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January 9th, 2018, 10:19 PM   #1
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Proving that P, Q, and R lie on the same straight line



Any help please?
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January 10th, 2018, 04:09 AM   #2
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What exactly do you expect of us? You gave us nothing useful to work with or to help you with?
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January 10th, 2018, 05:20 AM   #3
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Quote:
Originally Posted by Micrm@ss View Post
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)
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January 10th, 2018, 05:43 AM   #4
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A hint on what? You didn't tell us anything.
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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.
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January 10th, 2018, 05:17 PM   #6
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This is Pappus's hexagon theorem, and hence also a special case of Pascal's theorem.
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Last edited by skipjack; January 10th, 2018 at 08:42 PM.
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January 13th, 2018, 10:09 PM   #7
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Quote:
Originally Posted by Snair View Post
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.
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Last edited by skipjack; January 14th, 2018 at 01:25 AM.
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