My Math Forum Proving that P, Q, and R lie on the same straight line

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 January 9th, 2018, 10:19 PM #1 Member   Joined: Sep 2017 From: Saudi Arabia Posts: 37 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: 863 Thanks: 328 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
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Quote:
 Originally Posted by Micrm@ss What exactly do you expect of us? You gave us nothing useful to work with or to help you with?
Dear,

Just I need to know how to begin (a hint)

 January 10th, 2018, 05:43 AM #4 Senior Member   Joined: Oct 2009 Posts: 863 Thanks: 328 A hint on what? You didn't tell us anything.
January 10th, 2018, 07:42 AM   #5
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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
$\displaystyle P\left(\dfrac{ab(d-e)}{ad-eb};\dfrac{ed(a-b)}{ad-eb}\right)$.

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.
Attached Images
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 January 10th, 2018, 05:17 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 This is Pappus's hexagon theorem, and hence also a special case of Pascal's theorem. Thanks from greg1313, romsek and Hussain2629 Last edited by skipjack; January 10th, 2018 at 08:42 PM.
January 13th, 2018, 10:09 PM   #7
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From: Saudi Arabia

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Quote:
 Originally Posted by Snair 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 $\displaystyle P\left(\dfrac{ab(d-e)}{ad-eb};\dfrac{ed(a-b)}{ad-eb}\right)$. 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.
Thank you very much. í ¼í¼¹

Last edited by skipjack; January 14th, 2018 at 01:25 AM.

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