Parametrization of a line segment using angle as parameter I know this is probably elementary level for most people here, but I've been stuck on this problem for no less than 4 hours and I am completely clueless as to how to figure this out. Is it possible for anyone to point me to an explanation made for idiots?

 A: First you want a $y=m_1x+b_1$ equation for your line segment. Then a $y=m_2x+b_2$ equation for the line out from the origin. The second will be a function of $\theta$. Then you solve the simultaneous equations for $\theta$ and $y$ (eliminating $x$) and then for $\theta$ and $x$ (eliminating $y$).
A: OK, clearly $\theta$ will range from $0^\circ$ to $90^\circ$ because at the one end you have to point to $(12,0)$ on the $X$ axis and on the other end to $(6,0)$ on the $Y$ axis.
Now let's take some arbitrary value of $\theta$ in that range, and say the line at angle $\theta$ meets the line segment in blue at some point $P$ with coordinates $(x,y)$.  Drop a perpendicular line from $P$ to the $X$ axis, meeting the $X$ axis at $A$ which has coordinates  $(x,0)$.  If we call the origin of the coordinate system $O$, then we have created right triangle 
$OAP$.
In that right triangle, side $AP$ over side $AO$ is opposite over adjacent for angle $\theta$; therefore $\frac y x = \tan \theta$.  And since $P$ is on the blue line, which can be described as $y = 6 - \frac12 x$, we have two simultaneous equations:
$$
y = x \tan \theta//
y = 6 - \frac12 x
$$
To solve these, eliminate $y$ to find $x$, then plug $x$ back into either equation to find $y$.
$$
x \tan \theta =  6 - \frac12 x \implies x = \frac{12}{2\tan\theta + 1} \\
y = 6 - \frac{6}{2\tan\theta + 1}
$$
And this is the parameterization:

$$
x = \frac{12}{2\tan\theta + 1} \\
y = 6 - \frac{6}{2\tan\theta + 1} \\
0^\circ \leq \theta \leq 90^\circ
$$

A: $x$ and $y$ satisfies both
$$\frac x{12}+\frac y6 = 1\quad \wedge\quad \frac yx = \tan \theta$$
So e.g. to represent $x$ in terms of $\theta$, eliminate $y$:
$$\begin{align*}
y &= -\frac x2 + 6\\
\tan\theta &= \frac{-\frac x2 + 6}{x}\\
x\tan\theta &= -\frac x2+ 6\\
x &= \frac 6{\tan\theta + \frac12}\\
&= \frac{12}{2\tan\theta + 1}
\end{align*}$$
A: Use polar coordinates
$$x=r\cos\theta\ ,\quad y=r\sin\theta\ .$$
You need formulae in terms of $\theta$ only, so you want to find $r$ in terms of $\theta$.  To do this, find the equation of the line interms of $x$ and $y$, substitute the above into this equation, solve for $r$ in terms of $\theta$.
Don't forget that for a parametrisation you also need to specify the range of values of $\theta$.  This should be easy from the diagram.
