R-squared: X "explains" the percentage of variation of the Y values. Does axis order matter? Summary:  Am I correlating two independent variables?   Is that the problem?
Let's say I have data point for square footage of a house and the asking price.   Now, I can ask "Does square footage (x) determine price (y)?"  This is intuituve, and makes sense.  R-squared would end up saying "Square footage explains X% of the variation in price".  So far, so good.
But, what if I want to predict the square footage from the price?  That seems valid.  Now, I ask "Does price (x) determine square footage (y)?"  So far, it seems either can function as the independent or dependent variable.  However, the wording or r-squared seem off.  "Price explains X% of the variation in square footage".  Huh?  Square footage is not some sort of multi-factor variable.  It's more static.  Nothing "explains" the square footage, it just is.  Get what I'm saying?  Like if price only explains x% of square footage, what else would explain square footage?  Square footage is just square footage.  It's not like price, which can be determined by many things (square footage, renovations, size of yard, etc).
Another example can be age (x) and the mileage on a car (y).  With a regression equation, I can use one to predict the other.  Either order seems to work.  However does age "explain" the mileage, or does mileage "explain" the age?  In this case, both seem weird.  Both are just static independent variables.  Neither explains the other, if you ask me.
What am I missing here?  Thanks!
 A: Edited with added material in response to comments by @whuber
This is an answer based on probability theory, not statistical estimates, so your mileage may vary.
If random variables $X$ and $Y$ have correlation coefficient $\rho$, then the linear least-mean-square error estimate of $Y$ given the value of $X$ is 
$$\hat{Y} = \mu_Y + \rho\frac{\sigma_Y}{\sigma_X}(X - \mu_X),$$
and similarly,
the linear least-mean-square error estimate of $X$ given the value of $Y$ is 
$$\hat{X} = \mu_X + \rho\frac{\sigma_X}{\sigma_Y}(Y - \mu_Y).$$
Note that $\hat{Y}$ and $\hat{X}$ are  random variables that are linear functions of $X$ and $Y$ respectively. Their means are 
$$\begin{align*}
\mu_{\hat{Y}} &=E[\hat{Y}] = E\left[\mu_Y+\rho\frac{\sigma_Y}{\sigma_X}(X - \mu_X)\right] 
= \mu_Y+ \rho\frac{\sigma_Y}{\sigma_X}E[X - \mu_X] = \mu_Y\\
\mu_{\hat{X}} &= E[\hat{X}] =  E\left[\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(Y - \mu_Y)\right] 
= \mu_Y+ \rho\frac{\sigma_X}{\sigma_Y}E[Y - \mu_Y] = \mu_X
\end{align*}$$
while the variances are 
$$\begin{align*}
\sigma_{\hat{Y}}^2 &= E[(\hat{Y} - \mu_{\hat{Y}})^2]
= \frac{\rho^2\sigma_Y^2}{\sigma_X^2}E[(X-\mu_X)^2] = \rho^2\sigma_Y^2\\
\sigma_{\hat{X}}^2 &= E[(\hat{X} - \mu_{\hat{X}})^2]
= \frac{\rho^2\sigma_X^2}{\sigma_Y^2}E[(Y-\mu_Y)^2] = \rho^2\sigma_X^2
\end{align*}$$
Finally, the variances of the residual
errors $Y - \hat{Y}$ and $X - \hat{X}$ are $\sigma_Y^2(1-\rho^2)$ and
$\sigma_X^2(1-\rho^2)$ respectively.  One can think of these results as 
follows.

If we use the mean $\mu_Y$ as an estimate for $Y$, the mean-square error is  $\sigma_Y^2$, but if we know the value of $X$ and use $\mu_Y + \rho\frac{\sigma_Y}{\sigma_X}(X - \mu_X)$ as the estimate of $Y$,  the mean-square error is reduced
  to $\sigma_Y^2(1-\rho^2)$.
If we use the mean $\mu_X$ as an estimate for $X$, the mean-square error is  $\sigma_X^2$, but if we know the value of $Y$ and use $\mu_X + \rho\frac{\sigma_X}{\sigma_Y}(Y - \mu_Y)$ as the estimate of $X$,  the mean-square error is reduced
  to $\sigma_X^2(1-\rho^2)$.
In both cases, the mean-square error is reduced by the same fraction 
  $(1-\rho^2)$.


In terms of scatter plots (for discrete random variables or data)
on a plane with coordinate axes $x$ and $y$, 
we have two straight lines 
$$
\begin{align*}
y &= \mu_Y + \rho\frac{\sigma_Y}{\sigma_X}(x - \mu_X),\\
x &= \mu_X + \rho\frac{\sigma_X}{\sigma_Y}(y - \mu_Y),
\end{align*}
$$
of different slopes $\rho\sigma_Y/\sigma_X$ and
$\sigma_Y/\rho\sigma_X$ passing through the mean point
$(\mu_X,\mu_Y)$.  The reason for the different slopes is that 
we are choosing the slope to minimize the sum of the squares of
the  vertical
distances of the points from the line in the first case, and to minimize the
sum of the squares of the horizontal distances of the points from the 
line in the second case.  These sums of squared distances are 
$\sigma_Y^2(1-\rho^2)$ and $\sigma_X^2(1-\rho^2)$ respectively.
As a simple example, suppose that $(X,Y)$ takes on values $(0,0)$, $(0,1)$ and
$(1,1)$ with equal probability $\frac{1}{3}$ each or we have a scatter plot
with these three points.  One can grind through the calculations if desired,
but it should be intuitively obvious that we should estimate $Y$ as $\frac{1}{2}$
if $X = 0$ and as $1$ if $X = 1$, while we should estimate $X$ as $0$ if
$Y = 0$ and as $\frac{1}{2}$ if $Y = 1$, that is the two lines have different
slopes $\frac{1}{2}$ and $2$ (in fact, reciprocal slopes since $\sigma_X^2 = \sigma_Y^2 = \frac{2}{9}$ in this example).  This is what probability
theory gives.  But if you treat the three points as a small sample from
from an unknown population and use estimates of the population
means, variances and correlation coefficent, then your results may be
different.
A: Your wording is implying causality, which is not what the R^2 represents. "Does price (x) determine square footage (y)?" is implying causality which is not what is captured through a correlation. "Price explains X% of the variation in square footage" describes that there is a relationship between price and square footage, but not a causal one. This only implies that these variables vary together, not that price causes square footage. Its more akin to saying "In general, when price goes up X amount, square footage happens to go up Y amount"
A: Your example can legitimately be run the other way. Why not estimate square footage from price?  Suppose price data is publicly available, but square footage is not.  Yet you want to estimate square footage (to determine the carpet or furniture market, the likely heating cost, or whatever). It's perfectly valid to model square footage as a function of price.
In my opinion, you are getting hung up in the semantics of "independent" and "dependent" variables. Better to use "predictor" and "predicted".
