[quote
author=jackpot]Let A(x₁, y₁), B(x₂,
y₂) and C(x₃, y₃) be the vertices of
triangle ABC. Derive the formula for the coordinates of it's
circumcentre.


Please help!!!
Tags: benbuks, doubleDx, Laplacian, dejt4u AmazingAngel, STENON,
efficiencie.[/quote] Pls i will only attach its solutiin because I may not be able to type it

did i hear my name .?

( no she didn't call you , she only mentioned your name ) ok .. better

* goes back to what he was downloading * .

#math.clinic .hibernation mood........till further notice.



pray for him ooo.. he don't solve or study his books again this days ..kinda killing him softly .

hmm..

jackpot: Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of triangle ABC. Derive the formula for the coordinates of it's circumcentre.


Please help!!!
Tags: benbuks, doubleDx, Laplacian, dejt4u AmazingAngel, STENON, efficiencie.

general equation of a circle, x²+y²-2ax-2by+c=0, the centre is (a,b), three points on the circumference are given. This is enough!!

[quote
author=Laplacian]
general equation of a circle, x²+y²-2ax-2by+c=0,
the centre is (a,b), three points on the circumference are given. This
is enough!![/quote] Sorry Pls....Is that the same for coordinate of Circumcentre?

Laplacian:
general equation of a circle, x²+y²-2ax-2by+c=0, the centre is (a,b), three points on the circumference are given. This is enough!!

nice hint/try sir

@ jackpot leme c what can be done , gana have the formula in matrix form or otherwise .

i'll try it . might get bk 2 u .

ur really tempting me mathematically.

HABRUZZY: thanks alot bro have been able to sort my way round with some online guides thanks for the concern for now the only giving me headache is

In an examination in statistics Ade's mark in
standard units is 0.10.Marks are assumed to be
normally distributed.What percentage of the students
would be expected to score higher than Ade.

still here pls

jackpot: Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of triangle ABC. Derive the formula for the coordinates of it's circumcentre.


Please help!!!
Tags: benbuks, doubleDx, Laplacian, dejt4u AmazingAngel, STENON, efficiencie.

what i gt ws somehow long but i will try to type it here..

I got this by 1st finding the equation of a circle that passes through the 3points..

Let (m,n) represents the coordinates of the circumcenter..

Therefore,
m = 1/2 [(x₁² + y₁²)(y₃ - y₂) + (x₂² + y₂²)(y₁ - y₃) + (x₃² + y₃²)(y₂ - y₁)] / [x₁(y₃ - y₂) + x₂(y₁ - y₃) + x₃(y₂ - y₁)]

for the 2nd coordinate,
n = 1/2 [(x₁² + y₁²)(x₃ - x₂) + (x₂² + y₂²)(x₁ - x₃) + (x₃² + y₃²)(x₂ - x₁)] / [y₁(x₃ - x₂) + y₂(x₁ - x₃) + y₃(x₂ - x₁)]

benbuks:
your friend is ryt bro


medicine cobo. is eng , phy , bio & chem ,

unless you wanna study engr. den combo is eng. math , phy , & chem.

that's it .


don't you have a brochure ?

Nahhhh..........thanks anyway.

jackpot: Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of triangle ABC. Derive the formula for the coordinates of it's circumcentre.


Please help!!!
Tags: benbuks, doubleDx, Laplacian, dejt4u AmazingAngel, STENON, efficiencie.

Draw lines from one vertex to a mid point. You'll do
this three times. Then get equations for each line and
solve the three simultaneous equations.

any mathematics guru in da house may please help me with this:
If α and β are the roots of equation 3x²+5x-8=0
Find the value of α^4-ß^4

sundaylinus: any mathematics guru in da house may please help me with this:
If α and β are the roots of equation 3x²+5x-8=0
Find the value of α^4-ß^4

do it dis way,
α⁴-ß⁴ = (α²)²-(ß²)² = (α²-ß²)(α²+ß²)...diff. Of 2 squares;

simplifying further,
(α-ß)(α+ß)((α+ß)²-2αß)..

