I've been working my way through "Linear algebra and Geoemtry a second course" by Kaplansky and I must say the section on affine geometry is giving me quite a bit of trouble! I'm trying to work through the first two exercises (stated below) at the end of chapter 3 section 1. I'm sure they are very basic but I'm having trouble proving them... The affine setting is quite new to me! If somebody could hold me hand and walk me through these I'd really appreciate it. Thank you!
Let $V$ be a vector space of all ordered pairs of elements of a field $K$.
Let $p,q,r \in K$ be given with $p$ and $q$ not both $0$. I'm trying to show that the set of all $(x,y)$ satisfying $px+qy+r=0$ is a line, and also that every line has this form.
Show that the lines given by $px+qy+r=0$ and $p_1x+q_1y+r_1=0$ are parallel or identical if and only if $pq_1-p_1q=0$
Let $\mathbb A(E)$ be an affine space with translation space the $K$-vector space $E$, and suppose that $\dim(\mathbb A(E)):=\dim(E)=m$.
A (non empty) subset $S$ of $\mathbb A(E)$ is an affine subspace of $\mathbb A(E)$ iff, by definition:
$$ \vec F := \left\{Q-P\;:\;P,Q\in S\right\} \tag1$$
is a vector-subspace of $E$. In this case, we set $S=Q+\vec F,\;$ if $Q$ is any point of $S$ $\;\big[$the subspace passes through $Q$ $\big]$. Warning: in many textbooks, $\;Q-P\;$ is written $\;\vec{PQ}$.
Coordinate systems. Let $\;e:=(e_i)_{i=1,2,\ldots,m}\;$ be an ordered basis of $E$, and $\;x:=(x^i)_{1=1,2,\ldots,m}\;$ the corresponding (Cartesian) coordinate system on $E$, that is
$$ \forall v\in E,\qquad v=x^1(v)e_1+x^2(v)e_2+\ldots+x^m(v)e_m = \sum_{i=1}^m x^i(v)e_i. \tag2$$
Starting from $\;x\;$, we can define a similar (Cartesian) coordinate system $X=(X^i)_{i=1,2,\ldots,m}\;$ on $\mathbb A(E)$, if we choose and fix a point $O\in\mathbb A(E)$ $\;\big[$Origin of the (affine) coordinate system$\big]$ and define
$$ \forall P\in \mathbb A(E),\qquad X^i(P):=x^i(P-O)\qquad(i=1,2,\ldots,m). \tag3$$
Equations of a subspace. Let $\;S=Q+\vec F\;$ be an affine $n$-dimensional subspace of $\mathbb A(E)\quad(n\leq m).$ We have
$$ P\in S \iff P-Q\in \vec F. \tag4$$
Consider an ordered basis $(\varepsilon_1,\varepsilon_2,\ldots,\varepsilon_n)$ of $\vec F$. It is well known that this ordered basis can be extended to an ordered basis $\varepsilon=(\varepsilon_1,\ldots,\varepsilon_n,\varepsilon_{n+1},\ldots,\varepsilon_m)$ of $E$, and let denote by $y=(y^j)_{j=1,2,\ldots,m}$ the corresponding coordinate system.
Then, for each vector $v\in E$ we have:
\begin{align} v&=\sum_{i=1}^m x^i(v)e_i=\\[1ex] &=\sum_{i=1}^m x^i(v)\sum_{j=1}^m M^i_j \varepsilon_j=\tag5\\[1ex] &=\sum_{j=1}^m\sum_{i=1}^m M^i_jx^i(v)\varepsilon_j \end{align}
where $M=\big[M^i_j\big]$ is the $m\times m$-matrix that achieves the transition from the base $\varepsilon$ to the base $e$; matrix which, remember, is invertible.
