Question

Let
\[
3 x + 4 y = 0\\
1 + 5 x + 12 y = 0\\
\]
be two intersecting lines. Le I be the point of intersection of these two
lines.
Find the equations of the two bisectors lines of the two angles formed by the
the two above points at the point I.

Answer 1

@rational

Answer 2

Interesting.... perhaps putting them in y=mx+b form might lead me somewhere :D

\[\Large y = -\frac{4}{3}x\]\[\Large y = -\frac{5}{12}x-1\]

Answer 3

@terenzreignz , I got different equations.

Answer 4

oops...
\[\Large y = -\frac{5}{12}x-\frac1{12}\]

Answer 5

OK. This is the equation of one of the line and not the answer.

Answer 6

mhmm :3
and now... I'm thinking.

Answer 7

Here is a graph of the lines and their two bisectors. The red and green are the the two lines, the others are their bisectors.

Answer 8

Do you already have the answer? I have a feeling they'll be looking far from pretty....

Answer 9

Not really, it involves only rational numbers.

Answer 10

That's interesting... and encouraging... let me see :3

Answer 11

I guess just take the mean values of the slopes?

Answer 12

Look at the following for a hint:
http://openstudy.com/users/eliassaab#/updates/52d8eefee4b05ee05437b571

Answer 13

scratch the last one...

Answer 14

Crud... I hate it when I go wrong... for messing up details...
it should have been
\[\Large y = -\frac34x\]and\[\Large y=-\frac{5}{12}x-1\]

Answer 15

Then, translate:
\[\Large \bar x := x- \frac14\]\[\Large \bar y :=y+\frac3{16}\]

Answer 16

Equations become
\[\Large \bar y = -\frac34 \bar x\]\[\Large \bar y = -\frac5{12}\bar x\]

Answer 17

Find a point on the first line which is 1 unit away from the solution (centre).
\[\Large \sqrt{\bar x^2 + \frac{9}{16}\bar x^2}=\frac54\bar x = 1\implies \bar x= \frac45; \bar y = -\frac35\]
Similarly, on the second line, a point on the line which is 1 unit away from the solution is
\[\Large \bar x = \frac{12}{13};\bar y = -\frac{5}{13}\]

Answer 18

The midpoint of these two points is
\[\Large \bar x =\frac{56}{65};\bar y =-\frac{32}{65}\]

Answer 19

This and the point \[\Large \bar x = 0;\bar y = 0\] determine a line which is an angle bisector of the original two lines:
That line is given by\[\Large \bar y=-\frac{4}{7}\bar x\]
The other angle bisector is the line through the centre which is perpendicular to this line:
\[\Large \bar y = \frac74x\]

Answer 20

Reversing the translation and then simplifying, the lines are:
\[\Large y =-\frac47x-\frac5{112}\]\[\Large y=\frac74x-\frac58\]

@eliassaab ?

Answer 21

Oh, and I messed up my earlier correction (-_- what is wrong with me today)
\[\Large y = -\frac34x\]\[\Large y = -\frac5{12}\color{red}{(x+1)}\]

anyway, that aside, what do you think @eliassaab ?

Answer 22

@terenzreignz your equations are correct. That is what I got

Answer 23

using the hint from previous problem :

every point on angle bisector is equi distant from both the arms

say, P(x, y) be any point on the bisector, then this satisfies :-
\(\large \frac{3x+4y}{\sqrt{3^2+4^2}} = \pm \frac{1+5x+12y}{\sqrt{5^2+12^2}}\)

\(\large \frac{3x+4y}{5} = \pm \frac{1+5x+12y}{13}\)

\(\large y = \frac{1}{8} (14x-5)\)
\(\large y = \frac{1}{112}(-64x-5)\)

Answer 24

@rational, this is how I did it.