geometry
So I'm having a bit of a mathematical dilemma but I'm hoping someone can help me.
I want to know the equation of a line $g(x)$, and it shares an intersection $(9,3)$ with line $y=-0.25x + 5.25$. The intersection makes an angle of $38^\circ$. How might I work out the equation for $g(x)?$ Is there even enough to work with to make this possible?
You can assume that both lines are straight.
In the figure:
from the configuartion of the eclipse we know:
$\overline{AE}=r$
$\overline{AD}=l$
$\angle EDA=\alpha$
and we can find $\beta= \angle AED$ using the sin rule: $$ \frac{r}{\sin \alpha}=\frac{l}{\sin\beta} $$
Now we can find $\angle EAD=\pi-\alpha-\beta$ and , from this we find $EF$ and/or the arc $EG$.
Your argument is indeed good: in space you can write the equation of a straight line with only a point and a vector called the direction vector. Your problem doesn't ask for a specific line so, with only the direction vector, you've generated a set of parallel lines which all satisfy your requirement on the angles.
For completness, the parametric equation of a line crossing a point $P(x_p,y_p,z_p)$ with direction vector $\vec{v}=(a,b,c)$ is: $$\left\{\begin{matrix}x=x_p+at\\y=y_p+bt\\z=z_p+ct\end{matrix}\right.$$
Best Answer
The easiest way to do this is to use translation-rotation to transform the mobile endpoint first.
See this question (which I answered, so you should know how to prove the result): On the rotation of points issue
That gives you the method of rotating points about the origin. To make that apply here, first translate the endpoints so that the fixed endpoint would "become" the origin. So the mobile endpoint would become $(5-9,4-3) = (-4,1)$
Now apply that counterclockwise rotation transform to this new point:
$(-4,1)$ would become $(-4\cos 38^{\circ} - \sin 38^{\circ}, -4 \sin 38^{\circ} + \cos 38^{\circ})$
And then reverse that linear (translation) transform to get everything back into its rightful place on the plane:
$(-4\cos 38^{\circ} - \sin 38^{\circ}, -4 \sin 38^{\circ} + \cos 38^{\circ})$ would become $(9-4\cos 38^{\circ} - \sin 38^{\circ}, 3-4 \sin 38^{\circ} + \cos 38^{\circ})$
Finally, you have to do is find the equation of the line between that point and $(9,3)$
For that you can simply use $\displaystyle \frac{y-y_1}{x-x_1} = \frac{y- y_2}{x-x_2} $