How to calculate the difference of two slopes? Is there a method to understand if two lines are (more or less) parallel? I have two lines generated from linear regressions and I would like to understand if they are parallel. In other words, I would like to get the different of the slopes of those two lines.
Is there an R function to calculate this?
EDIT:
... and how can I get the slope (in degrees) of a linear regression line?
 A: The first question is actually from geometry. If you have two lines of the form:
$$y=a_1x+b_1$$
$$y=a_2x+b_2$$
then they are parallel if $a_1=a_2$. So if the slopes are equal then then the lines are parallel. 
For the second question, use the fact that $\tan \alpha=a_1$, where $\alpha$ is the angle the line makes with $x$-axis, and $a_1$ is the slope of the line. So 
$$\alpha=\arctan a_1$$
and to convert to degrees, recall that $2\pi=360$. So the answer in the degrees will be
$$\alpha=\arctan a_1\cdot \frac{360}{2\pi}.$$
The R function for $\arctan$ is called atan. 
Sample R code:
> x<-rnorm(100)
> y<-x+1+rnorm(100)/2
> mod<-lm(y~x)
> mod$coef
    (Intercept)           x 
      0.9416175   0.9850303 
    > mod$coef[2]
        x 
0.9850303 
> atan(mod$coef[2])*360/2/pi
       x 
44.56792 

The last line is the degrees.
Update. For the negative slope values conversion to degrees should follow different rule. Note that the angle with the x-axis can get values from 0 to 180, since we assume that the angle is above the x-axis. So for negative values of $a_1$, the formula is:
$$\alpha=180-\arctan a_1\cdot \frac{360}{2\pi}.$$
Note. While it was fun for me to recall high-school trigonometry, the really useful answer is the one given by Gavin Simpson. Since the slopes of regression lines are random variables, to compare them statistical hypothesis framework should be used.
A: ... following up on @mpiktas' answer, here's how you would extract slope from a lm object and apply the above formula.
# prepare some data, see ?lm
ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group <- gl(2,10,20, labels=c("Ctl","Trt"))
weight <- c(ctl, trt)

lm.D9 <- lm(weight ~ group)
# extract the slope (this is also used to draw a regression line if you wrote abline(lm.D9)
coefficients(lm.D9)["groupTrt"] 
      groupTrt 
   -0.371 
# use the arctan*a1 / (360 / (2*pi)) formula provided by mpiktas
atan(coefficients(lm.D9)["groupTrt"]) * (360/(2 * pi)) 
 groupTrt 
-20.35485 
180-atan(coefficients(lm.D9)["groupTrt"]) * (360/(2 * pi))
 groupTrt 
200.3549 

