<s>[INST] Implicit line equation Trying to understand parametric and implicit line equations but I'm at a complete halt now. I have a line that goes through P(0,1) and Q(3,2), I need to find the implicit equation N((x,y)- Z) = 0 such that Z is a point on the line and N is a vector perpendicular to the line. Please point me in the right direction. Thank you.
 [/INST] Given the points:
$$\begin{gathered}
  P(0,1) \hfill \\
  Q(3,2) \hfill \\ 
\end{gathered}$$
Build the vector pointing from $P$ to $Q$:
$$\overrightarrow {PQ}  = Q - P = \left( {\begin{array}{*{20}{c}}
  3 \\ 
  1 
\end{array}} \right)$$
Set parametrization:
$$\left( {\begin{array}{*{20}{c}}
  x \\ 
  y 
\end{array}} \right) = P + t\overrightarrow {PQ} = \left( {\begin{array}{*{20}{c}}
  0 \\ 
  1 
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
  3 \\ 
  1 
\end{array}} \right)$$
Observe that 
$$N = \left( {\begin{array}{*{20}{c}}
  { - 1} \\ 
  3 
\end{array}} \right)$$
is orthogonal for $\left( {\begin{array}{*{20}{c}}
  3 \\ 
  1 
\end{array}} \right)$. With:
$$\left( {\begin{array}{*{20}{c}}
  x \\ 
  y 
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
  0 \\ 
  1 
\end{array}} \right) = t\left( {\begin{array}{*{20}{c}}
  3 \\ 
  1 
\end{array}} \right)$$
we get from dot product:
$$N \cdot \left( {\begin{array}{*{20}{c}}
  x \\ 
  y 
\end{array}} \right) - N \cdot \left( {\begin{array}{*{20}{c}}
  0 \\ 
  1 
\end{array}} \right) = t \cdot N \cdot \left( {\begin{array}{*{20}{c}}
  3 \\ 
  1 
\end{array}} \right)$$
so:
$$ - x + 3y - 3 = 0$$
Now we set:
$$f(x,y) =  - x + 3y - 3$$
and verify:
$$\begin{gathered}
  f(P) =  - 0 + 3 - 3 = 0 \hfill \\
  f(Q) =  - 3 + 6 - 3 = 0 \hfill \\ 
\end{gathered} $$
Function $f$ implicitly defines a line.
</s>