Calculate coordinates on a line (or ray) If I have a ray, that starts at the origin, and has a slope of 0.5, how would I calculate the coordinates of a point at length 3 away from the origin (that's on the ray)?
This isn't homework; I learned this a long time ago, but now it's gone, and I find myself embarrassed at asking something so simple. 
 A: You have a right triangle as follows:
$\hskip2in$ 
The slope of the hypotenuse is
$$m=\frac{y-0}{x-0}=\frac{y}{x}.$$ 
You know that $m$ is equal to $\frac{1}{2}$, so $x=2y$. The Pythagorean theorem says that 
$$x^2+y^2=3^2=9.$$
Thus 
$$(2y)^2+y^2=5y^2=9$$
and therefore $$y=\sqrt{\frac{9}{5}}=\frac{3}{\sqrt{5}}\qquad\text{ and }\qquad x=2y=\frac{6}{\sqrt{5}}.$$
A: Let the coordinates of the point in the first quadrant (note there are two points 3 units away, the other in the third quadrant) be $(x,y)$. 
Since the slope of the ray is $1\over2$, and since slope is "rise/run":
$$
\tag{1}{1\over2}={y\over x}.
$$
By the Pythagorean Theorem
$$
\tag{2}x^2+y^2=9.
$$
We need to solve the system of equations (1) and (2).
Solving (1) for $y$ gives
$$
\tag{3}y={x\over 2}.
$$
Replacing $y$ in (2) with ${x\over 2}$ gives
$$
x^2+\bigl({\textstyle{x\over2}}\bigr)^2=9;
$$
or
$$
{5x^2\over 4}=9.
$$
Solving the above for positive  $x$ gives $x^2={9\cdot4 \over 5}$; whence $x=6/\sqrt5$. And then from (3), $y=3/\sqrt5$ (the   point in the third quadrant is  $x=-6/\sqrt5$, $y=-3/\sqrt5$).
A: Find any non-zero point on the ray.  In this case, for example, $(2,1)$ will do. Then the point you are looking for has the shape $\lambda(2,1)=(2\lambda,\lambda)$ where $\lambda$ is positive (because we are dealing with a ray, not a line).
Now use the distance formula (aka the Pythagorean Theorem) to conclude that $4\lambda^2+\lambda^2=9$, so
$$\lambda=\frac{3}{\sqrt{5}}.$$
