Answer
Proof given below.
Work Step by Step
The statement is equivalent to $ {\bf v}$ being perpendicular to the line,
if both slopes are defined.
Let P be a point on the line with x-coordinate of $x_{p}$
If $b\neq 0,\ \quad y_{p}=-\displaystyle \frac{a}{b}x_{p}-\frac{c}{b}$.
Let Q be a point on the line with x-coordinate of $x_{q}$
If $b\neq 0,\ \displaystyle \quad y_{q}=-\frac{a}{b}x_{q}-\frac{c}{b}$
The directed segement $\displaystyle \overrightarrow{PQ}=\langle x_{q}-x_{p},\ \ \frac{a}{b}(x_{p}-x_{q}) \rangle$
$\displaystyle \overrightarrow{PQ}\cdot {\bf v}=a( x_{q}-x_{p})+b\cdot\frac{a}{b}(x_{p}-x_{q}) =a( x_{q}-x_{p})-a( x_{q}-x_{p})=0$
So they are perpendicular.
So, for $b\neq 0, {\bf v}$ is perpendicular to the line.
If they are perpendicular, their slopes are negative reciprocals.
Note:
If $b=0,$ then ${\bf v}=\langle a,\ \ 0 \rangle$,
So the line is vertical, with equation $x=\displaystyle \frac{c}{a}$, slope undefined.
Any two points on it have the same x, so
vector $\overrightarrow{PQ}=\langle 0,\ \ y_{q}-y_{p} \rangle$
The dot product is again zero, $\qquad a(0)+0(y_{q}-y_{p})$
so ${\bf v}$ is perpendicular to the line, but the slope of the line is undefined.
(The statement does not apply to this case, as it includes defined slopes, but if restated in terms of ${\bf v}$ being perpendicular to the line, it works for this case as well.)
