cdot \sqrt{13}}{13} + \dfrac{\sqrt{37}}{37}\right) \cdot x + \left(\dfrac{3 \cdot \sqrt{13}}{13} + \dfrac{6 \cdot \sqrt{37}}{37}\right) \cdot y - \left(\dfrac{5\cdot \sqrt{13}}{13} + \dfrac{8 \cdot \sqrt{37}}{37}\right) = 0$

If you solve this equation for y and use approximate values you'll get:

$y = - 0.3954474799\cdot x + 1.485853875$