\documentclass[10pt,envcountsect]{beamer}

\mode<presentation> {
  \usetheme{Warsaw}
}

\usepackage[english]{babel}
\usepackage[latin1]{inputenc}
\usepackage{times}
\usepackage[T1]{fontenc}
\usepackage{pstricks}
\usepackage{multicol}



\begin{document}

\title{Episode 17 -- Different types of line equations}
\subtitle{European section, season 01}
\date{}

\begin{frame}
  \titlepage
\end{frame}

\begin{frame}{General form}

\Large  $$ax+by+c=0$$

Any line has an equation of this form, and in fact many as it's always possible to apply a coefficient to the
whole equation.

\end{frame}


\begin{frame}{Slope-intercept form}

\Large 

\begin{multicols}{2}
$$y=mx+p$$

{\large \begin{itemize}
\item $m$ is the slope ;
\item $p$ is the intercept.
\end{itemize}

\smallskip

This form exists and is unique for any line not parallel to the $y$-axis.}

\includegraphics[scale=1]{images/slopeintercept.eps}
\end{multicols}



\end{frame}


\begin{frame}{Point-slope form}

\begin{multicols}{2}

\Large  $$y-y_0=m(x-x_0)$$

{\large \begin{itemize}
\item $m$ is a the slope ;
\item $(x_0,y_0)$ are the coordinates of a point on the line.
\end{itemize}

\smallskip

This form exists for any line not parallel to the $y$-axis. It's not unique as any point can be used.}

\includegraphics[scale=1]{images/slopepoint.eps}
\end{multicols}

\end{frame}

\begin{frame}{Intercept form}

\begin{multicols}{2}

{\Large  $$\frac{x}{q}+\frac{y}{p}=1$$}

\large

\begin{itemize}
\item $p$ is the $y$-intercept ;
\item $q$ is the $x$-intercept.
\end{itemize}

\smallskip

This form exists and is unique for any line not parallel to an axis.

\includegraphics[scale=1]{images/intercept.eps}

\end{multicols}


\end{frame}

\begin{frame}{Part 1}
\begin{center}
\Large Part 1
\end{center}
\end{frame}

\begin{frame}{Part 2}
\begin{center}
\begin{tabular}{|c|c|c|c|c|} \hline
 & GF & SIF & PIF & IF \rule{0pt}{15pt}\\[5pt] \hline
$d_1$ & $3x-3y+3=0$ &  &  &  \rule{0pt}{15pt}\\[5pt] \hline
$d_2$ & &  &  & $\frac{x}{3}+\frac{y}{-3}=1$ \rule{0pt}{15pt}\\[5pt] \hline
$d_3$ & &  &  & $\frac{x}{5}+\frac{y}{2}=1$ \rule{0pt}{15pt}\\[5pt] \hline
$d_4$ & & $y=6x-11$ & & \rule{0pt}{15pt}\\[5pt] \hline
\end{tabular}
\end{center}
\end{frame}



\end{document}
