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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?

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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:

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

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  • $\begingroup$ Thanks to everyone for answering but you are the f***ing man for breaking it down like that. Exactly what I needed....now let's hope the next problem makes sense. :\ $\endgroup$ – blizz Apr 18 at 0:25
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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 $y$).

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$x$ 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 $y$:

$$\begin{align*} y &= -\frac x2 + 6\\ \tan\theta &= \frac{-\frac x2 + 6}{x}\\ x\tan\theta &= -\frac x2+ 6\\ x &= \frac 6{\tan\theta + \frac12}\\ &= \frac{12}{2\tan\theta + 1} \end{align*}$$

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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.

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