Relationship between Ax + By = C and by + ax = ab The general equation of line is $Ax + By = C$. Furthermore, the x and y intercepts respectively are $C/A$ and $C/B$. What I don't understand is this second equation which I need to solve a problem: $by + ax = ab$ where $b$ is the x-intercept and $a$ is the y-intercept. 
For context, the problem I'm trying to solve involves the fact that the area under the tangent line to $1/x$ is always 2. i.e, $Area = 1/2(2x)(2y)$. Therefore $xy = 4$. So the claim is that $x_1y + y_1x = x_1y_1$ where the subscripted ones are intercepts form a line.
Unfortunately I can't proceed without understanding the relationship between $Ax + By = C$ and $by + ax = ab$. Notice that $b$ and $a$ are flipped (that is $(b, 0)$ is the x-intercept in $by + ax = ab$) and that $C$ does not seem to equal $AB$, which I've tried to verify using wolfram alpha.
In other words, how does the general equation of the line lead to the lowercase equation?
 A: Let $f(x) = \frac{1}{x}$. Then the equation of a tangent line to this curve at some point $x_0$ is given by
\begin{align}
y &= f'(x_0)(x-x_0) + f(x_0)
\\
&= -\frac{1}{x_0^2}(x-x_0) + \frac1{x_0}
\\
&= -\frac{x}{x_0^2} + \frac2{x_0}
\end{align}
We can verify the area under the tangent line in the first quadrant is $2$ by
$$A = \frac12 x_{int} y_{int} = \frac12 (2x_0)\left(\frac2{x_0}\right) = 2.$$
This took a different approach, but I hope it helped. Also, comparing with your given equation $$by+ax=ab \quad\leftrightarrow\quad x_0y+\frac{x}{x_0} = 2$$
Without seeing the original prompt, I'm guessing that the problem is to find the tangent line equation using calculus and note that it has the form $by+ax=ab$. Your hangup may have to do with attempting to equate the general form of a line $Ax+By=C$ with the given form. I don't see the use in this. Just note that instead of any run-of-the-mill general line, the tangent line in this problem has an equation defined by two numbers $a,b$ and not three $A,B,C$ such that $A = b, B = a, C = ab$.
A: If $Ax+By=C$ and $by+ax=ab$ represent the same lines, then:
$$(A/C,B/C)=(1/b,1/a)$$
So, given $(a,b)$, we can take $(A,B,C)=(t/b,t/a,t)$ for any $t\neq 0$.
Likewise, given $(A,B,C)$, we can take $(a,b)=(C/B,C/A)$.