You can continue from there..
Note that: (α-ß) = (α²+ß²-2αß)^1/2 = ((α+ß)²-4αß)^1/2

dejt4u:
do it dis way,
α⁴-ß⁴ = (α²)²-(ß²)² = (α²-ß²)(α²+ß²)...diff. Of 2 squares;

simplifying further,
(α-ß)(α+ß)((α+ß)²-2αß)..

You can continue from there..
Note that: (α-ß) = (α²+ß²-2αß)^1/2 = ((α+ß)²-4αß)^1/2

dejt4u:
do it dis way,
α⁴-ß⁴ = (α²)²-(ß²)² = (α²-ß²)(α²+ß²)...diff. Of 2 squares;

simplifying further,
(α-ß)(α+ß)((α+ß)²-2αß)..

You can continue from there..
Note that: (α-ß) = (α²+ß²-2αß)^1/2 = ((α+ß)²-4αß)^1/2

dejt4u:
do it dis way,
α⁴-ß⁴ = (α²)²-(ß²)² = (α²-ß²)(α²+ß²)...diff. Of 2 squares;

simplifying further,
(α-ß)(α+ß)((α+ß)²-2αß)..

You can continue from there..
Note that: (α-ß) = (α²+ß²-2αß)^1/2 = ((α+ß)²-4αß)^1/2

thanks bro. But am confuse i dnt have idear on this, its an assigment pls put more light

sundaylinus: thanks bro. But am confuse i dnt have idear on this, its an assigment pls put more light

3x²+5x-8=0
Find the value of α^4-ß^4

do it dis way,
α⁴-ß⁴ = (α²)²-(ß²)² = (α²-ß²)(α²+ß²)...diff. Of 2 squares;

simplifying further,
(α-ß)(α+ß)((α+ß)²-2αß)....eqn*

You can continue from there..
Note that: (α-ß) = (α²+ß²-2αß)^1/2 = ((α+ß)²-4αß)^1/2...eqn**


putting eqn** in eqn*
(α-ß) = (α²+ß²-2αß)^1/2 = ((α+ß)²-4αß)^1/2 ...eqn**
(α-ß)(α+ß)((α+ß)²-2αß)...eqn*
= ((α+ß)²-4αß)^1/2(α+ß)((α+ß)²-2αß) ...eqn+

ur question, 3x²+5x-8=0
a=3, b=5, c=-8,
but (α+ß) = b/a = 5/3
and (αß) = -c/a = -(-8/3) = 8/3
subtituting the values above in eqn+,
=[(α+ß)²-4αß]^1/2(α+ß)[(α+ß)²-2αß)]
=[(5/3)²-4(8/3)]^1/2 x 5/3[(5/3)²-2(8/3)]
you can finish that up.. Looking at this, it may lead u to complex no.. Jst note dat i² = -1

AlphaMaximus:
Complex numbers? All u have to do is to find the roots of the given equation which are -8/3 and 1, find their fourth powers and the difference between their fourth powers! I don't see how we'll land in any "complex numbers".

yh my boss..u ar right.. But believe me if they asked you questions like this in my school nd u do it dis way.. You are jst paying with ZERO..
A question like this was asked in MTH101 exam wen i ws in part1.. And that very question contributed to the great number of F's in dat very course..
May God bless Professor S.S Okoya

jackpot: Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of triangle ABC. Derive the formula for the coordinates of it's circumcentre.


Please help!!!
Tags: benbuks, doubleDx, Laplacian, dejt4u AmazingAngel, STENON, efficiencie.

The circumcentre of a triangle is the point in the triangle equidistant from all three vertices/the point where any two perpendicular bisectors of any two of the three sides of a triangle intersect. Therefore to find the circumcentre of a triangle, we have to obtain the midpoints of two sides, their gradients and find the equations of each of their perpendicular bisectors and equate them to solve simultaneously for the x and y coordinates of the circumcenter.
To find midpoint of side AB:
(X1+X2/2 , Y1+Y2/2)<====midpoint of AB
The gradient ====>(y2-y1/x2-x1)
Therefore the equation of the parallel bisector of AB is found using:
Y-y1=m(x-x1)
Y-y1=[y2-y1/x2-x1](x-x1)........eq1
Now repeating the same procedure for line BC:
Midpoint====>(x2+x3/2,y2+y3/2)
Gradient=====>(y3-y2/x3-x2)
Equation of perpendicular bisector===>
Y-y2=m(x-x2)
Y-y2=[y3-y2/x3-x2](x-x2).......eq2
After both equations have been obtained , you simply solve simultaneously and obtain the values of x and y which are the x and y coordinates of the circumcentre.

