Endpoint of a line knowing slope, start and distance

In a Cartesian system, I've got the slope, start point and distance of a line segment. What's the formula to find the endpoint?

An equivalent way to Arturo's answer is as follows: from the slope $m$, you can determine the cosine and the sine of the angle from the horizontal axis of a line with that slope:

$$c=\frac{1}{\sqrt{1+m^2}} \qquad s=\frac{m}{\sqrt{1+m^2}}$$

(exercise: verify that they are the cosine and sine of a certain angle)

From this construction, you can easily determine the two points at a distance r from your starting point $(h,k)$ as $(h,k)\pm r(c,s)$.

If the point is $(a,b)$, then the distance from $(a,b)$ to $(x,y)$ is $$\sqrt{(x-a)^2 + (y-b)^2}.$$

If the point is $(a,b)$, then the points that lie on the line through $(a,b)$ with slope $m$ are the points of the form $$(x,y) = (a,b) + k(1,m)$$ where $k$ is a constant.

Putting the two together, if you know the start point $(a,b)$, and the slope $m$, and the distance $d$, then find the (two) values of $k$ that will give you a distance of $d$ by plugging in and solving for $k$.

This gives us the following formula for $k$ (where $d$ is the distance): $$k = \pm \frac{d}{\sqrt{1+m^2}}$$ When putting this in the above formula, we find $(x,y)$.