Since you wanted to see how this is done algebraically, we'll need to look at a particular property of quadratic functions. As I've said in a comment, if $r > 0$, then $ f(x) = rx^2 + sx + t \ $ represents an "upward-opening" parabola. By completing the square, we can write this as
$$f(x) \ = \ r \cdot (x^2 \ + \ \frac{s}{r} \cdot x \ + \ \frac{s^2}{4r^2}) \ + \ (t \ - \ r \cdot \frac{s^2}{4r^2}) \ = \ r \cdot ( \ x + \frac{s}{2r} \ )^2 \ + \ ( \ t - \frac{s^2}{4r} \ ) \ . $$
The vertex of this parabola lies at $ \ x = -\frac{s}{2r} \ $ , with the y-coordinate of the vertex being the minimum value of the function $ \ f(-\frac{s}{2r}) \ = \ t - \frac{s^2}{4r} .$ We will exploit this to reduce the amount of writing we will need to do.
We will label the closest point on the line to $ \ (x_1 , y_1) \ $ as $ \ (X,Y) \ $ . The distance-squared function is
$$D^2 \ = \ ( X - x_1 )^2 \ + \ (Y - y_1)^2 \ = \ ( X - x_1 )^2 \ + \ ( \ [-\frac{AX + C}{B} \ ] - y_1)^2 \ $$
$$= \ ( X - x_1 )^2 \ + \ ( \frac{A}{B} \cdot X \ + \ \frac{C}{B} + \ y_1)^2 \ = \ ( X - x_1 )^2 \ + \ ( \frac{A}{B} \cdot X \ + \ K \ )^2 \ , $$
where we will make a temporary substitution, $ \ K = \frac{C+By_1}{B} \ $ , to further save some writing. (This is often a good idea when working out complicated algebraic expressions, as it reduces the risk of embarrassing, time-wasting errors.) We will see that we can hold out for some time before needing to replace $K$ .
From here, we have
$$D^2 \ = \ ( \ X^2 \ - \ 2x_1X \ + \ x_1^2 \ ) \ + \ ( \ \frac{A^2}{B^2} \cdot X^2 \ + \ 2\frac{A}{B} K \cdot X \ + \ K^2 \ ) \ $$
$$= \ ( \ 1 + \frac{A^2}{B^2} \ ) \cdot X^2 \ + \ 2 ( \ \frac{A}{B} K - x_1 \ ) \cdot X \ + \ ( \ K^2 \ + x_1^2 \ ) , $$
which now has the "general form" for a quadratic function. Since we already know how to convert this to "standard form" by completing the square, we can simply use the results we discussed back at the start. The vertex of the parabola described by this equation lies at
$$X \ = \ -\frac{s}{2r} \ = \ -\frac{2 ( \ \frac{A}{B} K - x_1 \ )}{2( \ 1 + \frac{A^2}{B^2} \ )} \cdot \frac{B^2}{B^2} \ = \ \frac{ B^2 x_1 \ - \ AB K}{A^2 \ + \ B^2 \ } $$
$$= \ \frac{ B^2 x_1 \ - \ AB \cdot (\frac{C+By_1}{B})}{A^2 \ + \ B^2 \ } \ = \ = \ \frac{ B^2 x_1 \ - \ A \cdot (C \ + \ By_1)}{A^2 \ + \ B^2 \ } \ . $$
So here is the x-coordinate of the point on the line that is closest to $(x_1,y_1)$, confirming the stated result without the use of calculus.
