# 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?

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:

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

• 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. :\ – blizz Apr 18 at 0:25

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 $$y$$).

$$x$$ 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*}

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.