Equation of Bisector of Two Lines

[Raghav Gupta]

Homework Statement

Equations of the bisectors of the lines $3x - 4y + 7 =0$ and $12x + 5y - 2 = 0$
are ?

Homework Equations

Line equation is y = mx + c
where m is slope and c is y intercept

The Attempt at a Solution

Bisector is a line cutting equal angles between those 2 lines.
But what to do or approach that problem?

 

[BvU]

Make a drawing.
What do you know of the line you are looking for ? 1. some point it has to go through 2. something about the slope. That should be sufficient !

 

[Raghav Gupta]

Make a drawing.
What do you know of the line you are looking for ? 1. some point it has to go through 2. something about the slope. That should be sufficient !

Drawing seems difficult to me in which slopes are in fraction and y intercept is not 0.
I found the point of bisector, by solving that 2 equations which is (-3/7,10/7).
How to find slope?

 

[BvU]

You should really learn how to draw such a line. Here's a recipe:
1. take x = 0 and calculate what y is
2. take y = 0 and calculate what x is
you now have two points, so you can draw the line.

 

[Raghav Gupta]

You should really learn how to draw such a line. Here's a recipe:
1. take x = 0 and calculate what y is
2. take y = 0 and calculate what x is
you now have two points, so you can draw the line.

Okay, I will do that but first what is the use of that when we can solve algebraically the points ?

 

[BvU]

You can visually check your answer, develop a feeling, etc. etc. Comes in very handy when things become more complicated and more abstract.

Now for the slope of the bisectors: if line 1 has slope $\alpha$ and line 2 has slope $\beta$, what is the slope of one of the the bisectors ?

 

[Raghav Gupta]

Did the drawing and it gives me bad feeling,

 

[BvU]

Gives me a bad feeling too:

"I found the point of bisector, by solving that 2 equations which is (-3/7,10/7)"
-- but that point is clearly not where these two lines intersect!​

but not all that bad: small error only in $12x - 2 = 0 \Rightarrow x = 1/6$, not $-1/6$.

Now for the slope of the bisectors: if line 1 has slope $\alpha$ and line 2 has slope $\beta$ , what is the slope of one of the the bisectors ?

 

[Raghav Gupta]

Now for the slope of the bisectors: if line 1 has slope $\alpha$ and line 2 has slope $\beta$ , what is the slope of one of the the bisectors ?

Don't know.
If I know that, then I can easily find the equation of bisector.
So can you give a hint?

 

[Mark44]

Okay, I will do that but first what is the use of that when we can solve algebraically the points ?

You can visually check your answer, develop a feeling, etc. etc. Comes in very handy when things become more complicated and more abstract.

I agree completely with BvU here. My understanding of how the brain works is that one hemisphere is primarily analytical, and the other is more visual. The best way to tackle mathematics problems is to come at them with both sides of your brain. Too often, inexperienced students become enamoured with manipulating symbols of equations, and due to an error in their logic, come up with an incorrect solution. If they had also engaged the visual part of their brains, they would have been able to see that they had made a mistake.

 

[Raghav Gupta]

I agree completely with BvU here. My understanding of how the brain works is that one hemisphere is primarily analytical, and the other is more visual. The best way to tackle mathematics problems is to come at them with both sides of your brain. Too often, inexperienced students become enamoured with manipulating symbols of equations, and due to an error in their logic, come up with an incorrect solution. If they had also engaged the visual part of their brains, they would have been able to see that they had made a mistake.

But my point in this question is that if we know the diagram also, then also I think we cannot solve diagrammatically. ( Moreover we not get accurate diagrams).
We have to use algebra to solve.
Why to make a diagram? It is not giving me any info.
Without making drawing we can think that lines are intersecting.

 

[BvU]

RE your post #11: well, then we can still check whether our ultimate answer agrees with the picture we made.

Re your post #9: Don't know is fine with me, but not enough for you to bring the exercise to a successful conclusion. I could do the exercise for you, or give the answer to my own question, but that doesn't help you. That's why I asked in post #8.

Well, now we rummage through our toolkit of available relevant equations and come up with nothing at all, eh ? y = mx + c isn't good enough, so we'll have to develop a new tool somehow. To answer the question "if line 1 has slope $\alpha$ and line 2 has slope $\beta$ , what is the slope of one of the bisectors ?"

A first step could be to make it a bit simpler: what is a bisector of y= 0 and y = x ? Both through the origin, one slope zero, the other slope 1. Don't fall into the trap and claim "the line with slope 1/2!" because that's a little too easy and plain wrong. (Easily disproved: the bisector of y=0 and x=0 doesn't have slope $\infty/2$ but slope $1$)

Since the "bisector is a line cutting equal angles between those 2 lines" (should have had a place in the toolkit!) and one angle is 0 and the other is $\pi/4$ the line through the origin with a slope corresponding to an angle of $\pi/8$ seems a good candidate. So half the angle. Slope calculation awkward! Something with solving $\tan 2\phi = 1\rightarrow \tan\phi = ?$

 

[Mark44]

But my point in this question is that if we know the diagram also, then also I think we cannot solve diagrammatically. ( Moreover we not get accurate diagrams).

In post 1 you didn't say anything about the problem giving you a diagram. Also, your diagram doesn't have to be 100% accurate. It can give you a rough idea of what the solution looks like.

We have to use algebra to solve.
Why to make a diagram? It is not giving me any info.

As long as your diagram is reasonably accurate, it can give you lots of information. For example, in the diagram you show in post #7, I can see that one bisector will have a slope of about 1. The other bisector will have a slope of about -1. I can also get a reasonable idea of where the two lines intersect,

So from the drawing alone I can use the point of intersection and the two slopes, and write equations for the two bisectors. These won't be exactly correct, but they should be reasonably close, and should agree with the equations you get by using algebra alone.

If you make a mistake in your algebra calculation, you won't know that your work is incorrect. With a drawing, you have something to check against.

Without making drawing we can think that lines are intersecting.

 

[Raghav Gupta]

Since the "bisector is a line cutting equal angles between those 2 lines" (should have had a place in the toolkit!)

I have written that in attempt but not in relevant equation.

Something with solving $\tan 2\phi = 1\rightarrow \tan\phi = ?$

$\tan 2\phi = 1$
$$ ⇒ \frac{2tan\phi}{1-tan^2\phi} = 1 $$
Taking tanΦ = x

We get $2x = 1- x^2$

$⇒ x^2 + 2x - 1 = 0$

$⇒ x = \frac{-2 ± \sqrt{8}}{2}$

$⇒ tan\phi = -1 ± √2$
Now what is the next part?

 

[BvU]

Well done. So you know how to find the tangent of a halved angle.
If we go back to the exercise: first line has slope $m_1 = 3/4$, the second has slope $m_2 = -12/5$ .

(the notation $\alpha$ and $\beta$ for the slopes wasn't such a good idea , so I change to $m_1$ and $m_2$)​

If the angle of the first one is $\phi_1$ and the angle of the second one is $\phi_2$, what do you think you can expect for the angle of a bisector ?

 

[Raghav Gupta]

Well done. So you know how to find the tangent of a halved angle.
If we go back to the exercise: first line has slope $m_1 = 3/4$, the second has slope $m_2 = -12/5$ .

(the notation $\alpha$ and $\beta$ for the slopes wasn't such a good idea , so I change to $m_1$ and $m_2$)​

If the angle of the first one is $\phi_1$ and the angle of the second one is $\phi_2$, what do you think you can expect for the angle of a bisector ?

It would be $\frac{\phi_1 + \phi_2}{2}$ ? But I think it is not valid for all cases.
For this diagram it is true

We have only one dotted bi sector here?

For this it is not true

Sorry in second diagram, the angle $\phi_2$ should be on other side
and slope angle and between angles are different.

