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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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 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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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 Align Points.jpg (13.5 KB, 28 views) 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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 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. ������

Last edited by skipjack; January 14th, 2018 at 01:25 AM. Tags lie, line, proving, straight Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jl1974 Algebra 7 September 13th, 2017 11:55 PM poochie03 Algebra 3 November 5th, 2011 12:00 PM outsos Algebra 7 December 23rd, 2010 07:15 PM Kiranpreet Algebra 3 August 10th, 2008 07:45 AM Kiranpreet Algebra 2 August 9th, 2008 09:29 AM

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