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January 9th, 2018, 10:19 PM  #1 
Newbie Joined: Sep 2017 From: Saudi Arabia Posts: 29 Thanks: 1  Proving that P, Q, and R lie on the same straight line Any help please? 
January 10th, 2018, 04:09 AM  #2 
Senior Member Joined: Oct 2009 Posts: 439 Thanks: 147 
What exactly do you expect of us? You gave us nothing useful to work with or to help you with?

January 10th, 2018, 05:20 AM  #3 
Newbie Joined: Sep 2017 From: Saudi Arabia Posts: 29 Thanks: 1  
January 10th, 2018, 05:43 AM  #4 
Senior Member Joined: Oct 2009 Posts: 439 Thanks: 147 
A hint on what? You didn't tell us anything.

January 10th, 2018, 07:42 AM  #5 
Newbie Joined: Jan 2018 From: Tunis, Tunisia Posts: 10 Thanks: 4 Math Focus: Analysis 
Hi ! You can consider the frame $\displaystyle (O,x,y)$ where the straight line $\displaystyle l_1$ is the $\displaystyle x$axis and the straight line $\displaystyle l_2$ is the $\displaystyle y$axis and $\displaystyle O$ is the point of intersection $\displaystyle l_1$ and $\displaystyle l_2$. So we can say that the coordinates of points $\displaystyle A,B,C,D,E$ and $\displaystyle F$ can be written as : $\displaystyle A(a,0)~;~B(b,0)~;~C(c,0)~,~D(0,d)~;~E(0,e)$ and $\displaystyle F(0,f)$. Now you can find an equation to the straight line $\displaystyle (AE)$ and another for the straight line $\displaystyle (BD)$. I found them like this : $\displaystyle (AE)~;~y=\dfrac eax+e$ $\displaystyle (BD)~;~y=\dfrac dbx+d$ Now because $\displaystyle \{P\}=(AE)\cap(BD)$ we can compute the coordinates of $\displaystyle P$ by solving a linear system. I found that $\displaystyle P\left(\dfrac{ab(de)}{adeb};\dfrac{ed(ab)}{adeb}\right)$. By the same way we can compute the coordinates of $\displaystyle Q$ and $\displaystyle R$. And finally we can compute the coordinates of the vectors $\displaystyle \vec{PQ}$ and $\displaystyle \vec{PR}$ and we should find that these vectors are collinear. 
January 10th, 2018, 05:17 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,505 Thanks: 1741 
This is Pappus's hexagon theorem, and hence also a special case of Pascal's theorem.
Last edited by skipjack; January 10th, 2018 at 08:42 PM. 
January 13th, 2018, 10:09 PM  #7  
Newbie Joined: Sep 2017 From: Saudi Arabia Posts: 29 Thanks: 1  Quote:
Last edited by skipjack; January 14th, 2018 at 01:25 AM.  

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