We do not bother substituting this into either the quadratic function or the equation for the line, since we know from the standard form of the quadratic function that its minimal value is
$$D^2_{min} \ = \ t - \frac{s^2}{4r} \ = \ ( \ K^2 \ + \ x_1^2 \ ) \ - \ \frac{[ \ 2 ( \ \frac{A}{B} K - x_1 \ ) \ ]^2}{4( \ 1 + \frac{A^2}{B^2} \ )} \cdot \frac{B^2}{B^2} $$
$$= \ ( \ K^2 \ + \ x_1^2 \ ) \ - \ \frac{ ( \ A K \ - \ Bx_1 \ )^2}{ A^2 + B^2} \ = \ \frac{( \ K^2 \ + \ x_1^2 \ )(A^2 + B^2) \ - \ ( \ A K - Bx_1 \ )^2}{ A^2 + B^2} $$
$$= \ \frac{( \ A^2K^2 \ + \ B^2K^2 \ + \ A^2x_1^2 \ + \ B^2x_1^2 \ ) \ - \ ( \ A^2K^2 \ - \ 2ABKx_1 \ + \ B^2x_1^2 \ )}{ A^2 + B^2} $$
$$= \ \frac{ A^2x_1^2 \ + \ 2ABKx_1 \ + \ B^2K^2 \ }{ A^2 + B^2} \ = \ \frac{ ( \ Ax_1 \ + \ BK \ )^2 }{ A^2 + B^2}$$
$$= \ \frac{ ( \ Ax_1 \ + \ B \cdot [\frac{C+By_1}{B}] \ )^2 }{ A^2 + B^2} \ = \frac{ ( \ Ax_1 \ + \ By_1 \ + C \ )^2 }{ A^2 + B^2} \ .$$
Since $ \ \sqrt{x^2} = |x| \ , $ we at last have the relation for the minimum distance,
$$D_{min} \ = \ \frac{ | \ Ax_1 \ + \ By_1 \ + C \ | }{ \sqrt{A^2 + B^2}} \ .$$
It took me a few goes to find a reasonably clean way to develop this by algebra alone. The straight answer to your question about how "an expert" would set this up and proceed is that they wouldn't, but instead would use a more efficient method (such as the ones I linked to in my other answer). That may also explain why one doesn't have much luck finding an approach like this by looking around the internet. In any event, I hope this is what you were inquiring about.
Best Answer
I happen to have an explicit calculation written down, so here goes.
The key result to be familiar with is the fundamental theorem of symmetric polynomials. By this I mean not the result but with the process of converting any symmetric polynomial into expressions in the elementary symmetric polynomials. In our case, the coefficients of the given cubic are $-\sigma_1$, $\sigma_2$, and $-\sigma_3$, via the expansion of $(X-x_1)(X-x_2)(X-x_3)$.
I'm afraid that this will get (even more) unwieldy unless I condense notation. I will use $\Sigma x_1^ax_2^bx_3^c$ to refer to the sum of the elements in the orbit of the monomial $x_1^ax_2^bx_3^c$ under the action of the symmetric group. For instance $\Sigma x_1 = x_1 + x_2 + x_3 = \sigma_1$ and $\Sigma x_1^3x_2^2x_3 = x_1^3x_2^2x_3+x_1^3x_2x_3^2+x_1^2x_2^3x_3+x_1^2x_2x_3^3+x_1x_2^3x_3^2+x_1x_2^2x_3^3$.
First, I expand $\Delta$, which is not symmetric:
$\Delta = (x_1-x_2)(x_1-x_3)(x_2-x_3) = x_1^2x_2 - x_1^2x_3 - x_2^2x_1 + x_2^2x_3 - x_3^3x_2 + x_3^2x_1$.
Now we square it! Using the condensed notation I check directly
$\Delta^2 = \Sigma x_1^4x_2^2 - 2\Sigma x_1^4x_2x_3 - 2 \Sigma x_1^3x_2^3 +2\Sigma x_1^3x_2^2x_3 - 6 x_1^2x_2^2x_3^2$.
Now, it only remains to write this symmetric polynomial as a polynomial in $\sigma_1, \sigma_2, \sigma_3$. This is pretty tedious, but entirely mechanical. The leading terms justify why I'm choosing these particular expansions. As an aside, I compute only the coefficients of one monoial per orbit, and I also plug in $x_1=x_2=x_3=1$ as a double-check.
$\sigma_1^3\sigma_3 = \Sigma x_1^4x_2x_3 + 3\Sigma x_1^3x_2^2x_3 + 6 x_1^2x_2^2x_3^2$
$\sigma_2^3 = \Sigma x_1^3x_2^3 + 3\Sigma x_1^3x_2^2x_3 + 6x_1^2x_2^2x_3^2$
$\sigma_1\sigma_2\sigma_3 = \Sigma x_1^3x_2^2x_3 + 3x_1^2x_2^2x_3^2$
$\sigma_3^2 = x_1^2x_2^2x_3^2$
Now, it's just a matter of finding the right linear combination
Replace $\sigma_1,\sigma_2, \sigma_3$, with $-b, c, -d$, and you have the discriminant of $x^3+bx^2+cx+d$.