2
$\begingroup$

Why does a rearrangement of the equation lead to it missing a point at $(x_1,y_1)$? I understand the solution is indeterminate, but why does this happen? Why does rearrangement lead to a "loss in information" of the equation?

$\endgroup$
2
  • $\begingroup$ They are different equations. $\endgroup$
    – copper.hat
    Dec 22, 2013 at 8:55
  • $\begingroup$ The same thing happens if you compare $y=x^2$ to $y/x=x$, the latter has a hole at $(0,0)$. Basically the division means one must avoid zero denominator. $\endgroup$
    – coffeemath
    Dec 22, 2013 at 8:56

1 Answer 1

2
$\begingroup$

By convention, when you're given an equation which implicitly describes a function of $x$ like that, and you're not given any information about the domain, you take the largest possible domain. So in the first equation, we are assuming that $y=f(x)$ is a function defined implicitly by $y-y_1=m(x-x_1)$—that is to say, $y=f(x)$ satisfies that equality for every value of $x$ in the domain of $f$. From this information, it is easy to check that there are no values of $x$ which make the function undefined, so we can safely assume that $f$ is a function from the real numbers to the real numbers.

In your second example, again we assume that $y=f(x)$ is a function defined implicitly by $m=\frac{y-y_1}{x-x_1}$, and we want to find the largest possible domain for $f$ where this equation will hold. But a function that satisfies that equation for all $x$ in its domain must not be defined at $x=x_1$. So the largest domain that we can take here is $\mathbb{R}\setminus\{x_1\}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .