<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}


{\bf Question}

State Pappus' theorem and prove it using vectors.

\vspace{.25in}

{\bf Answer}

\setlength{\unitlength}{.3in}
\begin{picture}(8,8)
\put(0,1){\vector(1,0){8}}

\put(0.5,0.5){\vector(1,1){7}}

\put(3,3.4){\makebox(0,0){$A$}}

\put(5,5.4){\makebox(0,0){$B$}}

\put(7,7.4){\makebox(0,0){$C$}}

\put(3,0.5){\makebox(0,0){$A'$}}

\put(5,0.5){\makebox(0,0){$B'$}}

\put(7,0.5){\makebox(0,0){$C'$}}

\put(3,1){\line(1,2){2}}

\put(3,1){\line(2,3){4}}

\put(5,1){\line(-1,1){2}}

\put(5,1){\line(1,3){2}}

\put(7,1){\line(-2,1){4}}

\put(7,1){\line(-1,2){2}}

\put(8.25,1){\makebox(0,0){${\bf y}$}}

\put(7.7,7.7){\makebox(0,0){${\bf x}$}}

\put(3.25,2.2){\makebox(0,0){$R$}}

\put(4.5,2.6){\makebox(0,0){$Q$}}

\put(6.2,3.5){\makebox(0,0){$P$}}

\put(.8,1.25){\makebox(0,0){$0$}}

\put(9,4.5){\makebox(0,0)[l]{Pappus states that}}

\put(9,4){\makebox(0,0)[l]{$P,Q,R$ are collinear.}}

\end{picture}



Let the points have position vectors: $$A: \alpha{\bf x}
\hspace{.2in}B: \beta{\bf x} \hspace{.2in}C: \gamma{\bf x}
\hspace{.2in} A': \alpha'{\bf y} \hspace{.2in}B': \beta'{\bf y}
\hspace{.2in}C': \gamma'{\bf y}$$

Then $P$: $\ds t = \beta{\bf x} - (1-t)\gamma {\bf y} = s\gamma
{\bf x} + (1-s)\beta'{\bf y} $

So $t\beta = s\gamma$ and $(1-t)\gamma' = (1-s)\beta'$.

Solving these gives: $$s = \frac{\beta(\gamma' -
\beta')}{\gamma\gamma'-\beta\beta'} \hspace{.2in} 1-s =
\frac{\gamma'(\gamma-\beta)}{\gamma\gamma'-\beta\beta'}$$

So $${\bf P} : \frac{\beta\gamma(\beta' -
\gamma')}{\beta\beta'-\gamma\gamma'}{\bf x} +
\frac{\beta'\gamma'(\beta-\gamma)}{\beta\beta'-\gamma\gamma'}{\bf
y}$$

Similarly $${\bf Q} : \frac{\alpha\gamma(\gamma' -
\alpha')}{\gamma\gamma'-\alpha\alpha'}{\bf x} +
\frac{\gamma'\alpha'(\gamma-\alpha)}{\gamma\gamma'-\alpha\alpha'}{\bf
y}$$

$${\bf R} : \frac{\alpha\beta(\alpha' -
\gamma')}{\alpha\alpha'-\gamma\gamma'}{\bf x} +
\frac{\alpha'\beta'(\alpha-\beta)}{\alpha\alpha'-\beta\beta'}{\bf
y}$$

${}$

Thus $\alpha\alpha'(\beta\beta'-\gamma\gamma') {\bf P} +
\beta\beta'(\gamma\gamma'-\alpha\alpha'){\bf Q} +
\gamma\gamma'(\alpha\alpha'-\beta\beta'){\bf R} = {\bf 0}$

and $\alpha\alpha'(\beta\beta'-\gamma\gamma') +
\beta\beta'(\gamma\gamma'-\alpha\alpha') +
\gamma\gamma'(\alpha\alpha'-\beta\beta')= 0$

Thus $P Q R$ are collinear. 


\end{document}
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