 

[BvU]

In other words: since the correct angle is indeed $180^\circ - \phi_2$ (your $\phi_2$) the expression $\phi = {\phi_1+\phi_2\over 2}$ ($\phi_2$ in this picture) is correct also for your second picture.

Why do you draw arrows ?

And: now we are at it, and have pictures at hand (how useful !) once you have one bisector, what about the other ?

 

[Raghav Gupta]

We are supposed to draw arrows on line as they are lines but not line segments.
Otherwise our fraction of marks are cut for not showing arrows.

Don't know the angle phi of other
See this,

 

[BvU]

OK, I understand the arrows. And two arrows on each line is a bit much, so apparently you have settled on one per line .

Don't know the angle phi of other (bisector)

Well, try out a few more pictures, preferably a bit more accurate ! There's a simple discovery waiting for you !

 

[Raghav Gupta]

And two arrows on each line is a bit much, so apparently you have settled on one per line .

Sorry for that will take care in future.
Getting $\phi=\frac{\phi_1 + \phi_2 - 180}{2}$

 

[BvU]

And for the other one ?

 

[Raghav Gupta]

And for the other one ?

The angles of bisectors with positive x-axis are,
(Φ1 + Φ2)/2 and for other (Φ1+Φ2-180°)/2

 

[BvU]

Yes, so the "discovery" is that the bisectors are perpendicular. Line angles (angles between lines) add up to 180 degrees, so half-angles to 90.

 

[Raghav Gupta]

Yes, so the "discovery" is that the bisectors are perpendicular. Line angles (angles between lines) add up to 180 degrees, so half-angles to 90.

(Φ1+Φ2-180°)/2

Shouldn't that be (180 - phi1 - phi2)/2 ?

 

[BvU]

Check #24 with a drawing

 

[Raghav Gupta]

Check #24 with a drawing

That is coming true what I am saying in post 22 by drawing( I've already uploaded so many images).

 

[BvU]

Indeed #22 is correct.
(Φ1 + Φ2)/2 and for other (Φ1+Φ2)/2 + π/2 (or (Φ1+Φ2)/2 - π/2, same line).
Not π/2 - (Φ1 + Φ2)/2

 

[Raghav Gupta]

Indeed #22 is correct.
(Φ1 + Φ2)/2 and for other (Φ1+Φ2)/2 + π/2 (or (Φ1+Φ2)/2 - π/2, same line).
Not π/2 - (Φ1 + Φ2)/2

It's now becoming a complicated calculation.

Getting, $tan(\phi_1 + \phi_2) = \frac{m_1 + m_2}{1-m_1m_2} = -33/56$

let $tan( \frac{\phi_1 + \phi_2}{2} ) = m$

So $-33/56 = \frac{2m}{1-m^2}$

$⇒ 33m^2 -112m -33 = 0$

It is getting quadratic and finding roots are also becoming difficult.

 

[BvU]

Yes -- it's getting quadratic.
No -- finding roots of a quadratic equation should be second nature for you. And they go out of their way in making it easy by picking very nice numbers . This you can check by cheating: calculate $\tan(\phi_1/2)$ and $\tan(\phi_2/2)$ on your calculator.

So you doin't want to go via $\tan(\phi_1 + \phi_2)$ but via $\tan(\phi_1/2)$ and $\tan(\phi_2/2)$; much more economic.

 

[Raghav Gupta]

But we are not allowed to use calculators in our examination.
Is there some other way?
And if we have to use calculator, then we can also easily find roots of that quadratic also.

 

[BvU]

I wouldn't worry too much about exams. Doing exercises develops your insight and experience and with that exams are a breeze.

Without a calculator, you can note that $\tan = 3/4$ and $\tan = -12/5$ (especially in combination) remind you of the simplest Pythagorean rectangular triangles 3,4,5 and 12,5,13 !