dejt4u:
3x²+5x-8=0
Find the value of α^4-ß^4
do it dis way,
α⁴-ß⁴ = (α²)²-(ß²)² = (α²-ß²)(α²+ß²)...diff. Of 2 squares;
simplifying further,
(α-ß)(α+ß)((α+ß)²-2αß)....eqn*
You can continue from there..
Note that: (α-ß) = (α²+ß²-2αß)^1/2 = ((α+ß)²-4αß)^1/2...eqn**
putting eqn** in eqn*
(α-ß) = (α²+ß²-2αß)^1/2 = ((α+ß)²-4αß)^1/2 ...eqn**
(α-ß)(α+ß)((α+ß)²-2αß)...eqn*
= ((α+ß)²-4αß)^1/2(α+ß)((α+ß)²-2αß) ...eqn+
ur question, 3x²+5x-8=0
a=3, b=5, c=-8,
but (α+ß) = b/a = 5/3
and (αß) = -c/a = -(-8/3) = 8/3
subtituting the values above in eqn+,
=[(α+ß)²-4αß]^1/2(α+ß)[(α+ß)²-2αß)]
=[(5/3)²-4(8/3)]^1/2 x 5/3[(5/3)²-2(8/3)]
you can finish that up.. Looking at this, it may lead u to complex no.. Jst note dat i² = -1

yea think u made a slight mistake . (alpha+beta)=-b/a and alpha*beta=c/a

masperano:


yea think u made a slight mistake . (alpha+beta)=-b/a and alpha*beta=c/a

correct..thanks my guy..its bin very long that i do things like dis last

AlphaMaximus:
Complex numbers? All u have to do is to find the roots of the given equation which are -8/3 and 1, find their fourth powers and the difference between their fourth powers! I don't see how we'll land in any "complex numbers".

excellent!! straightforward.

dejt4u:
correct..thanks my guy..its bin very long that i do things like dis last

yea i noticed it. you were clever to break down the expansion, that is the most challenging part in the question

jackpot: Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of triangle ABC. Derive the formula for the coordinates of it's circumcentre.
Please help!!!
Tags: benbuks, doubleDx, Laplacian, dejt4u AmazingAngel, STENON, efficiencie.

Stennon how are you? and Jackpot why did u steal my profile pic . lol. I love clever splakss(chicks). where is Doubledx(a great guy). and what of efficiencie(wanted to ask what you studied. mathematics or engineering or physics) you got the age question i threw in. you indeed have a rigorous mathematical mind. To all the generals masperano sends his greetings. between am one hell of a lazy dude else i would have been typing in solutions . may be when am less busy will start contributing. You guyz are really doing a great job. Kudos!!!

benbuks:

Draw lines from one vertex to a mid point. You'll do
this three times. Then get equations for each line and
solve the three simultaneous equations.

Sir, is that not a referrence to the centroid?

AlphaMaximus:
The circumcentre of a triangle is the point in the triangle equidistant from all three vertices/the point where any two perpendicular bisectors of any two of the three sides of a triangle intersect. Therefore to find the circumcentre of a triangle, we have to obtain the midpoints of two sides, their gradients and find the equations of each of their perpendicular bisectors and equate them to solve simultaneously for the x and y coordinates of the circumcenter.
To find midpoint of side AB:
(X1+X2/2 , Y1+Y2/2)<====midpoint of AB
The gradient ====>(y2-y1/x2-x1)
Therefore the equation of the parallel bisector of AB is found using:
Y-y1=m(x-x1)
Y-y1=[y2-y1/x2-x1](x-x1)........eq1
Now repeating the same procedure for line BC:
Midpoint====>(x2+x3/2,y2+y3/2)
Gradient=====>(y3-y2/x3-x2)
Equation of perpendicular bisector===>
Y-y2=m(x-x2)
Y-y2=[y3-y2/x3-x2](x-x2).......eq2
After both equations have been obtained , you simply solve simultaneously and obtain the values of x and y which are the x and y coordinates of the circumcentre.

