parametric circle (1 Viewer)

[dawso]

hey all, got this question off a past school assessment task and im just wondering what you guys get for it (i think there is a flaw in the question)

Points P (x,y), Q(1,2) and R(3,-4) are such that QP perp. to RP
i) show that the locus of P is a circle
ii) find the centre and radius of the circle

the 'flaw' is only small but does anyone else find any problems with this proof??

once again, thankz in advance

-dawso

 

[acmilan]

I did it, and i think you are right. (was your error in (i) or (ii))?

 

[Slidey]

(2-y)/(1-x)=m1
(-4-y)/(3-x)=m2
m2.m1=-1
(2-y)(-4-y)=-(1-x)(3-x)
(y-2)(y+4)=-(x-1)(x-3)
y^2+2y-8=-(x^2-4x+3)
y^2+2y+x^2-4x=5
(y+1)^2 - 1 + (x-2)^2 - 4
(y+1)^2 + (x-2)^2=10
circle centre (2,-1) and radius sqrt10

 

[dawso]

well i think its obvious that my problem is in part i) cause theyre really aint anything that could be wrong with part ii) anyway, pm me what u think the problem is m8.... (i dont wanna post it in hear, we can make this the challenge of true maths people)

 

[dawso]

see now slide rule, u hav just done thsi algebraically, but if u do it by geometry and common sense, u see that in fact, it is not quite a circle....

 

[Slidey]

You think its an ellipse?

 

[dawso]

no, its in the shape of a circle, its just not QUITE a circle, try drawing it and look at it from a simple, basic point of view, not as an algebraic, mathematical problem

 

[Slidey]

I'm in English mode. I'm psyched for an assessment tomorrow worth 25%. You're going to have to spell this one out to me.

EDIT: For those wondering, there's an algabraic way to figure out why it isn't really a circle, too.

 

[Templar]

no, its in the shape of a circle, its just not QUITE a circle, try drawing it and look at it from a simple, basic point of view, not as an algebraic, mathematical problem

It is a locus problem, and it's meant to be seen from an algebraic, maths problem.

But yes, I think I see the flaw.

 

[dawso]

It is a locus problem, and it's meant to be seen from an algebraic, maths problem.

But yes, I think I see the flaw.

you are obviously one of these people that just learns how to do a question by what you are taught in class and doesnt actually think about it, did u know that algebra can be used to prove that 1+1=0, and anyway, the flaw, as proven by slide rule, can also be proved algebraically so there, anyway, dont just say u see the flaw, if anyone really wants my respect and sees the flaw, pm me to win a prize....(not a joke, u will win

 

[Templar]

you are obviously one of these people that just learns how to do a question by what you are taught in class and doesnt actually think about it, did u know that algebra can be used to prove that 1+1=0

Bad assumption.

And enlighten me with your proof of 1+1=0.

 

[Slidey]

To all those still wondering, here's the proof again:

(2-y)/(1-x)=m1
(-4-y)/(3-x)=m2
m2.m1=-1
(2-y)(-4-y)=-(1-x)(3-x)
(y-2)(y+4)=-(x-1)(x-3)
y^2+2y-8=-(x^2-4x+3)
y^2+2y+x^2-4x=5
(y+1)^2 - 1 + (x-2)^2 - 4
(y+1)^2 + (x-2)^2=10
circle centre (2,-1) and radius sqrt10

But from:
(2-y)/(1-x)=m1
(-4-y)/(3-x)=m2

x can't be 1 or 3, as this it isn't defined here. So the locus does not include any points on the lines x=1 and x=3. Since the circle has some points on these lines, it is not a complete circle.

The discluded points are:
(y+1)^2 + (x-2)^2=10
y+1=+/-3
(1,2), (1,-4), (3,2), (3,-4)
You'll note that two of these points are the original points Q and R.

Further, let it be known that under the axiom sets we use, 1+1=1'=2. It does not and will not ever equal zero.

 

[Templar]

Further, let it be known that under the axiom sets we use, 1+1=1'=2. It does not and will not ever equal zero.

Good. I thought arithmetic is complete.

 

[withoutaface]

you are obviously one of these people that just learns how to do a question by what you are taught in class and doesnt actually think about it, did u know that algebra can be used to prove that 1+1=0, and anyway, the flaw, as proven by slide rule, can also be proved algebraically so there, anyway, dont just say u see the flaw, if anyone really wants my respect and sees the flaw, pm me to win a prize....(not a joke, u will win

The algebra used in such a proof either relies on dividing by zero or not using +/- when you're square rooting.

 

[dawso]

The algebra used in such a proof either relies on dividing by zero or not using +/- when you're square rooting.

a, very good son, c, just because u end up with a solution at the end it doesnt mean that it is the correct answer, beleive it or not, by following the proof given be slide rule and not realising that x cannot equal 1 or 3, you are in fact dividing both sides of the equation by 0. As withoutaface said, this just destroys your results. eg:

. 1+1=0
. (x0) 0 =0

as 0=0, 1+1=0, waiting........no m8

therefore, the problem here is not a mathematical one, but a common sense one when u hav to look at the validity of results, anyway, enuf sounding like a smart arse, sick of maths, no ass tasks 4 3 weeks now, yayness

-dawso

 

[Slidey]

I have to disagree there: Yes, use common sense and inspection in maths, but also realise that if you follow correct mathematical procedure, you will always be correct. In this case it was a matter of noting that in your algebra what x could not equal.

 

[Templar]

I have to disagree there: Yes, use common sense and inspection in maths, but also realise that if you follow correct mathematical procedure, you will always be correct.

Not necessarily always, but it's a very good point. The mathematics being dealt with in this forum will rarely, if ever, be inconsistent.

 

[Slidey]

Not necessarily always, but it's a very good point. The mathematics being dealt with in this forum will rarely, if ever, be inconsistent.

I know I'm fumbling around in a dark cave here, but I would disagree in that even an inconsistent system is not merely 'incorrect'.

EDIT: Nor 'correct', so I concede your point.

 

[Templar]

This is highly irrelevant to the original topic, but for clarification:

Slide Rule, what you're talking about is undecidability, where the problem is neither formally proveable nor unproveable. However if you start with axioms and apply correct mathematical procedures, you'll still be correct, you'll just never reach what you're trying to prove.

In an inconsistent system, your axioms themselves lead to contradiction (ie you can prove something is both true and not true). Then what you have is incorrect.

Actually I probably have no idea what I'm going on about, so if any of the above makes no sense, just forget about it.