We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

Hey guys I'm sorry this is the third time that I post this exercice , any advice or answers can help mesmiley


be ABC a triangle non flat,we define the isobarycentre of the triangle ABC  the point G such that :


the point G is the centre of gravity of the triangle ABC .




1.the respective points I,J and k are placed in the segments (AB),(BC) and (AC). express the vector AJ  according to the vector(AB) and the vector(AC) ?.


2.using the expression: vector(AG)+vector(BG)+vector(CG)=0, express the vector(AG) accrording to the vector(AB) and the vector(AC) ? .


3.show that we can write vector(AG)=Kvector(AJ) "(AG=KAJ) " , for a certain k to be determined ?.

 Jan 4, 2019

In 1. what do you mean by retrospectively ?

Is J placed so that it lies on the vector AG or is it at some random point on BC, (and then, in which case, how can AJ be a scalar multiple of AG ? )

 Jan 5, 2019

1.respectively that the point I is placed in the middle of the segment AB and j in the middle of the segment BC and k in the middle of the segment AC.

- for J in the statement he says it's a respective point of segmnet BC so I have to put it.

Then for AJ and AG .I think I should use the relation of chasle .

I want to put an image so you can see the the drawing of the triangle but I can't let me .

 Jan 5, 2019
edited by Guest  Jan 5, 2019

So J is the mid-point of BC, why couldn't you say that first time round ?


Since J is the mid point of BC, we can say that CJ = - BJ, (equal in magnitude and opposite in direction).


Triangle law , twice.

    AJ = AB + BJ ,

    AJ = AC + CJ .


Add them and divide by 2,  AJ = (AB + AC)/2 , (the other two vectors have cancelled each other).


Now do a similar thing to get an expression for AG , (using  BG + CG =  - AG on the way).


Combine the two results, (you should find that k = 2/3).

 Jan 5, 2019

For AG  i did:







I used GA+GB+GC=0 instead of AG+BG+CG=0

 Jan 5, 2019
edited by Guest  Jan 5, 2019

And if we want to deduce that the points A,G and k are aligned how we can prove it ?

Guest Jan 5, 2019

It would have been better to have started with  AG = AB + BG and AG = AC + CG etc., 

but your final equation is correct and you can make use of it by multiplying throughout by a negative sign.

 - GA = - (- 1/3 AB - 1/3 AC) , so   AG = (1/3)( AB + AC ).

Now you should see that AJ and AG are both scalar multiples of the same vector AB  + AC

in which case A G and J must be aligned.

Finally, work out the value of k in order that  AG = k.AJ .

Guest Jan 6, 2019

So I said :

if AB and AC are collinear ABC are aligned

show that AG and AJ are collinear

goal: we want to get AG =KAJ 

mean to express AG and AC. .with the help of AB(AI=AB) and AC(AK=AC)


AJ=IA+AB+1/2BC (since BJ=1/2 of BC)

⇔1/3AB+AB+1/2(BA+1/2AC) since BA+AC=BC

⇔1/3AB+AB+1/2AB+1/2AC(because we have multiplied by 1/2 for AB and AC)

⇔5/6AB+1/2AC( I did 1/3+1-1/2)

now for AG I did 



 Now I am blocked. 

Is it correct for AJ?

 Jan 6, 2019

For k I said:





so 1/2 × 2/3 =1/3 

We have 2/3 × AJ = AG 

which prove that AG=2/3AJ

 Jan 6, 2019

Hi ,

Never mind for my answer I found it 




SO we have AG+AG+AB+AG+AC

which give:AG=1/3(AB+AC) 

Since J is the middle of BC we can write that AB+AC=AJ

So AG=1/3(AJ)=2/3AJ


 So A,G,J are aligned  

Thank you very much for your help appreciate it ❤😀

 Jan 6, 2019

9 Online Users