A rod of length L slides between two mutually perpendicular axes. At the end of the rods two lines making angle $60°$ and $ 30°$ with the rod are drawn. The locus of the point of intersection of the lines is ?
The problem I faced while doing this question is that I could not get started on how I could form the equation. Since the rod slides, taking the intercepts of Axis as $ l\cos\theta$ and $l\sin \theta$ and trying to form the equations of the lines, it's really getting complicated and I couldn't really get anywhere from it. Please Help.
If the angle of the rod with the horizontal axis is denoted by $\theta$, the intersections of the rod with the axes are
$$A=A_{\theta}=\binom{L \cos \theta}{0} \ \ \ \text{and} \ \ \ B=B_{\theta}=\binom{0}{ L \sin \theta}.$$
Let $N$ be the intersection point of the lines you have described. Triangle $ANB$ is a half-equilateral triangle. Let us call $M$ the vertex of the completed equilateral triangle $AMB$, i.e., such that $N$ is the midpoint of $[BM]$.
Let $I=\binom{\frac12 L \cos \theta}{\frac12 L \sin \theta}$ be the midpoint of $[AB].$
The altitude $IM$ of equilateral triangle $AMB$ is such that :
$$\vec{IM} = \dfrac{\sqrt{3}}{2}\binom{L \sin \theta}{L \cos \theta }$$
(recall : the altitude of an equilateral triangle is $\dfrac{\sqrt{3}}{2}$ times the length of its sides).
As
$$\vec{OM}=\vec{OI}+\vec{IM},$$
the locus of point $M=(x_M;y_M)$ is given by :
$$\begin{cases}x_M&=&\frac12 L \cos \theta + \dfrac{\sqrt{3}}{2}L \sin \theta \\ y_M&=&\frac12 L \sin \theta+ \dfrac{\sqrt{3}}{2} L \cos \theta \end{cases}$$
Therefore, the locus of midpoint $N=\dfrac12 (M+B)$ is given by parametric equations :
$$\begin{cases}x_N&=&\dfrac14 L \cos \theta + \dfrac{\sqrt{3}}{4}L \sin \theta \\ y_N&=& \dfrac{\sqrt{3}}{4} L \cos \theta + \dfrac34 L \sin \theta \end{cases} \ (0 \leq \theta \leq \dfrac{\pi}{2}).$$
Fig. 1 : Representation of many different positions of the rod $AB$ and the attached equilateral triangle $AMB$; the locus of point $N$ is materialized by the set of little black circles.