From $(5)$ we deduce, by the uniqueness of the vector expansion in a given basis, that
$$ v\in \vec F \qquad \iff\qquad \sum_{i=1}^m M^i_j x^i(v)= 0\quad\text{ for }\; j=n+1,\ldots,m, \tag6 $$
i.e. $\;v\in\vec F\;$ iff its $\;e$-coordinates satisfy the system of equations
$$\left\{ \begin{array}{l} M^1_{n+1}x^1+M^2_{n+1}x^2+\ldots+M^m_{n+1}x^m=0\\ M^1_{n+2}x^1+M^2_{n+2}x^2+\ldots+M^m_{n+2}x^m=0\\ \ldots \tag7\\[.5ex] M^1_{m}x^1+M^2_{m}x^2+\ldots+M^m_{m}x^m=0,\\ \end{array} \right.$$
which represent a (Cartesian) system of equations of the vector subspace $\vec F$ in the coordinate system $(x^1,x^2,\ldots, x^m)$. This system has rank $\;m-n\;$ since $M$ is an invertible matrix.
Finally, from that we can obtain a (Cartesian) system of equations for the affine subspace $S=Q+\vec F\;$ by observing that
$$ X^i(P)-X^i(Q) = x^i(P-O)-x^i(Q-O) = x^i\Big((P-O)-(Q-O)\Big) = x^i(P-Q), \tag 8$$
so $(4)$ and $(6)$ give: $\;P\in S=Q+\vec F\; \iff \; P-Q\in \vec F\;\iff\;$ the ($x$-coordinates of the) vector $P-Q$ satisfy the system $(6)$ [or $(7)$], i.e. iff for each $j=n+1,\ldots,m$:
$$ \sum_{i=1}^m M^i_jx^i(P-Q)=0, $$
that is
$$ \sum_{i=1}^m M^i_jX^i(P)=\sum_{i=1}^m M^i_jX^i(Q), \tag9$$
therefore $P\in S\;\iff\;$ its $X$-coordinates satisfy the system
$$\left\{ \begin{array}{l} M^1_{n+1}X^1+M^2_{n+1}X^2+\ldots+M^m_{n+1}X^m=Q_{n+1}\\ M^1_{n+2}X^1+M^2_{n+2}X^2+\ldots+M^m_{n+2}X^m=Q_{n+2}\\ \ldots \tag{10}\\[.5ex] M^1_{m}X^1+M^2_{m}X^2+\ldots+M^m_{m}X^m=Q_m,\\ \end{array} \right.$$
where
$$ Q_j := \sum_{i=1}^m M^i_jX^i(Q)\qquad[j=n+1,\ldots,m]. $$
The system $(10)$ is the desired system of equations of $S$. Note that this system has rank $m-n$.
Vice versa. Consider the system
$$ \sum_{i=1}^m M^i_jX^i=C_j\qquad[j=1,\ldots,t]. \tag{11}$$
and suppose that it is compatible, i.e. there exists $Q\in\mathbb A(E)$ such that
$$ \sum_{i=1}^m M^i_jX^i(Q)=C_j\qquad[j=1,\ldots,t]. \tag{12}$$
If $S$ is the set of solutions of $(11)$, then
$$ P\in S \;\iff\; \sum_{i=1}^mM^i_jX^i(P)=C_j\qquad[j=1,\ldots,t] \tag{13} $$
and comparing with $(12)$, it turns out that
$$ P\in S\;\iff\; \sum_{i=1}^m M^i_j\big[X^i(P)-X^i(Q)\big] = \sum_{i=1}^m M^i_j x^i(P-Q) = 0 \qquad[j=1,\ldots,t]$$
i.e. $\;P-Q\;$ belongs, for each $j=1,\ldots,t$, to the kernel $\;\vec F_j\;$ of the linear function $\;\displaystyle \sum_{i=1}^m M^i_jx^i,\;$ so $\;S=Q+\vec F$, where $\;\vec F = \displaystyle\bigcap_{j=1}^t \vec F_j. $
We observe that the rank $\varrho$ of the system $(11)$ is $\leq t$, and then the dimension of $\;\vec F\;$ is $\;m-\varrho\geq m-t$.
Conclusion In your case, we have $m=2$ and $n=1$. the matrix of the system (which has a unique equation), that is, in the notations you used, the matrix $\;[p\;\,q]$, has rank $1$ iff $\;p,q\;$ aren't both zero.