Yes, that's the basic. I need a compact formula. The type that you will just slot in the vertices and the answer will come out.

masperano:



Stennon how are you? and Jackpot why did u steal my profile pic . lol. I love clever splakss(chicks). where is Doubledx(a great guy). and what of efficiencie(wanted to ask what you studied. mathematics or engineering or physics) you got the age question i threw in. you indeed have a rigorous mathematical mind. To all the generals masperano sends his greetings. between am one hell of a lazy dude else i would have been typing in solutions . may be when am less busy will start contributing. You guyz are really doing a great job. Kudos!!!

mine had been there since 2012, me thinks.

sundaylinus: any mathematics guru in da house may please help me with this:
If α and β are the roots of equation 3x²+5x-8=0
Find the value of α^4-ß^4

Clearly.

dejt4u: yh my boss..u ar right.. But believe me if they asked you questions like this in my school nd u do it dis way.. You are jst paying with ZERO..
A question like this was asked in MTH101 exam wen i ws in part1.. And that very question contributed to the great number of F's in dat very course..
May God bless Professor S.S Okoya

I see that as an injustice. I am sticking with Alpha Maximus on this one. He should demand for a methodological solution in his statement. Such questions can be posed this way. . .

''using the theory of quadratic equations, . . .''


Or


''Use algebra to evaluate. . .''

masperano:



excellent!! straightforward.

Surely. I write it that way and you give me an F, only God will judge you.

Why will I choose a rigorous path when there is a shorter, flawless and concise path?

jackpot: Surely. I write it that way and you give me an F, only God will judge you.

Why will I choose a rigorous path when there is a shorter, flawless and concise path?

exactly..they ar so wicked in my school..more than 70% of my dept mate failed MTH in year 1 nd 2.. Yellow house suck ma GP small, na God eventually save me from them ooo..

dejt4u:

what i gt ws somehow long but i will try to type it here..

I got this by 1st finding the equation of a circle that passes through the 3points..

Let (m,n) represents the coordinates of the circumcenter..

Therefore,
m = 1/2 [(x₁² + y₁²)(y₃ - y₂) + (x₂² + y₂²)(y₁ - y₃) + (x₃² + y₃²)(y₂ - y₁)] / [x₁(y₃ - y₂) + x₂(y₁ - y₃) + x₃(y₂ - y₁)]

for the 2nd coordinate,
n = 1/2 [(x₁² + y₁²)(x₃ - x₂) + (x₂² + y₂²)(x₁ - x₃) + (x₃² + y₃²)(x₂ - x₁)] / [y₁(x₃ - x₂) + y₂(x₁ - x₃) + y₃(x₂ - x₁)]

That's what I got too. The final answer is also

((|BC|²)(AB.AC)+(|AC|²)(BA.BC)+(|AB|²)(CA.CB))/(2|AB x AC|²)

Please, can you express your answer in this form?

NOTE:
x is cross-product
. is dot-product

dejt4u:
exactly..they ar so wicked in my school..more than 70% of my dept mate failed MTH in year 1 nd 2.. Yellow house suck ma GP small, na God eventually save me from them ooo..

shiiit happens. I had my fair share too. When an intelligent student is on the receiving end, it is prejudicial.

Laplacian:
general equation of a circle, x²+y²-2ax-2by+c=0, the centre is (a,b), three points on the circumference are given. This is enough!!

confirm, but you see, Sir, i just need a direct proof leading to a compact formula. Haven't proved one completely yet.

masperano:



Stennon how are you? and Jackpot why did u steal my profile pic . lol. I love clever splakss(chicks). where is Doubledx(a great guy). and what of efficiencie(wanted to ask what you studied. mathematics or engineering or physics) you got the age question i threw in. you indeed have a rigorous mathematical mind. To all the generals masperano sends his greetings. between am one hell of a lazy dude else i would have been typing in solutions . may be when am less busy will start contributing. You guyz are really doing a great job. Kudos!!!

I'm here oga mi; it's been a long time bruv! Hope you are doing great...1luv