Exercise 1 page 247 of "Basic Mathematics" by S.Lang.
Question: Is mine $\eqref{2}$ or the author's $\eqref{1}$ solution correct?
Problem statement: Find the ordinary equation of the line $\{P + tA\}_{\ t\ in\ \mathbb R}$ in the following case
$$P = (3, 1), A = (7, -2)$$
The author's solution is $$\bbox[5px,border:2px solid red] {3x + 4y = 13} \tag{1}\label{1}$$
Which I belive is wrong, and could've possibly come from using points $P$ and $A$ as points of this same line, i.e. considering the line $\{P + t(A - P)\}_{\ t\ in\ \mathbb R}$, yielding the following equation:
$$(3, 1) + t(7 - 3, -2 - 1) = (3, 1) + t(4, -3)$$ Whose parametric representation is \begin{cases} x = 3 + 4t \\ y = 1 -3t \end{cases}
Following the procedure suggested by the author, we can eliminate the parameter $t$ by multiplying expressions $x$ and $y$ by some amount, and adding them together: \begin{cases} 3x = 9 + 12t \\ 4y = 4 -12t \end{cases}
It follows
\begin{align} 3x + 4y & = 9 + 12t + 4 -12t \\ & = 9 + 4 \\ & = 13 \end{align}
My interpretation of $\{P + tA\}_{\ t\ in\ \mathbb R}$ is all $t$ dillations of $A$ followed by a translation by $P$; the line parallel to $A$, passing through $P$.
With this interpretation I get \begin{align} P + tA & = (3, 1) + t(7, -2) \\ & = (3 + 7t, 1 -2t) \end{align}
Which can be represented parametrically as \begin{cases} x = 3 + 7t \\ y = 1 -2t \end{cases}
With the same procedure as before, we get \begin{cases} 2x = 6 + 14t \\ 7y = 7 -14t \end{cases}
It follows \begin{align} 2x + 7y & = 6 + 14t + 7 -14t \\ & = 6 +7 \\ & = 13 \end{align}
Yielding the ordinary equation $$\bbox[yellow,5px] {2x + 7y = 13} \tag{2}\label{2}$$
Calculate a normal vector for the line $\vec n = \hat A=\binom{2}{7}$. Now pick a fixed point on the line $P = (3,1)$. Now a equation for the line is: \begin{align*} \vec n \cdot \binom{x-3}{y-1} = 0 \Leftrightarrow 2(x-3)+7(y-1)=0 \Leftrightarrow 2x+7y = 13 \end{align*}
That is - you are right, and the book is wrong.