If you want it spelled out: Let $\tan (\alpha_1/2) = m$ when $\tan (\alpha_1) = 3/4$, then $${2m\over 1-m^2}=3/4 \ \Rightarrow \ m^2 + {8\over 3} m -1 = 0 \\ \Rightarrow \ (m+{4\over 3})^2 - (1+{16\over 9}) = 0 \ \Rightarrow \ m = -{4\over 3} \pm {5\over 3}$$and you want the positive one: $m = {1\over 3}$. Cute, eh ?

The other one yields an equally nice $m_2$ and then $m+ m_2\over 1-m m_2$ is pretty rational too. If you have one, you have the other from $-1/$ and a line with slope p through (-3/7, 10/7) is easily parametrized !

 

[Raghav Gupta]

I wouldn't worry too much about exams. Doing exercises develops your insight and experience and with that exams are a breeze.

Without a calculator, you can note that $\tan = 3/4$ and $\tan = -12/5$ (especially in combination) remind you of the simplest Pythagorean rectangular triangles 3,4,5 and 12,5,13 !

If you want it spelled out: Let $\tan (\alpha_1/2) = m$ when $\tan (\alpha_1) = 3/4$, then $${2m\over 1-m^2}=3/4 \ \Rightarrow \ m^2 + {8\over 3} m -1 = 0 \\ \Rightarrow \ (m+{4\over 3})^2 - (1+{16\over 9}) = 0 \ \Rightarrow \ m = -{4\over 3} \pm {5\over 3}$$and you want the positive one: $m = {1\over 3}$. Cute, eh ?

The other one yields an equally nice $m_2$ and then $m+ m_2\over 1-m m_2$ is pretty rational too. If you have one, you have the other from $-1/$ and a line with slope p through (-3/7, 10/7) is easily parametrized !

Hmm, this is okay
I was searching on google and found a very different approach and fast way
See link ditutor[\URL]
How that formula is derived?

 

[BvU]

Ha, you found something on the internet. At least that means you don't worry about exams that much any more (or do you have internet available then ?)

Now you can memorize that doubtlessly useful expression, or you can develop some insight. We've become familiar enough that I can safely assume the latter of the two. And in an edit you even ask for the derivation. Bear in mind that this is well beyond the call of duty wrt PF: they help with exercises but aren't a substitute for universities, schools or other means of teaching !

Reason I do it anyway is because it so nicely builds on your recently acquired knowledge of the normal equation for a line. (to me that also means that your curriculum and the exercises are well thought-through) !

The derivation: they write down the distances of P to the lines and set them equal. That's all.

To understand, you need an expression for the distance of a point to a line.

Unit normal equation for $Ax+By+C=0$ is $\displaystyle {Ax+By+C\over \sqrt{A^2+B^2}} =0$ .

Distance of line to origin is $|C|\over \sqrt{A^2+B^2}$ .


Unit normal equation for parallel line through $\vec P = (x_P, y_P)$ is $\displaystyle{Ax_P+By_P+D\over \sqrt{A^2+B^2}} =0$.

Distance of line to origin is $\displaystyle|D|\over \sqrt{A^2+B^2}$ .


Distance of P to line is $\displaystyle |D-C| = {|{Ax_P+By_P+C}|\over \sqrt{A^2+B^2} }$. Hey !​

 

[Raghav Gupta]

The derivation: they write down the distances of P to the lines and set them equal. That's all.

Well I am practicing for exam, so net is available to me now but was enchanted by that formula.
I know the distance formula from a line to point.
Understand now, that since it's a angle bisector the distances must be equal.

Thanks and a very much sorry from me.
It must have been painful for you to write that much text.

 

[BvU]

Well I am practicing for exam, so net is available to me now but was enchanted by that formula.
I know the distance formula from a line to point.
Understand now, that since it's a angle bisector the distances must be equal.

Thanks and a very much sorry from me.
It must have been painful for you to write that much text.

No need to apologize: you aren't forcing me, I do it all because I learn a bit too, and it's enjoyable. Good luck with your studies !