Question

Find the equation of the line that bisects the obtuse angles between the lines 11x+2y-7=0 and x+2y+2=0.

Answer 1

There are two kinds of bisectors. One bisecting the acute angle, and the other bisecting the obtuse one.\[\dfrac{a_1 x + b_1 y + c_1}{\sqrt{a_1 ^2 + b_1^2}} = \pm \dfrac{a_2 x + b_2 y + c_2}{\sqrt{a_2 ^2 + b_2 ^2}}\]The above will give you the bisectors for \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\). The underlying reason is that for any point \((x,y ) \) on the bisector, the distance of the point from both lines would be equal.

In the next post, I'll explain how you can distinguish between the two bisectors...

Answer 2

To distinguish between the two bisectors, follow these steps:

1. Make \(c_1\) and \(c_2\) positive.

Case 1: If \(a_1 a_2 + b_1 b_2 > 0\), then the "+" sign will give you the obtuse angle bisector.
Case 2: If \(a_1 a_2 + b_1 b_2 < 0\), then the "-" sign will give you the obtuse angle bisector.

Answer 3

So what values should I plug in? ...

Answer 4

1. Make \(c_1\) and \(c_2\) positive for both the lines.

Answer 5

So that would be 11x+2y+7=0?

Answer 6

Somewhat.\[11x + 2y - 7 = 0\]You're allowed to manipulate this in such a way that you make the constant term positive. I'm sure you know how you can do that.

Answer 7

-11x-2y+7=0?

Answer 8

Correct!

Step 2: Check the sign of \(a_1 a_2 + b_1 b_2\).

Careful with the values though. \(a_1 \) is not the coefficient of \(x\) in the original equation, but the new one.

Answer 9

It's negative..

Answer 10

Great, so the "-" sign will give you the equation of your bisector.\[\dfrac{\color{blue}{-11}x \color{blue}{-2}x + 7}{\sqrt{(-11)^2 + (-2)^2}} = - \dfrac{x + 2y + 2}{\sqrt{(1)^2 + (2)^2}}\]

Answer 11

x=5y/4+3/2

Answer 12

I'm not gonna check that, but I'm just gonna assume you're correct. =_=

Good going!

Answer 13

LMAO thank you! Can I ask you a few more questions?

Answer 14

Sure!

Answer 15

Find the equation of the line passing through (6,9) and is parallel to the line whose inclination is arctan (2).

Answer 16

Hint: If the inclination is \(\theta\), then the slope is \(\tan \theta\).

Answer 17

Oh, BTW, in the previous question, did you understand how that algorithm came into existence?

Answer 18

Encountered that for the first time to be honest.. hehe

Answer 19

OK, I'm gonna give you a full derivation of that once we're done with this question if you have the time for it. I was too lazy to type all of it out the last time.

Answer 20

If I already got the slope I can now use point slope to get the equation right??

Answer 21

Exactly.

Answer 22

Okay thank you! :)

Answer 23

Before I start with the derivation, how'd you have done the question? What method was taught to you?

Answer 24

We have to graph the line..

Answer 25

Yeah, but you need data to go with it too. Graphing the line would simply give you the way it'd look.

I think you can manage it with geometry too, but anyway, since we've begun already...