You can usually see how good your linear regression line is by looking at the residue plot. If you see the points randomly distributed, you're good. But if you see a pattern, it means there is something wrong with your model; you perhaps need a quadratic function instead of a linear function. I have the following points:
(10,21)(20,12)(30,10)(40,8)(50,7)(60,5.9)(70,6.3)(80,6.95)(90,7.57)(100,8.27)(110,9.03)(120,8.87)(130,10.79)(140,11.77)(150,12.83).
You can immediately see it's not properly described by a linear function, but we use our calculator to find the best line anyway: We get
$$y = -0.129785714x + 10.61028571$$ with an abysmal $r^2 = 0,0209426384$
If we look at the residue plot we can clearly see the points distributed in a quadratic function. That is also expected from your data (y keeps getting lower, than keeps getting higher). I wish I could show you a picture of the residue plot, but I don't know how.
So what I tried is using my calculator's (Ti84+) QuadReg option. We get a way better model:
$$ R^2 \approx 0.77$$
From the data, I expected this to be the best model for our data. However, if I do QuarticReg ($x^4$) we get
$$ R^2 \approx 0.95$$
While the residue plot of the data doesn't look like a quartic function to me, at all. So the higher the power gets, the better the function gets. If you have $x^n$, $R^2$ asymptotically approaches 1 as $n$ tends to $\infty$. Or so it seems, from this small test.
My problem with this is the obvious one: Which model should you take? The model with the highest $R^2$ (Quartic in this case, since my calculator can't go any higher), or should you just look at how the points are spread out in a residue plot (of the linear regression line) and choose the function that most resembles it (In this case I'd say Quadratic, but you can judge for yourself using the data).