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?
 A: 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 $(h,k)\pm r(c,s)$.
A: 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 $(x,y)$.
A: Y=mx+c is the equation of the line you have. (x1,y1) is the point and D is the distance. (x,y) is the point you don't know.
D= sqrt((x1-x)^2 +(y1-y)^2)
sub in for y
D= sqrt((x1-x)^2 +(y1-(mx+c))^2)
then solve for the only unknown, x. this is your x co-ord (2 values). then y=mx+c gives the y.
